Beginner.txt

Example : 
Solution :
# Python  
age=input(int("Age : "))
print(age);
                                                                                                                                               Maintainer : Gaurav-2803
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------
1. Write a program that asks the user  for their name and greets (Hi!, Hey!, Hello!, etc) them with their name. 
Example : (Hi!, Gaurav)
Solution : 
 #include <iostream>
 using namespace std;
 int main(){
    string name;
         cout<<" Enter Name  : ";
         cin>>name;
         cout<<"Hi "+name;
 return 0;
}

#Solved Q no 1 using python
#python
name=input("Enter your name: ")
print("Hello!,",name)

# rust
use std::io::{self, Write};
fn main() {
    let mut name = String::new();

    print!("Enter you name: ");
    io::stdout().flush().unwrap();
    io::stdin()
        .read_line(&mut name)
        .expect("Failed to read name");    
    println!("Hello, {}", name);
}

# Java

import java.util.Scanner;

public class Greet {
    static Scanner scanner = new Scanner(System.in);
    public static void greet(String name){
        System.out.println("Hello, "+name);
    }

    public static void main(String[] args) {
        System.out.print("Enter your name : ");
        String name = scanner.next();
        greet(name);
    }
}


2. Modify the previous program (i.e. Q.1) such that only the users Alice and Bob are greeted with their names.
Solution : 
 #include<iostream>
 using namespace std;
 int main(){
    string name;
         cout<<" Enter  Name : ";
         cin>>name;
         if(name=="Alice" || name=="bob")
         {
          cout<<"Hi "+name;
         }else{
          cout<<"Hi";
         }
        return 0;
}

#solved Q no 2 using python
# Python  
name = input("Enter name: ")
if(name=="Alice" or name=="Bob"):
    print("Hi!, "+name);

3. Write a program that asks the user for a number n and prints the sum of the numbers 1 to n
Solution : 
//include <iostream>
using namespace std;
int main()
{
    int n, a;
    a = 0;
    cin >> n;
    n++;
    while (n--)
    {
        a = a + n;
    }
    cout << a;
    return 0;
}

//another method 
include <iostream>
using namespace std;
int main()
{
    int n, a;
    a = 0;
   cin>>n;
    cout << (n*(n+1))/2;
    return 0;
}

#solved Q no 3 using python
#python
num = int(input("Enter numer n: "))

if num < 0:
   print("Enter a positive number")
else:
   sum = 0
   
   while(num > 0):
       sum += num
       num -= 1
   print("The sum is", sum)
   
# Java

import java.util.Scanner;

public class Sum {
    public static int sum(int n){
        if(n == 0){
            return 0;
        }
        else if (n == 1){
            return 1;
        }
        else {
            return sum(n-1)+(n);
        }
    }

    public static void main(String [] args){
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter the number : ");
        int n = scanner.nextInt();
        System.out.println(sum(n));
    }
}


4. Modify the previous program (i.e. Q.3) such that only multiples of three or five are considered in the sum, e.g. 3, 5, 6, 9, 10, 12, 15 for n=17
Solution : 
#include <iostream>
using namespace std;
int main()
{
    int n, ans;
    ans = 0;
    cin >> n;
    n++;
    while (n--)
    {
        if (n % 3 == 0 || n % 5 == 0)
        {
            ans = ans + n;
        }
    }
    cout << ans;
    return 0;
}

5. Write a program that asks the user for a number n and gives them the possibility to choose between computing the sum and computing the product of 1,…,n.
Solution : 

x=int(input('Enter a number :'))
y=input('Compute the sum or Compute the product(s/m) ? :')
sum=0
multi=1
for i in range(1,x+1):
    if y=='s':
        sum=sum+i
    if y=='m':
        multi=multi*i
if y=='s':
    print('Sum :',sum)    
if y=='m':
    print('Product :',multi)

======= In C++ ==========
#include <iostream>
using namespace std;
int main() {
    int n, a, b;
    a = 0;
    b = 1;
    cin >> n;
    n++;
    while (n--) {
        a = a + n;
        b = b * n;
    }
    cout << "Sum or Product (s/p) : ";
    char c;
    cin >> c;
    if (c == 's')
        cout << "Sum : " << a << endl;
    else if (c == 'p')
        cout << "Product : " << b << endl;
    else
        cout << "Invalid Input";
    return 0;
}
==========================

