Problem description:

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 in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.
Example:

Consider the following 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]
]
Given target = 5, return true.

Given target = 20, return false.

Solution:

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class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size();
        if (m == 0) return false;
        int n = matrix[0].size();

        //find in m*n array
        //pick a row then search each column position in it, then switch to other row
        int i = 0, j = n - 1; //notice that we start from last position in a row
        while (i < m && j >= 0) {
            if (matrix[i][j] < target)
                i++;
            else if (matrix[i][j] > target)
                j--;
            else
                return true;
        }
        return false;
    }
};

time complexity: $O(nlogn)$
space complexity: $O(1)$
reference: