LeetCode – Search a 2D Matrix (Java)

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has properties:

1) Integers in each row are sorted from left to right. 2) The first integer of each row is greater than the last integer of the previous row.

For example, consider the following matrix:

  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]

Given target = 3, return true.

Java Solution

This is a typical problem of binary search.

You may try to solve this problem by finding the row first and then the column. There is no need to do that. Because of the matrix's special features, the matrix can be considered as a sorted array. Your goal is to find one element in this sorted array by using binary search.

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix==null || matrix.length==0 || matrix[0].length==0) 
            return false;
        int m = matrix.length;
        int n = matrix[0].length;
        int start = 0;
        int end = m*n-1;
            int mid=(start+end)/2;
            int midX=mid/n;
            int midY=mid%n;
                return true;
        return false;
Category >> Algorithms  
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  • Melad Ezzat

    how could that be?

  • Maya

    Check first element and last element helps to eliminate unnecessary binary search

    if(target matrix[m-1][n-1]) return false;

  • RJ

    Can end also be m*n?

  • amsan7

    By iterating through all the rows our solution becomes O(m), while the provided solution is O(lg(mn))

  • Doug Pase

    A binary search across the entire array becomes more difficult if the matrix is sparse, such as a skyline matrix. In those cases it’s better to find the row (binary search) first, then the column (also binary search). The time complexity is the same when the matrix is square and dense, but it also works when the matrix has different length rows.

  • Vikrant Mishra

    Why is there a need to even perform a binary search if we just want to return a boolean. Since each row is sorted we can just compare whether our element is greater than the first element and less than the last element of a particular row and return true. If not we iterate through all the rows and keep checking for the aforementioned condition.

  • Bach

    Just want to point out that the time complexity for both your approach and “finding the row first and then the column” approach should be log(mn).