From 70da0c9be515187a44781677f218a721dabf6ef4 Mon Sep 17 00:00:00 2001
From: Pranav Sorte <pranavsorte@Pranavs-MacBook-Air.local>
Date: Mon, 1 Dec 2025 21:38:19 -0800
Subject: [PATCH 1/2] done

---
 KDiffPairs.java    | 43 +++++++++++++++++++++++++++++++++++++++++++
 PascalTriange.java | 37 +++++++++++++++++++++++++++++++++++++
 2 files changed, 80 insertions(+)
 create mode 100644 KDiffPairs.java
 create mode 100644 PascalTriange.java

diff --git a/KDiffPairs.java b/KDiffPairs.java
new file mode 100644
index 00000000..1220e2a6
--- /dev/null
+++ b/KDiffPairs.java
@@ -0,0 +1,43 @@
+// Time Complexity : O(n) + O(n) ~ O(n) where n is the number of elements 
+// 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
+/*
+I first counted the number of all the occurences of the numbers. The reason why I counted first is because the 2nd number might be encounter later than the first.
+The formula nums[i] - nums[j] == k is equivalent to nums[j] = nums[i] - k. So if k > 0, we  will find the nums[i] - k in the hashmap, if we find it which means the pair satisfies the condition and increment the count.
+One thing to note is that if k == 0, then if the number of occurences is greater than 2, it means we found a pair and increment the count.
+*/
+
+import java.util.HashMap;
+import java.util.Map;
+
+public class KDiffPairs {
+    public int findPairs(int[] nums, int k) {
+        HashMap<Integer, Integer> hmap = new HashMap<>();
+        int res = 0;
+
+        for(int num : nums) {
+            hmap.put(num, hmap.getOrDefault(num, 0) + 1);
+        }
+
+        for(Map.Entry<Integer,Integer> entry : hmap.entrySet()) {
+            int key = entry.getKey();
+            int val = entry.getValue();
+
+            if(k == 0) {
+                if(val > 1) {
+                    res++;
+                }
+            } else {
+                if(hmap.containsKey(key + k)) {
+                    res++;
+                }
+            }
+        }
+
+        return res;
+    }
+}
diff --git a/PascalTriange.java b/PascalTriange.java
new file mode 100644
index 00000000..0110f549
--- /dev/null
+++ b/PascalTriange.java
@@ -0,0 +1,37 @@
+// Time Complexity : O(n^2) where n is the numRows
+// 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
+/*
+I first stored the initial array of [[1]] in the result.
+I then started from row = 1 and then used the formula res[i-1][j-1] + res[i-1][j] to calculate the remaining columns.
+*/
+
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+public class PascalTriange {
+    public List<List<Integer>> generate(int numRows) {
+        List<List<Integer>> res = new ArrayList<>();
+        res.add(new ArrayList<>(Arrays.asList(1)));
+
+        for(int i = 1;i<numRows;i++) {
+            List<Integer> temp = new ArrayList<>();
+            for(int j = 0;j<=i;j++){
+                if(j == 0 || j == i) {
+                    temp.add(1);
+                } else {
+                    temp.add(res.get(i-1).get(j-1) + res.get(i-1).get(j));
+                }
+            }
+            res.add(temp);
+        }
+
+        return res;
+    }
+}

From 360dd5fd7a17cfdb21c5138ea68dacc46a28c684 Mon Sep 17 00:00:00 2001
From: Pranav Sorte <pranavsorte@Pranavs-MacBook-Air.local>
Date: Mon, 1 Dec 2025 21:38:44 -0800
Subject: [PATCH 2/2] done

---
 KDiffPairs.java | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/KDiffPairs.java b/KDiffPairs.java
index 1220e2a6..df875092 100644
--- a/KDiffPairs.java
+++ b/KDiffPairs.java
@@ -1,7 +1,7 @@
 // Time Complexity : O(n) + O(n) ~ O(n) where n is the number of elements 
 // Space Complexity : O(n)
 // Did this code successfully run on Leetcode : Yes
-// Any problem you faced while coding this : No
+// Any problem you faced while coding this : Yes, was not able to come up with the solution in the mock interview
 
 
 // Your code here along with comments explaining your approach