In this Java programming challenge, you are given two boolean parameters: weekday
and vacation
. The weekday
parameter is true
if it is a weekday, and the vacation
parameter is true
if you are on vacation. Your task is to write a Java method named sleepIn
that determines whether or not you can sleep in. The sleep-in condition is met if it is not a weekday or you are on vacation.
public boolean sleepIn(boolean weekday, boolean vacation) {
// Your code goes here
}
- If
sleepIn(false, false)
is called, it should returntrue
because it is not a weekday and you are not on vacation, so you can sleep in. - If
sleepIn(true, false)
is called, it should returnfalse
because it is a weekday, and you cannot sleep in. - If
sleepIn(false, true)
is called, it should returntrue
because even though it is a weekday, you are on vacation, so you can sleep in.
Implement the sleepIn
method in Java. Return the result of whether or not you can sleep in based on the given parameters using a return
statement.
Implement the following test cases in your Java program to verify the functionality of your sleepIn
method:
Test Case 1:
System.out.println(sleepIn(false, false)); // Expected output: true
Explanation: It is not a weekday (false) and you are not on vacation (false), so you can sleep in.
Test Case 2:
sleepIn(true, false); // Expected output: false
Explanation: It is a weekday (true), so you cannot sleep in.
Test Case 3:
sleepIn(false, true); // Expected output: true
Explanation: Even though it is a weekday (false), you are on vacation (true), so you can sleep in.
Test Case 4:
sleepIn(true, true); // Expected output: true
Explanation: It is a weekday (true), but you are on vacation (true), so you can sleep in.
Test Case 5:
sleepIn(false, false); // Expected output: true
Explanation: It is not a weekday (false) and you are not on vacation (false), so you can sleep in.
Ensure your sleepIn
method correctly handles these test cases.
You have two monkeys, monkey A and monkey B. For each monkey, you are provided with a boolean parameter in Java: aSmile
for monkey A and bSmile
for monkey B. Your task is to create a Java method named monkeyTrouble
. This method will determine whether we are in trouble based on the smiling status of the monkeys.
The trouble condition is defined as follows: you are in trouble if both monkeys are smiling or if neither of them is smiling. To put it simply, if both monkey A and monkey B are either smiling or not smiling, it signifies trouble. The method should return true
if we are in trouble, and false
otherwise.
Define the monkeyTrouble
method with the given signature and implement the logic as described.
Method Signature:
public boolean monkeyTrouble(boolean aSmile, boolean bSmile) {
// your code here
}
- Calling
monkeyTrouble(true, true)
should returntrue
, indicating trouble because both monkeys are smiling. - Calling
monkeyTrouble(false, false)
should returntrue
, indicating trouble because neither monkey is smiling. - Calling
monkeyTrouble(true, false)
should returnfalse
, indicating no trouble since only one monkey is smiling.
Implement the monkeyTrouble
method according to these rules. Use the return statement in Java to return the result.
Here are some test cases you can use to validate your monkeyTrouble
method:
monkeyTrouble(true, true)
should returntrue
. (Both monkeys are smiling, indicating trouble)monkeyTrouble(false, false)
should returntrue
. (Neither monkey is smiling, indicating trouble)monkeyTrouble(true, false)
should returnfalse
. (Only one monkey is smiling, indicating no trouble)monkeyTrouble(false, true)
should returnfalse
. (Only one monkey is smiling, indicating no trouble)monkeyTrouble(true, true)
should returntrue
. (Both monkeys are smiling, indicating trouble)
Ensure your monkeyTrouble
method produces the expected results for these test cases.
You are provided with two integers, "a" and "b". Your task is to create a Java method named sumDouble
. This method will calculate the sum of these two numbers. However, there's a special rule: if the two numbers are identical, you should double their sum before returning it.
Define the sumDouble
method in Java with the following signature and implement the logic as described.
Method Signature:
public int sumDouble(int a, int b) {
// your code here
}
- Calling
sumDouble(1, 2)
should return3
because the sum of 1 and 2 is 3. - Calling
sumDouble(3, 2)
should return5
because the sum of 3 and 2 is 5. - Calling
sumDouble(2, 2)
should return8
because the sum of 2 and 2 is 4, and it is then doubled to 8.
Your implementation of the sumDouble
method should follow these rules. Use the return statement in Java to return the result.
Consider these test cases to validate your sumDouble
method:
sumDouble(1, 2)
should return3
. (The sum of 1 and 2 is 3)sumDouble(3, 2)
should return5
. (The sum of 3 and 2 is 5)sumDouble(2, 2)
should return8
. (The sum of 2 and 2 is 4, doubled to 8)sumDouble(0, 0)
should return0
. (The sum of 0 and 0 is 0; the numbers are the same, but doubling zero still results in zero)sumDouble(-1, -1)
should return-4
. (The sum of -1 and -1 is -2, doubled to -4)
Ensure your sumDouble
method generates the correct outputs for these test cases.
You are provided with an integer "n". Your task is to create a Java method named diff21
. This method will calculate the absolute difference between "n" and 21, but with a twist:
- If "n" is less than 21, the method should return the absolute difference between "n" and 21.
- If "n" is greater than 21, the method should return double the absolute difference between "n" and 21.
Define the diff21
method in Java with the following signature and implement the logic as described.
