Done Competetive-Coding-3

Problem-1.py

@@ -0,0 +1,27 @@
+# Approach
+# since every index isdepend on the previous row i, i-1
+# recuurelation =  curr[i] =  prev[i]+prev[i-1]
+
+
+#Complexities
+# Time Complexity: O(N*N)
+# Space Complexity: O(N)
+
+def pascalTriangle(num):
+    num  = num+1
+    if num ==1:
+        return[1]
+
+    prev = [1]
+    for i in range(2,num+1):
+        curr = [1]
+        for j in range(1,i-1):
+
+            curr.append(prev[j-1]+prev[j])
+        curr.append(1)
+        prev = curr
+
+    return prev
+
+
+print(pascalTriangle(5))

Problem-2.py

@@ -0,0 +1,33 @@
+#Approach
+# Since We need to find the absolute difference is k . if num+ target or num-target is present in the hashSet then add thenum, num-target in the resukt
+
+
+#Complexities
+# Time Complexity: 0(N)
+# Space Complexity: O(N)
+
+
+
+
+from typing import List
+
+
+class Solution:
+    def findPairs(self, nums: List[int], k: int) -> int:
+        hashSet = set()
+        result = set()
+        for i in nums:
+            if i - k in hashSet:
+                if i > i - k:
+
+                    result.add((i, i - k))
+                else:
+                    result.add((i - k, i))
+            if i + k in hashSet:
+                if i > i + k:
+                    result.add((i, i + k))
+                else:
+                    result.add((i + k, i))
+            hashSet.add(i)
+
+        return len(set(result))