Easy
Count Negative Numbers in a Sorted Matrix — C++
Full explanation · Time O(m + n) · Space O(1)
// Time: O(m + n)
// Space: O(1)
class Solution {
public:
int countNegatives(vector<vector<int>>& grid) {
int result = 0, c = grid[0].size() - 1;
for (const auto& row : grid) {
while (c >= 0 && row[c] < 0) {
--c;
}
result += grid[0].size() - 1 - c;
}
return result;
}
};