Formatted question description: https://leetcode.ca/all/1351.html

1351. Count Negative Numbers in a Sorted Matrix (Easy)

Given a m * 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

Example 3:

Input: grid = [[1,-1],[-1,-1]]
Output: 3

Example 4:

Input: grid = [[-1]]
Output: 1

 

Constraints:

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

Related Topics:
Array, Binary Search

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/count-negative-numbers-in-a-sorted-matrix/

// Time: O(MN)
// Space: O(1)
class Solution {
public:
    int countNegatives(vector<vector<int>>& grid) {
        int ans = 0;
        for (auto &row : grid) {
            for (int c : row) {
                if (c < 0) ++ans;
            }
        }
        return ans;
    }
};

// OJ: https://leetcode.com/problems/count-negative-numbers-in-a-sorted-matrix/

// Time: O(MlogN)
// Space: O(1)
class Solution {
public:
    int countNegatives(vector<vector<int>>& grid) {
        int ans = 0;
        for (auto &row : grid) {
            ans += upper_bound(row.rbegin(), row.rend(), -1) - row.rbegin();
        }
        return ans;
    }
};

Solution 3. Search Break Points

Traverse from upper right to lower left.

// OJ: https://leetcode.com/problems/count-negative-numbers-in-a-sorted-matrix/

// Time: O(M+N)
// Space: O(1)
class Solution {
public:
    int countNegatives(vector<vector<int>>& grid) {
        int M = grid.size(), N = grid[0].size(), x = 0, y = N - 1, ans = 0;
        while (x < M) {
            while (y >= 0 && grid[x][y] < 0) --y;
            ans += N - 1 - y;
            ++x;
        }
        return ans;
    }
};

Java

class Solution {
    public int countNegatives(int[][] grid) {
        int count = 0;
        int rows = grid.length, columns = grid[0].length;
        for (int i = rows - 1; i >= 0; i--) {
            if (grid[i][columns - 1] >= 0)
                break;
            for (int j = columns - 1; j >= 0; j--) {
                if (grid[i][j] < 0)
                    count++;
                else
                    break;
            }
        }
        return count;
    }
}

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