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SetMatrixZeros.java
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SetMatrixZeros.java
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package Leetcode;
import java.util.HashSet;
import java.util.Set;
/**
* @author kalpak
*
* Given an m x n matrix. If an element is 0, set its entire row and column to 0. Do it in-place.
*
* 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?
*
*
* Example 1:
*
*
* Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]
* Output: [[1,0,1],[0,0,0],[1,0,1]]
* Example 2:
*
*
* Input: matrix = [[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]]
*
*
* Constraints:
*
* m == matrix.length
* n == matrix[0].length
* 1 <= m, n <= 200
* -2^31 <= matrix[i][j] <= 2^31 - 1
*
*/
public class SetMatrixZeros {
// Time Complexity: O(MxN)
// Space Complexity: O(M+N)
public static void setZeroes(int[][] matrix) {
Set<Integer> rows = new HashSet<>();
Set<Integer> cols = new HashSet<>();
for(int i = 0; i < matrix.length; i++) {
for(int j = 0; j < matrix[0].length; j++) {
if(matrix[i][j] == 0) {
rows.add(i);
cols.add(j);
}
}
}
for(int i = 0; i < matrix.length; i++) {
for(int j = 0; j < matrix[0].length; j++) {
if(rows.contains(i) || cols.contains(j)) {
matrix[i][j] = 0;
}
}
}
}
// Time Complexity: O(MxN)
// Space Complexity: O(1)
public static void setZeroesSpaceOptimized(int[][] matrix) {
boolean flag = false;
for(int i = 0; i < matrix.length; i++) {
if(matrix[i][0] == 0)
flag = true; // 1st column needs to be set to zero
for(int j = 1; j < matrix[0].length; j++) {
if(matrix[i][j] == 0) {
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
for(int i = 1; i < matrix.length; i++) {
for(int j = 1; j < matrix[0].length; j++) {
if(matrix[i][0] == 0 || matrix[0][j] == 0)
matrix[i][j] = 0;
}
}
// See if 1st row needs to be zeroed
if(matrix[0][0] == 0)
for(int j = 0; j < matrix[0].length; j++)
matrix[0][j] = 0;
// check if the 1st column needs to be zeroed
if(flag)
for(int i = 0; i < matrix.length; i++)
matrix[i][0] = 0;
}
}