# Search a 2D Matrix II LeetCode Solution

Here, We see Search a 2D Matrix II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:

• Integers in each row are sorted in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

Example 1:

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

## Search a 2D Matrix II LeetCode SolutionC++

``````class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int m = matrix.size(), n = m ? matrix[0].size() : 0, r = 0, c = n - 1;
while (r < m && c >= 0) {
if (matrix[r][c] == target) {
return true;
}
matrix[r][c] > target ? c-- : r++;
}
return false;
}
};```Code language: PHP (php)```

## Search a 2D Matrix II LeetCode SolutionJava

``````class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix == null || matrix.length < 1 || matrix[0].length <1) {
return false;
}
int col = matrix[0].length-1;
int row = 0;
while(col >= 0 && row <= matrix.length-1) {
if(target == matrix[row][col]) {
return true;
} else if(target < matrix[row][col]) {
col--;
} else if(target > matrix[row][col]) {
row++;
}
}
return false;
}
}```Code language: PHP (php)```

## Search a 2D Matrix II SolutionJavaScript

``````var searchMatrix = function(matrix, target) {
if(!matrix || !matrix.length) return false;
const rows = matrix.length;
const cols = matrix[0].length;
function hasTarget(startRow, endRow, startCol, endCol) {
if(startRow > endRow || startCol > endCol) return false;
const middleRow = Math.floor((endRow - startRow) / 2) + startRow;
const middleCol = Math.floor((endCol - startCol) / 2) + startCol;
if(matrix[middleRow][middleCol] === target) return true;
if (matrix[middleRow][middleCol] < target) {
return hasTarget(middleRow + 1, endRow, startCol, endCol) ||
hasTarget(startRow, middleRow, middleCol + 1, endCol);
} else {
return hasTarget(startRow, endRow, startCol, middleCol - 1) ||
hasTarget(startRow, middleRow - 1, middleCol, endCol);
}
}
return hasTarget(0, rows - 1, 0, cols - 1);
}```Code language: JavaScript (javascript)```

## Search a 2D Matrix II SolutionPython

``````class Solution(object):
def searchMatrix(self, matrix, target):
y, i, j = len(matrix), 0, len(matrix[0]) - 1
while i < y and ~j:
cell = matrix[i][j]
if cell == target: return True
elif cell > target: j -= 1
else: i += 1
return False```Code language: HTML, XML (xml)```
Scroll to Top