From d3b4de82b22de61cc3570d494efc1951a74a3f87 Mon Sep 17 00:00:00 2001
From: sreevardhan1099 <sreevardhan1099@gmail.com>
Date: Thu, 15 Jan 2026 18:25:52 -0500
Subject: [PATCH] Completed Competitive_Coding-3

---
 Problem-1.py | 38 ++++++++++++++++++++++++++++++++++++++
 Problem-2.py | 27 +++++++++++++++++++++++++++
 2 files changed, 65 insertions(+)
 create mode 100644 Problem-1.py
 create mode 100644 Problem-2.py

diff --git a/Problem-1.py b/Problem-1.py
new file mode 100644
index 00000000..4f604be2
--- /dev/null
+++ b/Problem-1.py
@@ -0,0 +1,38 @@
+#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
+        
+        
\ No newline at end of file
diff --git a/Problem-2.py b/Problem-2.py
new file mode 100644
index 00000000..25be5ab6
--- /dev/null
+++ b/Problem-2.py
@@ -0,0 +1,27 @@
+#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
+
+
+        
\ No newline at end of file