LeetCode in Kotlin

1351. Count Negative Numbers in a Sorted Matrix

Easy

Given a m x n matrix grid which is sorted in non-increasing order both row-wise and column-wise, return the number of negative numbers in grid.

Example 1:

Input: grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]

Output: 8

Explanation: There are 8 negatives number in the matrix.

Example 2:

Input: grid = [[3,2],[1,0]]

Output: 0

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 100
• -100 <= grid[i][j] <= 100

Follow up: Could you find an O(n + m) solution?

Solution

class Solution {
    fun countNegatives(grid: Array<IntArray>): Int {
        var count = 0
        for (row in grid) {
            for (v in row) {
                if (v < 0) {
                    count++
                }
            }
        }
        return count
    }
}

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