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.lengthn == grid[i].length1 <= m, n <= 100-100 <= grid[i][j] <= 100
Related Topics:
Array, Binary Search
// OJ: https://leetcode.com/problems/count-negative-numbers-in-a-sorted-matrix/
// Author: github.com/lzl124631x
// 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/
// Author: github.com/lzl124631x
// 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;
}
};Traverse from upper right to lower left.
// OJ: https://leetcode.com/problems/count-negative-numbers-in-a-sorted-matrix/
// Author: github.com/lzl124631x
// 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;
}
};