Method Signature:
public int diff21(int n) {
// your code here
}
- Calling
diff21(19)
should return2
, as the absolute difference between 19 and 21 is 2. - Calling
diff21(10)
should return11
, as the absolute difference between 10 and 21 is 11. - Calling
diff21(21)
should return0
, since the number is already 21, making the difference 0.
Implement the diff21
method according to these rules. Use the return statement in Java to return the result.
Here are some test cases you can use to validate your diff21
method:
diff21(25)
should return8
. (The absolute difference between 25 and 21 is 4, doubled to 8)diff21(30)
should return18
. (The absolute difference between 30 and 21 is 9, doubled to 18)diff21(50)
should return58
. (The absolute difference between 50 and 21 is 29, doubled to 58)diff21(21)
should return0
. (The number is 21, so the difference is 0)diff21(15)
should return6
. (The absolute difference between 15 and 21 is 6)
Make sure your diff21
method produces the expected outputs for these test cases.
You are presented with a scenario involving a loudly talking parrot. There are two parameters: talking
(a boolean) and hour
(an integer representing the hour in 24-hour format, ranging from 0 to 23). Your task is to create a Java method named parrotTrouble
. This method will determine if we are in trouble due to the parrot's talking.
We are in trouble if the parrot is talking and the hour is either before 7 AM or after 8 PM (inclusive). Specifically, if the parrot is talking (talking
is true) and the hour is less than 7 or greater than 20, then we are in trouble. The method should return true
if we are in trouble, and false
otherwise.
Define the parrotTrouble
method in Java with the following signature and implement the logic as described.
Method Signature:
public boolean parrotTrouble(boolean talking, int hour) {
// your code here
}
- Calling
parrotTrouble(true, 6)
should returntrue
, as the parrot is talking and the hour is before 7 AM. - Calling
parrotTrouble(true, 7)
should returnfalse
, as the parrot is talking but the hour is not before 7 AM. - Calling
parrotTrouble(false, 6)
should returnfalse
, as the parrot is not talking.
Implement the parrotTrouble
method according to these rules. Use the return statement in Java to return the result.
Consider these test cases to validate your parrotTrouble
method:
parrotTrouble(true, 21)
should returntrue
. (The parrot is talking after 8 PM)parrotTrouble(true, 12)
should returnfalse
. (The parrot is talking, but it's not before 7 AM or after 8 PM)parrotTrouble(false, 22)
should returnfalse
. (The parrot is not talking)parrotTrouble(true, 3)
should returntrue
. (The parrot is talking before 7 AM)parrotTrouble(false, 9)
should returnfalse
. (The parrot is not talking)
Ensure your parrotTrouble
method produces the correct outputs for these test cases.
You are given two integers, "a" and "b". Your task is to create a Java method named makes10
. This method will determine if either of the numbers is 10 or if their sum equals 10.
The method should return true
in the following cases:
- If either "a" or "b" equals 10.
- If the sum of "a" and "b" equals 10.
In all other cases, the method should return false
.
Define the makes10
method in Java with the following signature and implement the logic as described.
Method Signature:
public boolean makes10(int a, int b) {
// your code here
}
- Calling
makes10(9, 10)
should returntrue
, as one of the numbers is 10. - Calling
makes10(9, 9)
should returnfalse
, as neither of the numbers is 10 and their sum is not 10. - Calling
makes10(1, 9)
should returntrue
, as their sum equals 10.
Implement the makes10
method according to these rules. Use the return statement in Java to return the result.
Consider these test cases to validate your makes10
method:
makes10(10, 5)
should returntrue
. (One of the numbers is 10)makes10(3, 7)
should returntrue
. (The sum of the numbers equals 10)makes10(6, 6)
should returnfalse
. (Neither of the numbers is 10, and their sum is not 10)makes10(8, 2)
should returntrue
. (The sum of the numbers equals 10)makes10(11, -1)
should returntrue
. (One of the numbers is 10)
Ensure your makes10
method produces the correct outputs for these test cases.
You are given an integer "n". Your task is to create a Java method named nearHundred
. This method will determine if the number is within 10 units of either 100 or 200.
The method should return true
if the number "n" is within 10 units (inclusive) of 100 or 200. Otherwise, it should return false
.
In Java, you can calculate the absolute value of a number using the Math.abs(num)
method.
Define the nearHundred
method in Java with the following signature and implement the logic as described.
Method Signature:
public boolean nearHundred(int n) {
// your code here
}
- Calling
nearHundred(93)
should returntrue
, as 93 is within 10 units of 100. - Calling
nearHundred(90)
should returntrue
, as 90 is within 10 units of 100. - Calling
nearHundred(89)
should returnfalse
, as 89 is not within 10 units of 100 or 200.
Implement the nearHundred
method according to these rules. Use the return statement in Java to return the result.
Consider these test cases to validate your nearHundred
method:
nearHundred(100)
should returntrue
. (100 is exactly at 100)nearHundred(200)
should returntrue
. (200 is exactly at 200)nearHundred(105)
should returntrue
. (105 is within 10 units of 100)nearHundred(89)
should returnfalse
. (89 is not within 10 units of 100 or 200)nearHundred(190)
should returntrue
. (190 is within 10 units of 200)
Ensure your nearHundred
method produces the correct outputs for these test cases.
You are provided with two integers, "a" and "b", and a boolean parameter called "negative". Your task is to create a Java method named posNeg
. This method will determine if the two numbers meet specific conditions based on the value of "negative".
