The challenge

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],[23,30,34,50]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,50]], target = 13
Output: false

Example 3:

Input: matrix = [], target = 0
Output: false


  • m == matrix.length
  • n == matrix[i].length
  • 0 <= m, n <= 100
  • -10<sup>4</sup>&nbsp;<= matrix[i][j], target <= 10<sup>4</sup>

The solution in Java code

It’s fairly simple to resolve this challenge. All we need to do is perform an outer and then an inner loop through each of the array segments. If at any point we find our target value, we can return true. Otherwise at the very end, we simply return false.

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix.length==0) return false;
        for (int i=0; i<matrix.length; i++) {
            for (int j=0; j<matrix[i].length; j++) {
                if (matrix[i][j]==target) return true;
        return false;