6. Write a program that prints a multiplication table for numbers up to 12.
Solution : 
#include <iostream>
using namespace std;
int main()
{
    int a;
    for(int j=1; j<=12; j++)
    {
    for (int i = 1; i <= 10; i++)
    {
        a = j * i;
        cout << a << endl;
    }
}
    return 0;
}

# Java

public class Multiplication {

    public static void table(int n){
        for (int i = 1; i < 11; i++) {
            System.out.print(i*n+" ");
        }
        System.out.println();
    }

    public static void main(String[] args) {
        for (int i = 1; i < 13 ; i++){
            table(i);
        }
    }

}


7. Write a program that prints all prime numbers between 2 numbers. 
Solution : 
//cpp
class Solution {
  public:
  
  bool isprime(int n){
        if(n<=1)
            return false;
        for(int i=2;i<=sqrt(n);i++){
            if(n%i==0)
            return false;
        }
        return true;
  }
    void primeRange(int M, int N) {
        vector<int>ans;
        for(M;M<=N;M++){
            if(isprime(M))
                ans.push_back(M);
        }
        for(int i=0;i<ans.size();i++){
	cout<<ans[i]<<" ";
	}
    }
};

print('Enter two numbers :')
x=int(input())
y=int(input())
print('Prime numbers : ',end=' ')
for i in range(x,y+1):
    cnt=0
    for j in range(1,i+1):
        if i%j==0:
            cnt+=1
    if cnt==2:
        print(i,end=' ')



8. Write a program that prints the next 20 leap years.
Solution : 
Code:
#include<stdio.h>
int nr_leap_years(int y){
return( y/4 - y/100 + y/400 );
}

int is_leap_year(int y)
{
return(nr_leap_years(y) - nr_leap_years(y-1));
}

int main(void){
int nr,i;
nr = 0;
i = 2021;

while(nr<20){
if (is_leap_year(i)){
printf(“\n %d”, i);
nr++;
}
i++;
}
return(0);
}

---------------JAVA-----------------------
import java.util.*;
class LeapYear {
	public static void main (String[] args){
		Scanner sc = new Scanner(System.in);
		System.out.println("Enter the year to check for leap year.");
		int year = sc.nextInt();

		System.out.println("The next 20 leap years from the given year " + year+ " are: ");
		int count =0;
		while(count != 20){
		year = year+4;
		if((year%400==0) || (year%4==0 && year%100 != 0)){
		System.out.println(year);
		}
		count++;
		}
	]
]

9. Write a function that returns the largest element in a array.
Solution : 
class Main{
public static void main(String[] args){
int[] arr={1,2,3,4,5};
Arrays.sort(arr);
System.out.println(arr[arr.length-1]);
}
}

--------------------c++--------------------
#include<bits/stdc++.h>
using namespace std;

int largest(int arr[], int n){
    int max=arr[0];
    for(int i=1;i<n;i++){
        if(arr[i]>max)
            max=arr[i];
    }
    return max;
}

int main(){
    int n;
    cin >> n;
    
    int arr[n];
    for(int i=0;i<n;i++)
        cin >> arr[i];
    
    cout<<largest(arr,n);
    return 0;
}
-------------------------------------------

10. Write a function that checks whether an element occurs in a array.
Solution : 
#include <iostream>
using namespace std;

int main() {

  int i, n;
  float arr[100];

  cout << "Enter total number of elements(1 to 100): ";
  cin >> n;
  cout << endl;

  // Store number entered by the user
  for(i = 0; i < n; ++i) {
    cout << "Enter Number " << i + 1 << " : ";
    cin >> arr[i];
  }

  // Loop to store largest number to arr[0]
  for(i = 1;i < n; ++i) {

    // Change < to > if you want to find the smallest element
    if(arr[0] < arr[i])
      arr[0] = arr[i];
  }

  cout << endl << "Largest element = " << arr[0];

  return 0;
}

11. Write a function that returns the elements on odd positions in a list.
Solution : 
# Python code to return the elements on odd positions in a list.
a = [41, 19, 74, 107, 12309, -82, 64, 39, 501, 124, 70, 1111]
length_of_a = len(a)
for i in range(1, length_of_a, 2):
    print(a[i], end=" ")