Here's how the method should behave:
-
If "negative" is false, the method should return true if one of the numbers is positive and the other is negative. Specifically, if one number is greater than 0 and the other is less than 0, then the result should be true. However, if both numbers are either positive or both are negative, the result should be false.
-
If "negative" is true, the method should return true only if both numbers are negative. That is, if both numbers are less than 0, then the result should be true. Otherwise, it should be false.
Define the posNeg
method in Java with the following signature and implement the logic as described.
Method Signature:
public boolean posNeg(int a, int b, boolean negative) {
// your code here
}
- Calling
posNeg(1, -1, false)
should returntrue
, as one number is positive and the other is negative. - Calling
posNeg(-1, 1, false)
should returntrue
, as one number is negative and the other is positive. - Calling
posNeg(-4, -5, true)
should returntrue
, as both numbers are negative.
Implement the posNeg
method according to these rules. Use the return statement in Java to return the result.
Consider these test cases to validate your posNeg
method:
posNeg(2, -3, false)
should returntrue
. (One number is positive, the other is negative)posNeg(-4, 0, false)
should returnfalse
. (Both numbers are not of opposite signs)posNeg(-6, -8, true)
should returntrue
. (Both numbers are negative)posNeg(10, 5, true)
should returnfalse
. (Both numbers are positive)posNeg(0, -2, false)
should returntrue
. (One number is neither positive nor negative, and the other is negative)
Ensure your posNeg
method produces the correct outputs for these test cases.
You are given a string. Your task is to create a Java method named notString
that modifies the string according to specific rules.
The method should behave as follows:
-
If the given string does not begin with the word "not", the method should return a new string where the word "not" has been prepended to the front. For instance, if the input string is "candy", the method should return "not candy".
-
However, if the given string already begins with the word "not", the method should return the string unchanged. That is, if the string already starts with "not", return the original string as it is.
In Java, you can check if a string starts with a specific substring using the startsWith
method.
Define the notString
method in Java with the following signature and implement the logic as described.
Method Signature:
public String notString(String str) {
// your code here
}
- Calling
notString("candy")
should return"not candy"
, as the string does not start with "not", so "not" is added to the front. - Calling
notString("x")
should return"not x"
, as the string does not start with "not", so "not" is added to the front. - Calling
notString("not bad")
should return"not bad"
, as the string already starts with "not", so no modification is needed.
Implement the notString
method according to these rules. Use the return statement in Java to return the modified string or the original string based on the conditions.
Consider these test cases to validate your notString
method:
notString("programming")
should return"not programming"
. (The string does not start with "not")notString("notepad")
should return"notepad"
. (The string already starts with "not")notString("good")
should return"not good"
. (The string does not start with "not")notString("not good")
should return"not good"
. (The string already starts with "not")notString("")
should return"not "
. (The string is empty, so "not" is added to the front)
Ensure your notString
method produces the correct outputs for these test cases.
You are provided with a non-empty string and an integer "n". Your task is to create a Java method named missingChar
. This method will create a new string by removing the character at index "n" from the original string.
The value of "n" will always be a valid index within the original string, meaning "n" will be within the range of 0 to str.length() - 1
(inclusive).
Define the missingChar
method in Java with the following signature and implement the logic as described.
Method Signature:
public String missingChar(String str, int n) {
// your code here
}
- Calling
missingChar("kitten", 1)
should return"ktten"
, as the character at index 1 (the second character) is removed. - Calling
missingChar("kitten", 0)
should return"itten"
, as the character at index 0 (the first character) is removed. - Calling
missingChar("kitten", 4)
should return"kittn"
, as the character at index 4 (the fifth character) is removed.
Implement the missingChar
method according to these rules. Use the return statement in Java to return the modified string where the character at index "n" has been removed.
Consider these test cases to validate your missingChar
method:
missingChar("programming", 3)
should return"programing"
. (The 'g' at index 3 is removed)missingChar("apple", 0)
should return"pple"
. (The 'a' at index 0 is removed)missingChar("elephant", 6)
should return"elephan"
. (The 't' at index 6 is removed)missingChar("hello", 4)
should return"hell"
. (The 'o' at index 4 is removed)missingChar("goodbye", 2)
should return"gobye"
. (The 'o' at index 2 is removed)
Ensure your missingChar
method produces the correct outputs for these test cases.
You are given a string. Your task is to create a Java method named frontBack
. This method will create a new string by exchanging the first and last characters of the original string.
Specifically, the first character of the original string will become the last character in the new string, and the last character of the original string will become the first character in the new string.
Define the frontBack
method in Java with the following signature and implement the logic as described.
Method Signature:
public String frontBack(String str) {
// your code here
}
- Calling
frontBack("code")
should return"eodc"
, as the first ('c') and last ('e') characters are exchanged. - Calling
frontBack("a")
should return"a"
, as the string contains only one character, so no exchange is needed. - Calling
frontBack("ab")
should return"ba"
, as the first ('a') and last ('b') characters are exchanged.
Implement the frontBack
method according to these rules. Use the return statement in Java to return the modified string where the first and last characters have been exchanged.
