completed Competitive_Coding-3

problem1.cpp

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+/* Pascal's Triangle */
+
+// Time Complexity : O(n^2), where n = numRows
+// Space Complexity : O(1) auxiliary space, O(n^2) is the size of the output
+// 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
+// The first and second rows will always be [1] and [1,1] respectively
+// For each row, we need to calculate the elements between index 0 and index i-1, which is given by the formula: row[j] = triangle[i-1][j-1] + triangle[i-1][j]
+// After we fill a row, push it into the triangle matrix at the appropriate index
+
+class Solution {
+public:
+    vector<vector<int>> generate(int numRows) {
+        vector<vector<int>> triangle(numRows);
+        for(int i=0; i<numRows; i++) {
+            vector<int> row(i+1, 1);
+            for(int j=1; j<i; j++) {
+                row[j] = triangle[i-1][j-1] + triangle[i-1][j];
+            }
+            triangle[i] = row;
+        }
+        return triangle;
+    }
+};

problem2.cpp

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+/* K-Diff Pairs */
+
+// Time Complexity : O(nlogn)
+// 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
+// Sort the array and then start pointer i at index 0 and pointer j at index 1
+// Check the difference nums[j] - nums[i]: if the diff is too big then move i forward, else move j forward
+// When we find the pair, increment ans and then increment i & j to the next unique element
+// When we increment i & j to the next unique element, we might end up with i & j at the same index, so put an if condition at the start of the loop to handle this
+
+class Solution {
+public:
+    int findPairs(vector<int>& nums, int k) {
+        int n = nums.size();
+        if (n == 1) return 0;
+
+        sort(nums.begin(), nums.end());
+        int ans = 0;
+        int i = 0; int j = 1;
+
+        while(i < n && j < n) {
+            if (i == j) { // to ensure distinct indices
+                j++;
+                continue;
+            }
+            int a = nums[i];
+            int b = nums[j];
+            if(b - a == k) { // we found a pair
+                ans++;
+                while(i < n && nums[i] == a) i++; // to ensure no duplicate pairs
+                while(j < n && nums[j] == b) j++;
+            }
+            else if(b - a < k) { // diff is too small, increase j
+                j++;
+            } else { // diff is too big, increase i
+                i++;
+            }
+        }
+
+        return ans;
+    }
+};