Given a rows x cols
matrix mat
, where mat[i][j]
is either 0
or 1
, return the number of special positions in mat
.
A position (i,j)
is called special if mat[i][j] == 1
and all other elements in row i
and column j
are 0
(rows and columns are 0-indexed).
Example 1:
Input: mat = [[1,0,0], [0,0,1], [1,0,0]] Output: 1 Explanation: (1,2) is a special position because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0.
Example 2:
Input: mat = [[1,0,0], [0,1,0], [0,0,1]] Output: 3 Explanation: (0,0), (1,1) and (2,2) are special positions.
Example 3:
Input: mat = [[0,0,0,1], [1,0,0,0], [0,1,1,0], [0,0,0,0]] Output: 2
Example 4:
Input: mat = [[0,0,0,0,0], [1,0,0,0,0], [0,1,0,0,0], [0,0,1,0,0], [0,0,0,1,1]] Output: 3
Constraints:
rows == mat.length
cols == mat[i].length
1 <= rows, cols <= 100
mat[i][j]
is0
or1
.
Solution: Sum for each row and column
Brute force:
Time complexity: O(R*C*(R+C))
Space complexity: O(1)
We can pre-compute the sums for each row and each column, ans = sum(mat[r][c] == 1 and rsum[r] == 1 and csum[c] == 1)
Time complexity: O(R*C)
Space complexity: O(R+C)
C++
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class Solution { public: int numSpecial(vector<vector<int>>& mat) { const int rows = mat.size(); const int cols = mat[0].size(); vector<int> rs(rows); vector<int> cs(cols); for (int r = 0; r < rows; ++r) for (int c = 0; c < cols; ++c) { rs[r] += mat[r][c]; cs[c] += mat[r][c]; } int ans = 0; for (int r = 0; r < rows; ++r) for (int c = 0; c < cols; ++c) ans += mat[r][c] && rs[r] == 1 && cs[c] == 1; return ans; } }; |
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