=========c++=========
vector<int> oddPosition(vector<int> arr, int n) {
    vector<int> oddPos(n/2);
    for(int i = 1; i < n; i += 2) {
        oddPos.push_back(arr[i]);
    }
    return oddPos;
}

12. Write three functions that compute the sum of the numbers in a array: using a for-loop, a while-loop and recursion.
Solution :  
class code{
public static void main(String[] args){
int[] arr={1,2,3,4,5};
int ans1=usingfor(arr);
int ans2=usingwhile(arr);
int ans3=using rec(arr);
}
public int usingfor(int[] arr){
int ans=0;
for(int i=0; i<arr.lengthl i++){
ans=ans+arr[i];
}
return ans;
}
public int usingwhile(int[] arr){
int ans=0;
int i=0;
while(i<arr.length){
ans=ans+arr[i];
i++;
}
return ans;
}
public int usingrec(int[] arr, int i){
if(i=>arr.length){
return 0;
}
return arr[i]+usingrec(arr,i+1);
}
}


13. Write a function that concatenates two arrays. [a,b,c], [1,2,3] → [a,b,c,1,2,3]
Solution : 
public class Concat {

    public static void main(String[] args) {
        int[] array1 = {1, 2, 3};
        int[] array2 = {4, 5, 6};

        int aLen = array1.length;
        int bLen = array2.length;
        int[] result = new int[aLen + bLen];

        System.arraycopy(array1, 0, result, 0, aLen);
        System.arraycopy(array2, 0, result, aLen, bLen);

        System.out.println(Arrays.toString(result));
    }
}


14. Write a function that merges two sorted into a new sorted list. [1,4,6],[2,3,5] → [1,2,3,4,5,6]. You can do this quicker than concatenating them followed by a sort.
Solution : 

def merge(list1,list2):
    print(sorted(list1 + list2))
    
list1 = [1,2,3,4,5]
list2= [10,20,30,40,50,60]
merge(list1,list2)

15. Write a function that computes the list of the first n Fibonacci numbers. The first two Fibonacci numbers are 1 and 1. The first few are therefore 1, 1, 1+1=2, 1+2=3, 2+3=5, 3+5=8.
Solution : 
class Main {
  public static int fibonacci_numbers(int n)
  {
    if(n == 0){
      return 0;
    }
    else if(n == 1){
      return 1;
    }
    else{
      return fibonacci_numbers(n-2) + fibonacci_numbers(n-1);
    }
  }
  public static void main (String[] args) {
    int n = 7;
    for(int i = 0; i < n; i++){
      System.out.print(fibonacci_numbers(i)+ " ");
    }
  }
}

16. Write a function that return the Factorial of given number n.
Solution : 
# Python 
def factorial(n):
    if n==0:
        return 1
    else:
        return n*factorial(n-1)

17. Print the following pattern without using print in-built function.
i.      ii.     iii.
***     123     321
***     123     321
***     123     321

Solution : 
//cpp
for(int i=0;i<3;i++){
for(int j=0;j<3;j++){
cout<<"*";
}
}

for(int i=0;i<3;i++){
for(int j=1;j<=3;j++){
cout<<"j";
}
}

for(int i=0;i<3;i++){
for(int j=3;j>=0;j++){
cout<<"j";
}
}



18. Convert i. Minutes -> Seconds 
            ii. Fahrenheit -> Celsius -> Kelvin

Solution : #include<stdio.h>
#include<conio.h>
int main()
{
	 float celsius, fahr, kelvin;
	 clrscr();
	 printf("Enter temperature in celsius: ");
	 scanf("%f", &celsius);
	 fahr = 1.8 * celsius + 32.0;
	 kelvin = 273.15 + celsius;
	 printf("%0.2f Celsius = %0.2f Fahrenheit\n", celsius, fahr);
	 printf("%0.2f Celsius = %0.2f Kelvin",celsius, kelvin);
	 getch();
	 return(0);
}