Consider these test cases to validate your frontBack
method:
frontBack("hello")
should return"oellh"
. (The 'h' and 'o' are exchanged)frontBack("goodbye")
should return"eoodbyg"
. (The 'g' and 'e' are exchanged)frontBack("programming")
should return"gormmingp"
. (The 'p' and 'g' are exchanged)frontBack("cat")
should return"tatc"
. (The 'c' and 't' are exchanged)frontBack("open")
should return"npeo"
. (The 'o' and 'n' are exchanged)
Ensure your frontBack
method produces the correct outputs for these test cases.
You are given a string. Your task is to create a Java method named front3
. This method will create a new string by repeating the front of the original string three times.
To determine the "front" of the string, consider the first three characters. If the string's length is less than 3, then the entire string becomes the "front".
Define the front3
method in Java with the following signature and implement the logic as described.
Method Signature:
public String front3(String str) {
// your code here
}
- Calling
front3("Java")
should return"JavJavJav"
because the front ("Jav") is repeated three times. - Calling
front3("Chocolate")
should return"ChoChoCho"
because the front ("Cho") is repeated three times. - Calling
front3("abc")
should return"abcabcabc"
because the string length is less than 3, so the entire string is repeated three times.
Implement the front3
method according to these rules. Use the return statement in Java to return the modified string where the front is repeated three times.
Consider these test cases to validate your front3
method:
front3("Hi")
should return"HiHiHi"
. (The front "Hi" is repeated three times)front3("Python")
should return"PytPytPyt"
. (The front "Pyt" is repeated three times)front3("Elephant")
should return"EleEleEle"
. (The front "Ele" is repeated three times)front3("a")
should return"aaa"
. (The entire string "a" is repeated as it's less than 3 characters)front3("")
should return""
. (An empty string repeated is still an empty string)
Ensure your front3
method produces the correct outputs for these test cases.
You are given a string. Your task is to create a Java method named backAround
. This method will create a new string by taking the last character of the original string and adding it to the front and back.
Specifically, you need to take the last character of the string and concatenate it with the original string itself, at both the beginning and the end.
Define the backAround
method in Java with the following signature and implement the logic as described.
Method Signature:
public String backAround(String str) {
// your code here
}
- Calling
backAround("cat")
should return"tcatt"
, as the last character ('t') is added to the front and back of the original string. - Calling
backAround("Hello")
should return"oHelloo"
, as the last character ('o') is added to the front and back of the original string. - Calling
backAround("a")
should return"aaa"
, as the string contains only one character, and that character is added to the front and back.
Implement the backAround
method according to these rules. Use the return statement in Java to return the modified string where the last character is added to the front and back.
Consider these test cases to validate your backAround
method:
backAround("Programming")
should return"gProgrammingg"
. (The last character 'g' is added to the front and back)backAround("Goodbye")
should return"eGoodbyee"
. (The last character 'e' is added to the front and back)backAround("Z")
should return"ZZ"
. (The single character 'Z' is added to the front and back)backAround("12345")
should return"5123451"
. (The last character '5' is added to the front and back)backAround("")
should return""
. (An empty string remains unchanged)
Ensure your backAround
method produces the correct outputs for these test cases.
You are given a string. Your task is to create a Java method named front22
. This method will create a new string by taking the first two characters of the original string and adding them both at the front and back.
If the length of the string is less than 2, use whatever characters are there. In other words, if the string has only one character or is empty, the result will be the same as the original string, with those characters (or lack thereof) added at both the front and back.
Define the front22
method in Java with the following signature and implement the logic as described.
Method Signature:
public String front22(String str) {
// your code here
}
- Calling
front22("kitten")
should return"kikittenki"
because the first two characters ("ki") are added to the front and back of the original string. - Calling
front22("Ha")
should return"HaHaHa"
because the string has only two characters, so those characters ("Ha") are added to the front and back. - Calling
front22("abc")
should return"ababcab"
because the first two characters ("ab") are added to the front and back.
Implement the front22
method according to these rules. Use the return statement in Java to return the modified string where the first two characters are added to the front and back.
Consider these test cases to validate your front22
method:
front22("coding")
should return"cocodingco"
. (The first two characters "co" are added to the front and back)front22("H")
should return"HHH"
. (The single character "H" is repeated three times)front22("Hi")
should return"HiHiHi"
. (The two characters "Hi" are repeated three times)front22("hello")
should return"hehellohe"
. (The first two characters "he" are added to the front and back)front22("")
should return""
. (An empty string remains unchanged)
Ensure your front22
method produces the correct outputs for these test cases.
You are given a non-negative number. Your task is to create a Java method named or35
. This method will determine if the number is a multiple of 3 or a multiple of 5.
To check if a number is a multiple of another number, you can use the modulo operator %
in Java. The modulo operator returns the remainder when one number is divided by another. If the result of the modulo operation is 0, it means the number is divisible without any remainder, and thus it is a multiple of the divisor.
Define the or35
method in Java with the following signature and implement the logic as described.
Method Signature:
public boolean or35(int n) {
// your code here
}
- Calling
or35(3)
should returntrue
because 3 is a multiple of 3. - Calling
or35(10)
should returntrue
because 10 is a multiple of 5. - Calling
or35(8)
should returnfalse
because 8 is neither a multiple of 3 nor a multiple of 5.
Implement the or35
method according to these rules. Use the return statement in Java to return true
if the number is a multiple of 3 or 5, and false
otherwise.
