Search a 2D Matrix

Author: sachuverma

DSA Crack Sheet/README.md

@@ -44,6 +44,7 @@
 ### Matrix
 
 - [Spirally traversing a matrix](https://practice.geeksforgeeks.org/problems/spirally-traversing-a-matrix-1587115621/1# "view question") - [Cpp Solution](./solutions/Spirally%20traversing%20a%20matrix.cpp)  
+- [Search a 2D Matrix](https://leetcode.com/problems/search-a-2d-matrix/ "view question") - [Cpp Solution](./solutions/Search%20a%202D%20Matrix.cpp)  
 - []( "view question") - [Cpp Solution](./solutions/.cpp)  
 
 ### Strings

DSA Crack Sheet/solutions/Search a 2D Matrix.cpp

@@ -0,0 +1,44 @@
+/*
+Search a 2D Matrix
+==================
+
+Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
+
+Integers in each row are sorted from left to right.
+The first integer of each row is greater than the last integer of the previous row.
+
+Example 1:
+Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
+Output: true
+
+Example 2:
+Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
+Output: false
+
+Constraints:
+m == matrix.length
+n == matrix[i].length
+1 <= m, n <= 100
+-104 <= matrix[i][j], target <= 104
+*/
+
+class Solution
+{
+public:
+  bool searchMatrix(vector<vector<int>> &matrix, int target)
+  {
+    int n = matrix.size(), m = matrix[0].size();
+    int r = 0, c = m - 1;
+
+    while (r < n && c >= 0)
+    {
+      if (matrix[r][c] == target)
+        return true;
+      else if (matrix[r][c] < target)
+        r++;
+      else
+        c--;
+    }
+    return false;
+  }
+};