19. Design a quiz game which take answer as input to given question (atleast 5) and print final result(Total marks & percentage). +2 for Correct answer, -0.5 for Incorrect.
Solution : 
def new_game():

    guesses = []
    correct_guesses = 0
    question_num = 1

    for key in questions:
        print("-------------------------")
        print(key)
        for i in options[question_num-1]:
            print(i)
        guess = input("Enter (A, B, C, or D): ")
        guess = guess.upper()
        guesses.append(guess)

        correct_guesses += check_answer(questions.get(key), guess)
        question_num += 1

    display_score(correct_guesses, guesses)

# -------------------------
def check_answer(answer, guess):

    if answer == guess:
        print("CORRECT!")
        return 1
    else:
        print("WRONG!")
        return 0

# -------------------------
def display_score(correct_guesses, guesses):
    print("-------------------------")
    print("RESULTS")
    print("-------------------------")

    print("Answers: ", end="")
    for i in questions:
        print(questions.get(i), end=" ")
    print()

    print("Guesses: ", end="")
    for i in guesses:
        print(i, end=" ")
    print()

    score = int((correct_guesses/len(questions))*100)
    print("Your score is: "+str(score)+"%")

# -------------------------
def play_again():

    response = input("Do you want to play again? (yes or no): ")
    response = response.upper()

    if response == "YES":
        return True
    else:
        return False
# -------------------------


questions = {
 "Who created Python?: ": "A",
 "What year was Python created?: ": "B",
 "Python is tributed to which comedy group?: ": "C",
 "Is the Earth round?: ": "A",
    "Value of pi?: ": "C",
}

options = [["A. Guido van Rossum", "B. Elon Musk", "C. Bill Gates", "D. Mark Zuckerburg"],
          ["A. 1989", "B. 1991", "C. 2000", "D. 2016"],
          ["A. Lonely Island", "B. Smosh", "C. Monty Python", "D. SNL"],
          ["A. True","B. False", "C. sometimes", "D. What's Earth?"],
          ['1.24','2.34','3.14','3.68']]

new_game()

while play_again():
    new_game()

print("Byeeeeee!")

20. Design a tool which find area & perimeter of different shapes.
Solution : 	
# define a function for calculating
# the area of a shapes
def calculate_area(name):\

	# converting all characters
	# into lower cases
	name = name.lower()

	# check for the conditions
	if name == "rectangle":
		l = int(input("Enter rectangle's length: "))
		b = int(input("Enter rectangle's breadth: "))

		# calculate area of rectangle
		rect_area = l * b
		print(f"The area of rectangle is {rect_area}.")


	elif name == "square":
		s = int(input("Enter square's side length: "))

		# calculate area of square
		sqt_area = s * s
		print(f"The area of square is {sqt_area}.")


	elif name == "triangle":
		h = int(input("Enter triangle's height length: "))
		b = int(input("Enter triangle's breadth length: "))

		# calculate area of triangle
		tri_area = 0.5 * b * h
		print(f"The area of triangle is {tri_area}.")


	elif name == "circle":
		r = int(input("Enter circle's radius length: "))
		pi = 3.14

		# calculate area of circle
		circ_area = pi * r * r
		print(f"The area of circle is {circ_area}.")


	elif name == 'parallelogram':
		b = int(input("Enter parallelogram's base length: "))
		h = int(input("Enter parallelogram's height length: "))

		# calculate area of parallelogram
		para_area = b * h
		print(f"The area of parallelogram is
		{para_area}.")

	else:
		print("Sorry! This shape is not available")

# driver code
if __name__ == "__main__" :

print("Calculate Shape Area")
shape_name = input("Enter the name of shape whose area you want to find: ")

# function calling
calculate_area(shape_name)

21. Return first, middle & last element of array.
Solution :
#python
def func():
    arr = [1,2,3,4,5]
    ans = []
    middle = 5//2
    ans.append(arr[0])
    ans.append(arr[middle])
    ans.append(arr[-1])
    return ans
answer = func()
print(answer)


22. Largest subarray with 0 sum
Solution : 
class GfG
{
    
    int maxLen(int arr[], int n)
    {
        
        // Your code here
        HashMap<Integer,Integer>a=new HashMap<>();
        int s=0;
        int count=0;
        a.put(0,-1);
        for(int i=0;i<n;i++)
        {
            s=s+arr[i];
            if(s==0 || a.containsKey(s))
            {
                count=Math.max(count,i-a.get(s));
            }
            else
            {
                a.put(s,i);
            