Consider these test cases to validate your or35
method:
or35(15)
should returntrue
. (15 is both a multiple of 3 and a multiple of 5)or35(14)
should returnfalse
. (14 is neither a multiple of 3 nor a multiple of 5)or35(20)
should returntrue
. (20 is a multiple of 5)or35(9)
should returntrue
. (9 is a multiple of 3)or35(7)
should returnfalse
. (7 is neither a multiple of 3 nor a multiple of 5)
Ensure your or35
method produces the correct outputs for these test cases.
You are given a string. Your task is to create a Java method named startHi
. This method will determine if the string starts with the characters "hi".
To check if a string starts with a specific substring, you should compare the first few characters of the string with that substring. In this case, you need to compare the first two characters of the string with the string "hi".
Define the startHi
method in Java with the following signature and implement the logic as described.
Method Signature:
public boolean startHi(String str) {
// your code here
}
- Calling
startHi("hi there")
should returntrue
because the string starts with "hi". - Calling
startHi("hi")
should returntrue
because the string consists only of "hi". - Calling
startHi("hello hi")
should returnfalse
because the string does not start with "hi".
Implement the startHi
method according to these rules. Use the return statement in Java to return true
if the string starts with "hi", and false
otherwise.
Consider these test cases to validate your startHi
method:
startHi("hi friends")
should returntrue
. (The string starts with "hi")startHi("HI")
should returnfalse
. (The string does not start with "hi" due to case sensitivity)startHi("his")
should returnfalse
. (The string does not start with "hi" because it has additional characters after "hi")startHi("h")
should returnfalse
. (The string does not have enough characters to match "hi")startHi("Hi there")
should returnfalse
. (The string does not start with "hi" due to case sensitivity)
Ensure your startHi
method produces the correct outputs for these test cases.
You are given two temperatures. Your task is to create a Java method named icyHot
. This method will determine if one temperature is less than 0 and the other temperature is greater than 100.
To solve this challenge, compare the given temperatures and check if one of them is less than 0 and the other is greater than 100.
Define the icyHot
method in Java with the following signature and implement the logic as described.
Method Signature:
public boolean icyHot(int temp1, int temp2) {
// your code here
}
- Calling
icyHot(120, -1)
should returntrue
because one temperature (-1) is less than 0 and the other temperature (120) is greater than 100. - Calling
icyHot(-1, 120)
should returntrue
because one temperature (-1) is less than 0 and the other temperature (120) is greater than 100. - Calling
icyHot(2, 120)
should returnfalse
because both temperatures are either less than 0 or not greater than 100.
Implement the icyHot
method according to these rules. Use the return statement in Java to return true
if one temperature is less than 0 and the other temperature is greater than 100, and false
otherwise.
Consider these test cases to validate your icyHot
method:
icyHot(-5, 105)
should returntrue
. (One temperature is less than 0 and the other is greater than 100)icyHot(-2, 99)
should returnfalse
. (Both temperatures are either less than 0 or not greater than 100)icyHot(0, 101)
should returnfalse
. (Both temperatures are either less than 0 or not greater than 100)icyHot(50, -50)
should returntrue
. (One temperature is less than 0 and the other is greater than 100)icyHot(0, 0)
should returnfalse
. (Both temperatures are either less than 0 or not greater than 100)
Ensure your icyHot
method produces the correct outputs for these test cases.
You are given two integer values. Your task is to create a Java method named loneTeen
. This method will determine if one of the numbers is a "teen" number (in the range of 13 to 19, inclusive) while the other is not.
To solve this challenge, check if one of the numbers is in the range of 13 to 19 (inclusive), while the other number is not. If this condition is met, the method should return true
. However, if both numbers are in the teen range or both numbers are not in the teen range, the method should return false
.
Define the loneTeen
method in Java with the following signature and implement the logic as described.
Method Signature:
public boolean loneTeen(int a, int b) {
// your code here
}
- Calling
loneTeen(13, 99)
should returntrue
because one number (13) is in the teen range, while the other number (99) is not. - Calling
loneTeen(21, 19)
should returntrue
because one number (19) is in the teen range, while the other number (21) is not. - Calling
loneTeen(13, 13)
should returnfalse
because both numbers are in the teen range.
Implement the loneTeen
method according to these rules. Use the return statement in Java to return true
if one of the numbers is a teen while the other is not, and false
otherwise.
Consider these test cases to validate your loneTeen
method:
loneTeen(14, 16)
should returntrue
. (One number is in the teen range, while the other is not)loneTeen(15, 18)
should returnfalse
. (Both numbers are in the teen range)loneTeen(20, 21)
should returnfalse
. (Both numbers are not in the teen range)loneTeen(13, 20)
should returntrue
. (One number is in the teen range, while the other is not)loneTeen(17, 17)
should returnfalse
. (Both numbers are in the teen range)
Ensure your loneTeen
method produces the correct outputs for these test cases.
You are given a string. Your task is to create a Java method named delDel
. This method will check if the string contains the substring "del" starting at index 1. If it does, return a modified string where the "del" substring has been deleted. Otherwise, return the string unchanged.
To solve this challenge, examine the string to determine if the substring "del" starts at index 1. If it does, remove the "del" substring from the string and return the modified string. If not, return the original string without any changes.
Define the delDel
method in Java with the following signature and implement the logic as described.
