Merge pull request codeharborhub#3649 from vedang1/main

Create 0240-search-a-2d-matrix-ii.md

Author: ajay-dhangar

dsa-solutions/lc-solutions/0200-0299/0240-search-a-2d-matrix-ii.md

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+---
+id: search-a-2d-matrix-ii
+title: Search A 2D Matrix ii
+sidebar_label: 0240-Search A 2D Matrix ii
+tags:
+    - DP
+    - Leet code
+description: "Solution to leetocde 240"
+---
+
+## Approach
+The approach involves starting from the top-right corner of the matrix and iteratively narrowing down the search based on comparisons with the target.
+
+1. Initialize `m` and `n` to the number of rows and columns in the matrix, respectively. Also, initialize `r` to 0 (the topmost row) and `c` to `n - 1` (the rightmost column).
+2. Enter a while loop that continues as long as the current position is within the matrix (`r < m` and `c >= 0`).
+3. Inside the loop, compare the element at the current position (`r, c`) with the target.
+     - If they are equal, it means the target is found, and the function returns `true`.
+     - If the element is greater than the target, update `c` to move to the left, narrowing down the search.
+     - If the element is less than the target, update `r` to move downwards, narrowing down the search.
+4. The loop continues until the search range is exhausted or the target is found.
+5. If the loop completes without finding the target, the function returns `false`.
+
+## Time Complexity
+The time complexity of this code is `O(m + n)`, where `m` is the number of rows and `n` is the number of columns in the matrix. In each iteration of the while loop, either `r` is incremented or `c` is decremented, leading to a total of at most `m + n` iterations.
+
+## Space Complexity
+The space complexity of this code is `O(1)` because it uses only a constant amount of extra space for variables (`m`, `n`, `r`, `c`). The space usage does not depend on the size of the input matrix.
+
+```cpp
+class Solution {
+public:
+        bool searchMatrix(vector<vector<int>>& matrix, int target) {
+
+                int m = matrix.size(), n = matrix[0].size(), r = 0, c = n - 1;
+
+                while (r < m && c >= 0){
+
+                        if (matrix[r][c] == target)
+                                return true;
+                        else if(matrix[r][c] > target)
+                                c--;
+                        else
+                                r++;
+                }
+
+                return false;
+        }
+}
+```