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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;
    }
};

Solution 2. Binary Search

// 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
• C++
• Python
• Go
• TypeScript
• Javascript
• RenderScript

• 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;
      }
  }
  
  ############
  
  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;
      }
  }
  

• // 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;
      }
  };
  

• 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
  
  
  

• 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
  }
  

• 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;
  }
  
  

• /**
   * @param {number[][]} grid
   * @return {number}
   */
  var countNegatives = function (grid) {
      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;
  };
  
  

• impl Solution {
      pub fn count_negatives(grid: Vec<Vec<i32>>) -> i32 {
          let m = grid.len();
          let n = grid[0].len();
          let mut i = m;
          let mut j = 0;
          let mut res = 0;
          while i > 0 && j < n {
              if grid[i - 1][j] >= 0 {
                  j += 1;
              } else {
                  res += n - j;
                  i -= 1;
              }
          }
          res as i32
      }
  }
  
  

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