Method Signature:
public String delDel(String str) {
// your code here
}
- Calling
delDel("adelbc")
should return"abc"
because the string starts with "del" at index 1, so the "del" substring is deleted. - Calling
delDel("adelHello")
should return"aHello"
because the string starts with "del" at index 1, so the "del" substring is deleted. - Calling
delDel("adedbc")
should return"adedbc"
because the string does not start with "del" at index 1, so the string remains unchanged.
Implement the delDel
method according to these rules. Use the return statement in Java to return the modified string without the "del" substring if it starts at index 1. Otherwise, return the original string as it is.
Consider these test cases to validate your delDel
method:
delDel("adelicious")
should return"adelicious"
. (The string does not start with "del" at index 1)delDel("adello")
should return"alo"
. (The "del" substring is removed)delDel("abcdel")
should return"abcdel"
. (The string does not start with "del" at index 1)delDel("delfood")
should return"delfood"
. (The string does not start with "del" at index 1)delDel("delta")
should return"delta"
. (The string does not start with "del" at index 1)
Ensure your delDel
method produces the correct outputs for these test cases.
You are given a string. Your task is to create a Java method named mixStart
. This method will check if the string begins with the substring "ix", where the first character can be any character.
To solve this challenge, examine the string to determine if the substring "ix" appears at the beginning of the string, following any first character. If the string starts with a character followed by "ix", return true
. Otherwise, return false
.
Define the mixStart
method in Java with the following signature and implement the logic as described.
Method Signature:
public boolean mixStart(String str) {
// your code here
}
- Calling
mixStart("mix snacks")
should returntrue
because the string begins with "mix". - Calling
mixStart("pix snacks")
should returntrue
because the string begins with "pix", where 'p' replaces the first character. - Calling
mixStart("piz snacks")
should returnfalse
because the string does not begin with a character followed by "ix".
Implement the mixStart
method according to these rules. Use the return statement in Java to return true
if the string starts with any character followed by "ix", and false
otherwise.
Consider these test cases to validate your mixStart
method:
mixStart("9ix snacks")
should returntrue
. (The string begins with '9ix' where '9' replaces the first character)mixStart("hix snacks")
should returntrue
. (The string begins with 'hix' where 'h' replaces the first character)mixStart("fix snacks")
should returntrue
. (The string begins with 'fix' where 'f' replaces the first character)mixStart("nix snacks")
should returntrue
. (The string begins with 'nix' where 'n' replaces the first character)mixStart("mixing snacks")
should returntrue
. (The string begins with 'mix')
Ensure your mixStart
method produces the correct outputs for these test cases.
You are given a string. Your task is to create a Java method named startOz
. This method will create a new string using the first two characters of the original string, but with specific conditions: include the first character only if it is 'o', and include the second character only if it is 'z'.
To solve this challenge, examine the string and extract the first two characters. Include the first character in the new string if it is 'o'. Include the second character if it is 'z'. If these conditions are not met for each respective character, exclude them from the new string.
Define the startOz
method in Java with the following signature and implement the logic as described.
Method Signature:
public String startOz(String str) {
// your code here
}
- Calling
startOz("ozymandias")
should return"oz"
because both the first character ('o') and the second character ('z') meet the conditions. - Calling
startOz("bzoo")
should return"z"
because the first character is not 'o', but the second character is 'z'. - Calling
startOz("oxx")
should return"o"
because the first character is 'o', but the second character is not 'z'.
Implement the startOz
method according to these rules. Use the return statement in Java to return the new string that consists of the first two characters, following the given conditions.
Consider these test cases to validate your startOz
method:
startOz("booz")
should return""
. (Neither 'o' at the start nor 'z' as the second character)startOz("oz")
should return"oz"
. (Both characters 'o' and 'z' are included)startOz("zzz")
should return""
. (No 'o' at the start and 'z' is not the second character)startOz("ooz")
should return"o"
. (First character 'o' is included, but the second character is not 'z')startOz("zo")
should return"z"
. (Second character 'z' is included, but the first character is not 'o')
Ensure your startOz
method produces the correct outputs for these test cases.
You are given three integer values: a
, b
, and c
. Your task is to create a Java method named intMax
. This method will determine the largest value among the three integers.
To solve this challenge, compare the three integers and find the maximum value. In other words, identify the number that is larger than both the other numbers.
Define the intMax
method in Java with the following signature and implement the logic as described.
Method Signature:
public int intMax(int a, int b, int c) {
// your code here
}
- Calling
intMax(1, 2, 3)
should return3
because 3 is the largest value among 1, 2, and 3. - Calling
intMax(1, 3, 2)
should return3
because 3 is the largest value among 1, 3, and 2. - Calling
intMax(3, 2, 1)
should return3
because 3 is the largest value among 3, 2, and 1.
Implement the intMax
method according to these rules. Use the return statement in Java to return the largest value among the three integers.
Consider these test cases to validate your intMax
method:
intMax(5, 10, 7)
should return10
. (10 is the largest value among 5, 10, and 7)intMax(-1, -5, -3)
should return-1
. (-1 is the largest value among -1, -5, and -3)intMax(0, 0, 0)
should return0
. (0 is the largest value among 0, 0, and 0)intMax(100, 200, 150)
should return200
. (200 is the largest value among 100, 200, and 150)intMax(-10, 0, 10)
should return10
. (10 is the largest value among -10, 0, and 10)
Ensure your intMax
method produces the correct outputs for these test cases.
You are given two integer values: a
and b
. Your task is to create a Java method named close10
. This method will determine which value is closer to 10 and return that value. In the event of a tie, return 0.
