Completed Competitive_Coding-3

Problem-1.py

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+#Pascal's Triangle
+# Time Complexity :O(n2)
+# Space Complexity :O(1)
+# Did this code successfully run on Leetcode :yes
+# Any problem you faced while coding this :no
+
+
+# Your code here along with comments explaining your approach
+#start with first row of pascal triangle as [1] and then iterate through numsRows and keep adding arrays to result array by adding sum of the two numbers directly above and also parallely checking the two edges of each row
+#time - O(n2)
+#space - O(1)
+
+class Solution:
+    def generate(self, numRows: int) -> List[List[int]]:
+        result = [[]]
+        if(numRows==0):
+            return result
+        result = [[1]]
+        i=2
+        while(i<=numRows):
+            p1=0
+            p2=1
+            curr=[]
+            j=0
+            while(j<i):
+                if(j==0 or j==i-1):  #check if it is edges of each row
+                    curr.append(1)
+                    j=j+1
+                    continue
+                curr.append(result[i-2][p1]+result[i-2][p2]) #sum of the two numbers directly above 
+                j=j+1
+                p1=p1+1
+                p2=p2+1
+            result.append(curr)
+            i=i+1
+        return result
+
+

Problem-2.py

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+#K-diff Pairs in an Array
+# Time Complexity :O(n)
+# Space Complexity :O(n)
+# Did this code successfully run on Leetcode :yes
+# Any problem you faced while coding this :no
+
+
+# Your code here along with comments explaining your approach
+#maintain a frequency map, iterate through hashmap and if incase the diff is zero increase count by 1 at each element 
+#check if key+k is present in map dna increase +1 as pair is found
+# time - O(n)
+# space - O(n)
+class Solution:
+    def findPairs(self, nums: List[int], k: int) -> int:
+        hashMap = Counter(nums) # frequency map
+        count=0
+        for key, value in hashMap.items():
+            if k==0:
+                if value>1:
+                    count =count+1
+                continue
+            if key+k in hashMap:
+                count=count+1
+        return count
+
+
+