        }
       
    }
     return count;
}
}


23.Find duplicates in an array
Solution :
class Solution {
    public static ArrayList<Integer> duplicates(int arr[], int n) { 
        
        // code here
        int a[]= new int [n];
      int c=0;
      ArrayList<Integer> list = new ArrayList<>();
      for(int i=0;i<n;i++)
      {
          a[arr[i]]=a[arr[i]]+1;
      }
      for(int j=0;j<n;j++)
      {
          if(a[j]>1){
           list.add(j);
          c++;
              
          }
           else
               continue;
      }
      if(c==0)
      {
       list.add(-1);
           return list;
      }
      
       else
           return list;
    }
}

=======
#python
a = []
n=int(input("enter total number of elements to be added in array:"))
for i in range(0,n):
    
    x=input("Enter element to add in array:")
    a.append(x)
print("Array: ", a)
mid=(0+arr.length-1)/2;
print("first element is" ,a[0])
print("last element is", a[-1])
print("middle element is",a[mid])

=======
#Java
static int[] printArray(int[] arr){
    int mid=(0+arr.length-1)/2;
    int[] nums=new int[3];
    //Storing values in new array
    nums[0]=arr[0];
    nums[1]=arr[mid];
    nums[2]=arr[arr.length-1];
    System.out.println("First,middle,last element of array is");
    return nums;
}

24.Write a program to find the gcd of 2 numbers
Solution : 
#C++

int gcd(int a,int b)
{
	if(b==0)return a;

	return gcd(b,a%b);
}

int main()
{
  int a,b;
  cin>>a>>b;
  cout<<gcd(a,b)<<endl;
  //Time Complexity:O(log(min(a,b)))
}


25.Conversion of Binary number to its decimal
#C
Solution:
#include<stdio.h>

int main()
{
  int T,N,K,i,j,A,B;
  char s[100][100];

  scanf("%d",&T);

  for(i=0;i<T;i++)
  {
    scanf("%d %d %d",&N,&A,&B);

    scanf("%s",s[j]);
K=0;
    for(j=0;j<N;j++)
    {
     if(s[i][j]=='0')
     K=K+A;
     if(s[i][j]=='1')
     K=K+B;
    }

    printf("%d\n",K);
  }
  return 0;
}

26.Split the array into equal sum subarrays.
Solution:
sample = [2, 4, 5, 1, 2, 3]
 
sum = 0
 
sum_left=[]
for val in sample:
    sum += val
    sum_left.append(sum)
     
sum_right=[]
for val in sample:
    sum_right.append(sum)
    sum -= val

for i in range(len(sum_left)):
    if sum_left[i] == sum_right[i]:
        print(f"Matching index is {i}.")
        break
	}

27 the minimum numbers of operations required by Chef to make parity of all the elements same.
Solution: 
#include <iostream>
using namespace std;
#define ll long long 
int main() {
	// your code goes here
	int t;
	cin>>t;
	while(t--)
	{
	    int n;
	    cin>>n;
	    ll arr[n];
	    ll even=0,odd=0;
	    for(int i=0;i<n;i++)
	    {
	        cin>>arr[i];
	        if(arr[i]%2==1)
	        {
	            odd++;
	        }
	       
	    }
	    if(odd==n || odd== 0)
	    {
	        cout<<"0"<<endl;
	    }
	    else
	    {
	        cout<<n-odd<<endl;
	    }
	}
	return 0;	
	
28. Equilibrium index of an array
Solution:	
int findEquilibrium(int A[], int n)
{
  //Your code here
  double s=0;
  for(int i=0;i<n;i++){
      s=s+A[i];
      
  }
  double s1=0;
  for(int i=0;i<n;i++){
      s1=s1+A[i];
      if(((s-A[i+1])/2)==s1){
          return i+1;
          
      }
  }
  return -1;
  
}


29. Polynomials
Solution: 
import numpy as np
poly = [float(x) for x in input().split()]
x = float(input())
print(np.polyval(poly, x))