To solve this challenge, compare the absolute differences between 10 and the given integers, a
and b
. The number with the smaller absolute difference is closer to 10. If both absolute differences are the same, it's a tie, and you should return 0.
Define the close10
method in Java with the following signature and implement the logic as described.
Method Signature:
public int close10(int a, int b) {
// your code here
}
- Calling
close10(8, 13)
should return8
because the absolute difference between 8 and 10 is smaller than the absolute difference between 13 and 10. - Calling
close10(13, 8)
should return8
because the absolute difference between 8 and 10 is smaller than the absolute difference between 13 and 10. - Calling
close10(13, 7)
should return0
because both absolute differences, 3 and 3, are the same, resulting in a tie.
Implement the close10
method according to these rules. Use the return statement in Java to return the value that is closer to 10, or 0 if both values are equidistant from 10.
Consider these test cases to validate your close10
method:
close10(5, 12)
should return5
. (5 is closer to 10 than 12)close10(10, 20)
should return10
. (10 is exactly 10, while 20 is farther)close10(0, 10)
should return10
. (10 is exactly 10, while 0 is farther)close10(15, 8)
should return8
. (8 is closer to 10 than 15)close10(11, 12)
should return11
. (11 is closer to 10 than 12)
Ensure your close10
method produces the correct outputs for these test cases.
You are given two integer values: a
and b
. Your task is to create a Java method named in3050
. This method will check if both values are within a specific range and return true
if both values are either in the range of 30 to 40 (inclusive) or in the range of 40 to 50 (inclusive). Otherwise, it should return false
.
To solve this challenge, determine if both values are within the specified ranges. The ranges are inclusive, meaning that the boundary values are considered valid.
Define the in3050
method in Java with the following signature and implement the logic as described.
Method Signature:
public boolean in3050(int a, int b) {
// your code here
}
- Calling
in3050(30, 31)
should returntrue
because both values are within the range of 30 to 40 (inclusive). - Calling
in3050(30, 41)
should returnfalse
because one value is within the range of 30 to 40 (inclusive), but the other value is not. - Calling
in3050(40, 50)
should returntrue
because both values are within the range of 40 to 50 (inclusive).
Implement the in3050
method according to these rules. Use conditional statements to check if both values are within the specified ranges. Return true
if both conditions are met; otherwise, return false
.
Consider these test cases to validate your in3050
method:
in3050(35, 45)
should returntrue
. (Both values are within the range of 30 to 40)in3050(25, 35)
should returnfalse
. (One value is not within the range of 30 to 40)in3050(40, 55)
should returnfalse
. (One value is not within the range of 40 to 50)in3050(30, 50)
should returnfalse
. (Values are in different ranges)in3050(35, 55)
should returnfalse
. (One value is not within the range of 40 to 50)
Ensure your in3050
method produces the correct outputs for these test cases.
You are given two positive integer values: a
and b
. Your task is to create a Java method named max1020
. This method will determine the larger value between the two that falls within the range of 10 to 20 (inclusive). If neither value is within that range, return 0.
To solve this challenge, compare the values of a
and b
and determine which value is larger and within the range of 10 to 20 (inclusive). If both values are within the range, return the larger value. If only one value is within the range, return that value. If neither value is within the range, return 0.
Define the max1020
method in Java with the following signature and implement the logic as described.
Method Signature:
public int max1020(int a, int b) {
// your code here
}
- Calling
max1020(11, 19)
should return19
because both values are within the range of 10 to 20 (inclusive), and 19 is the larger value. - Calling
max1020(19, 11)
should return19
because both values are within the range of 10 to 20 (inclusive), and 19 is the larger value. - Calling
max1020(11, 9)
should return11
because one value is within the range of 10 to 20 (inclusive), and 11 is the larger value.
Implement the max1020
method according to these rules. Use conditional statements to check if each value is within the specified range. Compare the values and return the larger value if it falls within the range. If neither value is within the range, return 0.
Consider these test cases to validate your max1020
method:
max1020(15, 25)
should return15
. (Only 15 is within the range)max1020(9, 25)
should return0
. (Neither value is within the range)max1020(21, 5)
should return0
. (Neither value is within the range)max1020(17, 13)
should return17
. (Both are within the range, but 17 is larger)max1020(10, 20)
should return20
. (Both are within the range, but 20 is larger)
Ensure your max1020
method produces the correct outputs for these test cases.
You are given a string. Your task is to create a Java method named stringE
. This method will check if the string contains between 1 and 3 occurrences of the letter 'e'. Return true
if the string meets this condition; otherwise, return false
.
To solve this challenge, count the number of occurrences of the letter 'e' in the given string and check if it falls within the range of 1 to 3 (inclusive). If the count is within this range, return true
; otherwise, return false
.
Define the stringE
method in Java with the following signature and implement the logic as described.
Method Signature:
public boolean stringE(String str) {
// your code here
}
- Calling
stringE("Hello")
should returntrue
because there is one occurrence of the letter 'e' in the string. - Calling
stringE("Heelle")
should returntrue
because there are three occurrences of the letter 'e' in the string. - Calling
stringE("Heelele")
should returnfalse
because there are four occurrences of the letter 'e' in the string.
Implement the stringE
method according to these rules. Use a loop to count the occurrences of 'e' in the string. Then, check if the count is within the range of 1 to 3 (inclusive) and return true
or false
accordingly.
