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Set_Matrix_Zeroes.js
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Set_Matrix_Zeroes.js
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/*
Set Matrix Zeroes
https://leetcode.com/problems/set-matrix-zeroes/
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?
*/
/**
* @param {number[][]} matrix
* @return {void} Do not return anything, modify matrix in-place instead.
*/
var setZeroes = function (matrix) {
if (matrix.length === 0) return;
var pivotRow = -1;
var pivotCol = -1;
var iterRow = 0;
var iterCol = 0;
var found = false;
// Find a pivot
while (!found && iterRow < matrix.length) {
iterCol = 0;
while (!found && iterCol < matrix[0].length) {
if (matrix[iterRow][iterCol] === 0) {
found = true;
pivotRow = iterRow;
pivotCol = iterCol;
}
iterCol++;
}
iterRow++;
}
if (!found) return;
// Update the Column value
for (var i = 0; i < matrix.length; i++) {
if (i == pivotRow) continue;
for (var j = 0; j < matrix[0].length; j++) {
if (j == pivotCol) continue;
if (matrix[i][j] === 0) {
matrix[i][pivotCol] = 0;
matrix[pivotRow][j] = 0;
}
}
}
for (i = 0; i < matrix.length; i++)
if (matrix[i][pivotCol] === 0 && i !== pivotRow) fillRow(matrix, i);
for (i = 0; i < matrix[0].length; i++)
if (matrix[pivotRow][i] === 0 && i !== pivotCol) fillCol(matrix, i);
fillCol(matrix, pivotCol);
fillRow(matrix, pivotRow);
return matrix;
};
var fillRow = function (matrix, row) {
for (var i = 0; i < matrix[0].length; i++) matrix[row][i] = 0;
};
var fillCol = function (matrix, col) {
for (var i = 0; i < matrix.length; i++) matrix[i][col] = 0;
};
module.exports.setZeroes = setZeroes;