30. Implement Tower of Hanoi.
Solution:
def hanoi(disks, source, auxiliary, target):
    if disks == 1:
        print('Move disk 1 from peg {} to peg {}.'.format(source, target))
        return
    hanoi(disks - 1, source, target, auxiliary)
    print('Move disk {} from peg {} to peg {}.'.format(disks, source, target))
    hanoi(disks - 1, auxiliary, source, target) 
disks = int(input('Enter number of disks: '))
hanoi(disks, 'A', 'B', 'C')

31. Kadane's Algorithm (largest contiguous array sum)
Solution:
#include <bits/stdc++.h>
using namespace std;

int maxSubArraySum(int a[], int size)
{
	int max_so_far = INT_MIN, max_ending_here = 0;
	for (int i = 0; i < size; i++) {
		max_ending_here = max_ending_here + a[i];
		if (max_so_far < max_ending_here)
			max_so_far = max_ending_here;

		if (max_ending_here < 0)
			max_ending_here = 0;
	}
	return max_so_far;
}
int main()
{
	int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };
	int n = sizeof(a) / sizeof(a[0]);
	int max_sum = maxSubArraySum(a, n);
	cout << "Maximum sum is " << max_sum;
	return 0;
}

32. Print all possible subsets of a given array. (power set)
Solution:	
	class Solution {
	    public void powerSet(int[] arr, int i, ArrayList<Integer> ll, List<List<Integer>> ps){
		if(i>=arr.length){
		    ps.add(new ArrayList<>(ll));
		    return;
		}
		ll.add(arr[i]);
		powerSet(arr, i+1, ll, ps);
		ll.remove(ll.size()-1);
		powerSet(arr, i+1, ll, ps);
	    }
	    public List<List<Integer>> subsets(int[] nums) {
		List<List<Integer>> ps = new ArrayList<>();
		powerSet(nums, 0, new ArrayList<>(), ps);
		return ps;
	    }
	}

33.Implementation of the QuickSort Algorithm in C++ Language
#include <iostream>
using namespace std;

int partition(int arr[], int start, int end)
{

    int pivot = arr[start];

    int count = 0;
    for (int i = start + 1; i <= end; i++)
    {
        if (arr[i] <= pivot)
            count++;
    }
    int pivotIndex = start + count; //---->Placing the pivot element to its right position
    swap(arr[pivotIndex], arr[start]);

    // Sorting the subarrays left part and right part ot the pivot element
    int i = start;
    int j = end;

    while (i < pivotIndex && j > pivotIndex)
    {

        while (arr[i] <= pivot)
        {
            i++;
        }
        while (arr[j] > pivot)
        {
            j--;
        }
        if (i < pivotIndex && j > pivotIndex)
        {
            swap(arr[i++], arr[j--]);
        }
    }
    return pivotIndex;
}
void QuickSort(int arr[], int start, int end)
{
    if (start >= end)
    {
        return;
    }

    int p = partition(arr, start, end); //----->Partitioning the arr_data

    QuickSort(arr, start, p - 1); //---->Sorting the left subarray (left part)

    QuickSort(arr, p + 1, end); //---->Sorting the right subarray (right part)
}
int main()
{
    int arr_data[] = {99, 67, 4, 656, 7, 88};
    int size = 6;
    cout << "THE DATA BEFORE APPLYING QUICKSORT\n";
    for (int i = 0; i < size; i++)
    {
        cout << arr_data[i] << " ";
    }
    QuickSort(arr_data, 0, size - 1);
    cout << "\n\nTHE DATA AFTER APPLYING QUICKSORT\n";
    for (int i = 0; i < size; i++)
    {
        cout << arr_data[i] << " ";
    }
    return 0;

34. C Program to Calculate Standard Deviation.
Solution:
#include <stdio.h>
#include <math.h>

void main()
{
  float Price[50];
  int  i, Number;
  float Mean, Variance, SD, Sum=0, Differ, Varsum=0;

  printf("\nPlease Enter the N Value\n");
  scanf("%d", &Number);

  printf("\nPlease Enter %d real numbers\n",Number);
  for(i=0; i<Number; i++)
   {
     scanf("%f", &Price[i]);
   }

  for(i=0; i<Number; i++)
   {
     Sum = Sum + Price[i];
   }
  
  Mean = Sum /(float)Number;

  for(i=0; i<Number; i++)
   {
     Differ = Price[i] - Mean;
     Varsum = Varsum + pow(Differ,2);
   }
  
