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In this post, you will find the solution for the **Search a 2D Matrix** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Search a 2D Matrix– LeetCode Problem

Search a 2D Matrix– LeetCode Problem

**Problem:**

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 = 3Output:true

**Example 2:**

Input:matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13Output:false

**Constraints:**

`m == matrix.length`

`n == matrix[i].length`

`1 <= m, n <= 100`

`-10`

^{4}<= matrix[i][j], target <= 10^{4}

**Search a 2D Matrix– LeetCode Solutions**

Search a 2D Matrix in C++:

class Solution { public: bool searchMatrix(vector>& matrix, int target) { if (matrix.empty()) return false; const int m = matrix.size(); const int n = matrix[0].size(); int l = 0; int r = m * n; while (l < r) { const int mid = l + (r - l) / 2; const int i = mid / n; const int j = mid % n; if (matrix[i][j] == target) return true; if (matrix[i][j] < target) l = mid + 1; else r = mid; } return false; } };

Search a 2D Matrix in Java:

class Solution { public boolean searchMatrix(int[][] matrix, int target) { if (matrix.length == 0) return false; final int m = matrix.length; final int n = matrix[0].length; int l = 0; int r = m * n; while (l < r) { final int mid = l + (r - l) / 2; final int i = mid / n; final int j = mid % n; if (matrix[i][j] == target) return true; if (matrix[i][j] < target) l = mid + 1; else r = mid; } return false; } }

Search a 2D Matrix in Python:

class Solution: def searchMatrix(self, matrix: List[List[int]], target: int) -> bool: if not matrix: return False m = len(matrix) n = len(matrix[0]) l = 0 r = m * n while l < r: mid = l + (r - l) // 2 i = mid // n j = mid % n if matrix[i][j] == target: return True if matrix[i][j] < target: l = mid + 1 else: r = mid return False