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1351. Count Negative Numbers in a Sorted Matrix

Description

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?

Solutions

Solution 1: Start Traversing from the Bottom Left or Top Right

According to the characteristic that both rows and columns are arranged in non-increasing order, we can start traversing from the bottom left corner towards the top right direction.

When encountering a negative number, it indicates that all elements to the right of the current position in this row are negative. We add the number of remaining elements in this row to the answer, that is, $n - j$, and move up a row, that is, $i \leftarrow i - 1$. Otherwise, move to the right column, that is, $j \leftarrow j + 1$.

After the traversal is over, return the answer.

The time complexity is $O(m + n)$, where $m$ and $n$ are the number of rows and columns of the matrix, respectively. The space complexity is $O(1)$.

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class Solution:
    def countNegatives(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        i, j = m - 1, 0
        ans = 0
        while i >= 0 and j < n:
            if grid[i][j] < 0:
                ans += n - j
                i -= 1
            else:
                j += 1
        return ans

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class Solution {
    public int countNegatives(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int ans = 0;
        for (int i = m - 1, j = 0; i >= 0 && j < n;) {
            if (grid[i][j] < 0) {
                ans += n - j;
                --i;
            } else {
                ++j;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int countNegatives(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int ans = 0;
        for (int i = m - 1, j = 0; i >= 0 && j < n;) {
            if (grid[i][j] < 0) {
                ans += n - j;
                --i;
            } else
                ++j;
        }
        return ans;
    }
};

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func countNegatives(grid [][]int) int {
    m, n := len(grid), len(grid[0])
    ans := 0
    for i, j := m-1, 0; i >= 0 && j < n; {
        if grid[i][j] < 0 {
            ans += n - j
            i--
        } else {
            j++
        }
    }
    return ans
}

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function countNegatives(grid: number[][]): number {
    const m = grid.length,
        n = grid[0].length;
    let ans = 0;
    for (let i = m - 1, j = 0; i >= 0 && j < n; ) {
        if (grid[i][j] < 0) {
            ans += n - j;
            --i;
        } else {
            ++j;
        }
    }
    return ans;
}