  Variance = Varsum / (float)Number;
  SD = sqrt(Variance);
  
  printf("Mean               = %.2f\n", Mean);
  printf("Varience           = %.2f\n", Variance);
  printf("Standard deviation = %.2f\n", SD);
}


35. Getting the kth smallest element from the array
Example : Array = [1, 4, 5, 3, 2]
	  2nd smallest element : 2
# Java
Solution : 
	import java.lang.*;
	import java.util.*;

	class Demo
	{
		public static void main(String args[])
		{
			Scanner scan = new Scanner(System.in);
			
			System.out.print("Enter the size of array : ");
			int n = scan.nextInt();
			
			int[] arr = new int[n];
			
			System.out.print("Enter the array elements : ");
			for(int i = 0; i < n ; i++)
			{
				int temp = scan.nextInt();
				arr[i] = temp;
			}
			
			System.out.print("Enter the value of k to get the kth smallest value in  array : ");
			int k = scan.nextInt();
			
			Arrays.sort(arr);
			
			System.out.println("The kth smallest element is : " + arr[k-1]);
		}
	}

36. Implementation of Insertion sort in C
Example : Array = {23, 15, 29, 11, 1}
	  After insertion sort : {1, 11, 15, 23, 29}
# C language
Solution : 
	#include <stdio.h>

	void display(int array[], int size)
	{
	    for (int i = 0; i < size; i++)
	    {
		printf("%d ", array[i]);
	    }
	}

	void insertion_sort(int array[], int size)
	{
	    int hole, value, i;
	    for (i = 0; i < size; i++)
	    {
		value = array[i];
		hole = i;
		while (hole > 0 && array[hole - 1] > value)
		{
		    array[hole] = array[hole - 1];
		    hole = hole - 1;
		}
		array[hole] = value;
	    }
	}

	int main()
	{
	    int array[5] = {23, 15, 29, 11, 1};
	    int size = 5;

	    insertion_sort(array, size);
	    display(array, size);

	    return 0;
	}
	
37.Flatten binary tree to linked list
# java language
Solution : class Solution
{
    public static void flatten(Node root)
    {
        //code here
        if (root == null || root.left == null &&
	        root.right == null) {
    		return;
    	}
    	
    	Node t = root.right;

    	// if root.left exists then we have
    	// to make it root.right
    	if (root.left != null) {
    
    		// move left recursively
    		flatten(root.left);
    
    		// store the node root.right
    		Node tmpRight = root.right;
    		root.right = root.left;
    		root.left = null;
    
    		// find the position to insert
    		// the stored value
    		t = root.right;
    		while (t.right != null) {
    			t = t.right;
    		}
    
    		// insert the stored value
    		t.right = tmpRight;
    	}
    
    	// now call the same function
    	// for root.right
    	flatten(t);
    }
}

38.Sum of k smallest elements in BST
# java language
Solution : class Tree {
    int sum;
    int cnt;
    int sum(Node root, int k) { 
        sum=0;
        cnt=0;
        dfs(root,k);
        return sum;
        // Code here
    } 
    void dfs(Node root,int k){
        if(root==null)return;
        dfs(root.left,k);
        if(cnt<k){
            sum+=root.data;
            cnt++;
        }
        if(cnt<k){
            dfs(root.right,k);
        }
    }
}


39. Write a cpp program to print the sum of all nodes in Binaey Tree

Solution :
// CPP
/**********************************************************
	Binary Tree Node class structure

	template <typename T>
	class BinaryTreeNode {
    	public : 
    	T data;
    	BinaryTreeNode<T> *left;
    	BinaryTreeNode<T> *right;

    	BinaryTreeNode(T data) {
        	this -> data = data;
        	left = NULL;
        	right = NULL;
    	}
	};

***********************************************************/

// Solution.cpp
int getSum(BinaryTreeNode<int>* root) {
    if(root == NULL)
        return 0;
    return root->data + getSum(root->left) + getSum(root->right);
        
    
}

// main.cpp
#include <iostream>
#include <queue>

template <typename T>
class BinaryTreeNode {
   public:
    T data;
    BinaryTreeNode<T>* left;
    BinaryTreeNode<T>* right;