Consider these test cases to validate your stringE
method:
stringE("elephant")
should returntrue
. (Two occurrences of 'e')stringE("cat")
should returnfalse
. (No occurrence of 'e')stringE("developer")
should returnfalse
. (Four occurrences of 'e')stringE("e")
should returntrue
. (One occurrence of 'e')stringE("Elephant")
should returnfalse
. (The uppercase 'E' is not counted)
Ensure your stringE
method produces the correct outputs for these test cases.
You are given two non-negative integer values: a
and b
. Your task is to create a Java method named lastDigit
. This method will check if both values have the same last digit. Return true
if they have the same last digit; otherwise, return false
.
To solve this challenge, compare the last digits of the given values a
and b
. To extract the last digit of a number, use the modulo operator %
, which returns the remainder of the division operation. If the last digits of both values are equal, return true
; otherwise, return false
.
Define the lastDigit
method in Java with the following signature and implement the logic as described.
Method Signature:
public boolean lastDigit(int a, int b) {
// your code here
}
- Calling
lastDigit(7, 17)
should returntrue
because both values end with the digit 7. - Calling
lastDigit(6, 17)
should returnfalse
because the last digits of the two values are different. - Calling
lastDigit(3, 113)
should returntrue
because both values end with the digit 3.
Implement the lastDigit
method according to these rules. Use the modulo operator %
to extract the last digits of a
and b
. Compare the last digits and return true
if they are equal; otherwise, return false
.
Consider these test cases to validate your lastDigit
method:
lastDigit(12, 22)
should returntrue
. (Both values end with the digit 2)lastDigit(31, 92)
should returnfalse
. (Different last digits)lastDigit(50, 100)
should returntrue
. (Both values end with the digit 0)lastDigit(99, 999)
should returntrue
. (Both values end with the digit 9)lastDigit(123, 456)
should returnfalse
. (Different last digits)
Ensure your lastDigit
method produces the correct outputs for these test cases.
You are given a string. Your task is to create a Java method named endUp
. This method will modify the string by converting the last three characters to uppercase. If the string has fewer than three characters, convert the existing characters to uppercase. Return the modified string.
To solve this challenge, manipulate the string to convert the last three characters to uppercase. Use the toUpperCase()
method to convert a string or substring to uppercase. If the string has fewer than three characters, apply the conversion to the entire string.
Define the endUp
method in Java with the following signature and implement the logic as described.
Method Signature:
public String endUp(String str) {
// your code here
}
- Calling
endUp("Hello")
should return"HeLLO"
because the last three characters "llo" have been converted to uppercase. - Calling
endUp("hi there")
should return"hi thERE"
because the last three characters "ere" have been converted to uppercase. - Calling
endUp("hi")
should return"HI"
because the string has fewer than three characters, so the entire string is converted to uppercase.
Implement the endUp
method according to these rules. Use conditional statements to check the length of the string and determine the appropriate conversion. Apply toUpperCase()
to the relevant part of the string.
Consider these test cases to validate your endUp
method:
endUp("coding")
should return"codING"
. (Last three characters converted to uppercase)endUp("OpenAI")
should return"OpenAI"
. (Entire string converted as it's fewer than three characters)endUp("UPPER")
should return"UPPER"
. (Entire string converted as it's fewer than three characters)endUp("a")
should return"A"
. (Single character string converted to uppercase)endUp("Hi there!")
should return"Hi there!"
. (String unchanged as last three characters are not within the last three positions)
Ensure your endUp
method produces the correct outputs for these test cases.
You are given a non-empty string and a positive integer N
. Your task is to create a Java method named everyNth
. This method will create a new string by selecting specific characters from the original string, determined by the value of N
. Starting from the first character (index 0), include every N
th character from the original string in the new string.
To solve this challenge, iterate over the characters of the original string and select only those characters at indexes that are multiples of N
. For example, if N
is 3, include characters at indexes 0, 3, 6, and so on. Build a new string by concatenating these selected characters.
Define the everyNth
method in Java with the following signature and implement the logic as described.
Method Signature:
public String everyNth(String str, int n) {
// your code here
}
- Calling
everyNth("Miracle", 2)
should return"Mrce"
because you select characters at indexes 0, 2, 4, and 6 from the original string. - Calling
everyNth("abcdefg", 2)
should return"aceg"
because you select characters at indexes 0, 2, 4, and 6 from the original string. - Calling
everyNth("abcdefg", 3)
should return"adg"
because you select characters at indexes 0, 3, and 6 from the original string.
Implement the everyNth
method according to these rules. Use a loop or other iterative construct to iterate over the characters of the original string. Use conditional statements to select the characters that correspond to the multiples of N
. Build the new string by concatenating these selected characters.
Consider these test cases to validate your everyNth
method:
everyNth("Hello, world!", 5)
should return"H,w"
. (Characters at indexes 0 and 5)everyNth("Programming", 3)
should return"Pgmn"
. (Characters at indexes 0, 3, 6, and 9)everyNth("abcdefghij", 1)
should return"abcdefghij"
. (Every character is selected)everyNth("1234567890", 4)
should return"1470"
. (Characters at indexes 0, 4, 8)everyNth("abcdefg", 7)
should return"a"
. (Only the first character is selected)
Ensure your everyNth
method produces the correct outputs for these test cases.