From e57eb4721a326e1e2e41a2b939ecbba85dcddcbd Mon Sep 17 00:00:00 2001
From: Manisha Rana <manisha1195@gmail.com>
Date: Mon, 12 Jan 2026 20:09:47 -0500
Subject: [PATCH] CC-3 Done

---
 KDiffPairs.java     | 42 ++++++++++++++++++++++++++++++++++++++++++
 PascalTriangle.java | 44 ++++++++++++++++++++++++++++++++++++++++++++
 2 files changed, 86 insertions(+)
 create mode 100644 KDiffPairs.java
 create mode 100644 PascalTriangle.java

diff --git a/KDiffPairs.java b/KDiffPairs.java
new file mode 100644
index 00000000..f400b020
--- /dev/null
+++ b/KDiffPairs.java
@@ -0,0 +1,42 @@
+// Time Complexity : O(N)
+// Space Complexity : O(N), extra hashmap used
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+// Approach
+// Since we need unique pair, we dont want any value to be considered multiple times, causing duplicate pairs.
+// Need to handle edge case if k=0, then that occurs 2 or more times should be considered a pair.
+// We will keep a frequency map, go over each key, try to find if there exists key-K value in the map, if yes, increase
+// the result count.
+import java.util.HashMap;
+import java.util.Map;
+
+public class KDiffPairs {
+
+    public int findPairs(int[] nums, int k) {
+        // Frequency map makes sure a value is not part of multiple pairs
+        Map<Integer, Integer> temp = new HashMap<>();
+        int count = 0;
+        for(int num: nums){
+            if(temp.containsKey(num)){
+                temp.put(num, temp.get(num)+1);
+            }else {
+                temp.put(num, 1);
+            }
+           // temp.merge(num, 1, Integer::sum);
+        }
+
+        for(int key: temp.keySet()){
+            if(k==0){
+                // If no difference, then the same value can create a pair if it has occurred 2 or more times.
+                if(temp.get(key) > 1){
+                    count++;
+                }
+            }else if(temp.containsKey(key-k)){
+                count++;
+            }
+        }
+
+        return count;
+    }
+}
diff --git a/PascalTriangle.java b/PascalTriangle.java
new file mode 100644
index 00000000..7ca31bf7
--- /dev/null
+++ b/PascalTriangle.java
@@ -0,0 +1,44 @@
+// Time Complexity : O(NXN), where N is number of rows
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+// Approach
+// One approach is we can add 1 to beginning and ending of each row to avoid going beyond the boundaries.
+// Second approach check if i/j is beyond boundary, dont fetch them just use 0 and add it to [i-1],[j] value.
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class PascalTriangle {
+    public List<List<Integer>> generate(int numRows) {
+        List<List<Integer>> result = new ArrayList<>();
+        if(numRows == 0) return result;
+        // create first row
+        result.add(List.of(1));
+
+        for (int i = 1; i < numRows; i++) {
+            List<Integer> temp = new ArrayList<>();
+            for (int j = 0; j < numRows; j++) {
+                int top = 0;
+                // Check if [i-1][j] value exist by comparing j and size of previous row. Else use 0.
+                if (j < result.get(i-1).size()) {
+                    top = result.get(i - 1).get(j);
+                }
+                int topLeft = 0;
+                // Check if [i-1][j-1] value exists, this will fail when j is at the boundary, j==0. Else use 0.
+                if (j > 0) {
+                    topLeft = result.get(i - 1).get(j - 1);
+                }
+                temp.add(top + topLeft);
+                // As soon as required number of elements are processes break. 
+                if(j == i){
+                    break;
+                }
+            }
+            result.add(temp);
+        }
+
+        return result;
+    }
+}