    BinaryTreeNode(T data) {
        this->data = data;
        left = NULL;
        right = NULL;
    }
};

using namespace std;
#include "solution.cpp"

BinaryTreeNode<int>* takeInput() {
    int rootData;
    cin >> rootData;
    if (rootData == -1) {
        return NULL;
    }
    BinaryTreeNode<int>* root = new BinaryTreeNode<int>(rootData);
    queue<BinaryTreeNode<int>*> q;
    q.push(root);
    while (!q.empty()) {
        BinaryTreeNode<int>* currentNode = q.front();
        q.pop();
        int leftChild, rightChild;

        cin >> leftChild;
        if (leftChild != -1) {
            BinaryTreeNode<int>* leftNode = new BinaryTreeNode<int>(leftChild);
            currentNode->left = leftNode;
            q.push(leftNode);
        }

        cin >> rightChild;
        if (rightChild != -1) {
            BinaryTreeNode<int>* rightNode =
                new BinaryTreeNode<int>(rightChild);
            currentNode->right = rightNode;
            q.push(rightNode);
        }
    }
    return root;
}

int main() {
    BinaryTreeNode<int>* root = takeInput();
    cout << getSum(root);
}

41.Given an array A of size N of integers,find the sum of minimum and maximum element in the array?
# java language
Solution :class Solution
{ 
      
    /* Function to print the second largest 
    elements */
    public static int findSum(int arr[],  
                                     int arr_size) 
    { 
       int max,min;
    	int i;
    	if(arr_size % 2 == 1){
			max = arr[0];
			min = arr[0];
			i = 1;
		}
		else{
			if ( arr[0] < arr[1] ){
				max = arr[1];
				min = arr[0];
			}
			else
			{
				max = arr[0];
				min = arr[1];
			}
			i = 2;
		}
		while(i < arr_size){
			if ( arr[i] < arr[i+1] ){
				if ( arr[i] < min )
					min = arr[i];
				if ( arr[i+1] > max )
					max = arr[i+1];
			}
			else
			{
				if ( arr[i] > max )
					max = arr[i];
				if ( arr[i+1] < min )
					min = arr[i+1];
			}
			i = i + 2;
		}
		return min + max;
    } 
}

42. Find the Missing Number, Given an array arr[] of size N-1 with integers in the range of [1, N], the task is to find the missing number from the first N integers
using c++
solution - C++ program to Find the missing element

#include <bits/stdc++.h>
using namespace std;

void findMissing(int arr[], int N)
{
	int i;
	int temp[N + 1];
	for(int i = 0; i <= N; i++){
	temp[i] = 0;
	}

	for(i = 0; i < N; i++){
	temp[arr[i] - 1] = 1;
	}


	int ans;
	for (i = 0; i <= N ; i++) {
		if (temp[i] == 0)
			ans = i + 1;
	}
	cout << ans;
}

/* Driver code */
int main()
{
	int arr[] = { 1, 3, 7, 5, 6, 2 };
	int n = sizeof(arr) / sizeof(arr[0]);
	findMissing(arr, n);
}

43.Sort the elemnts in an array by using selection sort in c programming

#c language

#include<stdio.h>

//declaration of display function
void display(int array[], int size)
{
    for (int i = 0; i < size; i++)
    {
        printf("%d ", array[i]);
    }
}


void selection_sort(int array[], int size)
{
    int least, i, j, temp, m,l;
    
    for ( i = 0; i < size; i++)         
    {
        //least = array[i];   
        int m=i;
        for ( j = i + 1; j < size; j++)         
        {
            if (array[m]> array[j])           
            {
               // least = array[j];        
                m=j;         
            }   
        }
        //swap lopgic in least and ith element
        if(i!=m)
        {
        temp = array[m];
        array[m] = array[i];
        array[i] = temp;
        }
    }
}

int main()
{
    int array[20] ;
    int size ;

    printf("Enter the size of array :\n");
    scanf("%d",&size);
    
    printf("Enter the data that we have to sort :\n");
    for(int i=0;i<size;i++)
    {
        scanf("%d",&array[i]);   
    }

    //calling the selection_sort funtion
    selection_sort(array, size);
     
    //calling the display function
    display(array, size);

    return 0;
}