diff --git a/Exercise1.cpp b/Exercise1.cpp
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+++ b/Exercise1.cpp
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+// Problem - Pascal Triangle
+//  Time Complexity : O(n^2)
+//  Space Complexity : O(1)
+//  Did this code successfully run on Leetcode : Yes
+//  Any problem you faced while coding this : No
+
+// Explaination
+// 1. We initialize the result 2d array with 1 element our base condition
+// 2. We go through each row and if we have our at first or last element we just
+// add 1
+// 3. Otherwise we add previous_row above_col value and previous_row above_col -
+// 1 value
+
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution {
+public:
+  vector<vector<int>> generate(int numRows) {
+    vector<vector<int>> res = {{1}};
+    if (numRows == 1)
+      return res;
+    for (int i = 1; i < numRows; i++) {
+      vector<int> temp;
+      for (int j = 0; j <= i; j++) {
+        // base cond
+        if (j == 0 || j == i) {
+          temp.push_back(1);
+        } else {
+          // logic
+          temp.push_back(res[i - 1][j] + res[i - 1][j - 1]);
+        }
+      }
+      res.push_back(temp);
+    }
+
+    return res;
+  }
+};
diff --git a/Exercise2.cpp b/Exercise2.cpp
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+++ b/Exercise2.cpp
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+// Problem - K-diff pairs in an Array
+//  Time Complexity : O(n)
+//  Space Complexity : O(1)
+//  Did this code successfully run on Leetcode : Yes
+//  Any problem you faced while coding this : No
+
+// Explaination
+// 1. We first sort the array so that we can use two pinters to navigate
+// 2. We treat like a two sum problem skip duplicates and not using the same
+// index, and make the pointer forward based on current value < k or > k
+// 3. In the end we return the count
+
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution {
+public:
+  int findPairs(vector<int> &nums, int k) {
+    sort(nums.begin(), nums.end());
+    int n = (int)nums.size();
+    int count = 0;
+    int i = 0;
+    int j = 0;
+
+    while (i < n && j < n) {
+      if (i == j || nums[j] - nums[i] < k)
+        j++;
+      else if (nums[j] - nums[i] > k)
+        i++;
+      else {
+        i++;
+        j++;
+        count++;
+        while (i < n && nums[i] == nums[i - 1])
+          i++;
+      }
+    }
+
+    return count;
+  }
+};