Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.
Example 1:
Input: [ [1,1,1], [1,0,1], [1,1,1] ] Output: [ [1,0,1], [0,0,0], [1,0,1] ]
Example 2:
Input: [ [0,1,2,0], [3,4,5,2], [1,3,1,5] ] Output: [ [0,0,0,0], [0,4,5,0], [0,3,1,0] ]
Follow up:
- A straight forward solution using O(mn) space is probably a bad idea.
- A simple improvement uses O(m + n) space, but still not the best solution.
- Could you devise a constant space solution?
Solution 1
Use two arrays to track whether the i-th row / j-th column need to be zeroed.
Time complexity: O(mn)
Space complexity: O(m+n)
C++
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class Solution { public: void setZeroes(vector<vector<int>>& matrix) { const int m = matrix.size(); const int n = matrix[0].size(); vector<int> rows(m); vector<int> cols(n); for (int i = 0; i < m; ++i) for (int j = 0; j < n; ++j) { rows[i] |= (matrix[i][j] == 0); cols[j] |= (matrix[i][j] == 0); } for (int i = 0; i < m; ++i) for (int j = 0; j < n; ++j) if (rows[i] || cols[j]) matrix[i][j] = 0; } }; |
Solution 2
Use the first row / first col to indicate whether the i-th row / j-th column need be zeroed.
C++
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class Solution { public: void setZeroes(vector<vector<int>>& matrix) { const int m = matrix.size(); const int n = matrix[0].size(); bool col0 = false; bool row0 = false; for (int i = 0; i < m; ++i) col0 |= (matrix[i][0] == 0); for (int j = 0; j < n; ++j) row0 |= (matrix[0][j] == 0); for (int i = 1; i < m; ++i) for (int j = 1; j < n; ++j) if (matrix[i][j] == 0) matrix[0][j] = matrix[i][0] = 0; for (int i = 1; i < m; ++i) for (int j = 1; j < n; ++j) if (matrix[i][0] == 0 || matrix[0][j] == 0) matrix[i][j] = 0; if (row0) for (int j = 0; j < n; ++j) matrix[0][j] = 0; if (col0) for (int i = 0; i < m; ++i) matrix[i][0] = 0; } }; |
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