LeetCode – Set Matrix Zeroes (Java)

 

Given a m * n matrix, if an element is 0, set its entire row and column to 0.
Do it in place.

Analysis

This problem should be solved in place, i.e., no other array should be used. We can use the first column and the first row to track if a row/column should be set to 0.

Since we used the first row and first column to mark the zero row/column, the original values are changed.

Specifically, given, the following matrix
set-matrix-zero-1
this problem can be solved by following 4 steps:

Step 1:
First row contains zero = true;
First column contains zero = false;

Step 2: use first row and column to mark zero row and column.
set-matrix-zero-2

Step 3: set each elements by using marks in first row and column.
set-matrix-zero-3

Step 4: Set first column and row by using marks in step 1.
set-matrix-zero-4

Java Solution

public class Solution {
    public void setZeroes(int[][] matrix) {
        boolean firstRowZero = false;
        boolean firstColumnZero = false;
 
        //set first row and column zero or not
        for(int i=0; i<matrix.length; i++){
            if(matrix[i][0] == 0){
                firstColumnZero = true;
                break;
            }
        }
 
        for(int i=0; i<matrix[0].length; i++){
            if(matrix[0][i] == 0){
                firstRowZero = true;
                break;
            }
        }
 
        //mark zeros on first row and column
        for(int i=1; i<matrix.length; i++){
            for(int j=1; j<matrix[0].length; j++){
                if(matrix[i][j] == 0){
                   matrix[i][0] = 0;
                   matrix[0][j] = 0;
                }
            }
        }
 
        //use mark to set elements
        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;
                }
            }
        }
 
        //set first column and row
        if(firstColumnZero){
            for(int i=0; i<matrix.length; i++)
                matrix[i][0] = 0;
        }
 
        if(firstRowZero){
            for(int i=0; i<matrix[0].length; i++)
                matrix[0][i] = 0;
        }
 
    }
}

Related posts:

  1. LeetCode – Search a 2D Matrix II (Java)
  2. LeetCode – Spiral Matrix (Java)
  3. LeetCode – Longest Increasing Path in a Matrix (Java)
  4. LeetCode – Move Zeroes (Java)
Category >> Algorithms  
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  • Zaid Haq

    When does it get overwritten?

  • shekharsaxena

    import java.util.LinkedList;
    import java.util.List;

    public class MakeZerosInMatrix {

    public static void main(String[] args) {
    int[][] array = {{0, 2, 3}, {4, 5, 6}, {2, 5, 0}};

    for (int i = 0; i < array.length; i++) {
    for (int j = 0; j < array.length; j++) {
    System.out.print(array[i][j] + " ");
    }
    System.out.println(" ");
    }
    List visitedRowIndex = new LinkedList();
    List visitedColumnIndex = new LinkedList();
    for (int i = 0; i < array.length; i++) {
    for (int j = 0; j < array.length; j++) {
    if (!visitedRowIndex.contains(i) && !visitedColumnIndex.contains(j) && array[i][j] == 0) {
    visitedColumnIndex.add(j);
    visitedRowIndex.add(i);
    for (int k = 0; k < array.length; k++) {
    array[k][j] = 0;
    }
    for (int k = 0; k < array.length; k++) {
    array[i][k] = 0;
    }
    }
    }
    System.out.println(" ");
    }

    for (int i = 0; i < array.length; i++) {
    for (int j = 0; j < array.length; j++) {
    System.out.print(array[i][j] + " ");
    }
    System.out.println(" ");
    }
    }
    }

  • Nikhil !

    the last lines are getting messed up in code-editor Here it is:


    for(int n : coord)
    {
    int j=n%10;
    int i=n/10; int k=0;
    int l=0;
    while(k<row)

    {

    arr1[k][j]=0;

    k++;

    }

    while(l<column)

    {
    arr1[i][l]=0;
    l++;
    }
    }

  • Sha R Ath

    public class SetZeroMatrix {

    static Scanner in = new Scanner(System.in);

    public static void main(String[] args) {

    System.out.println(“Enter size of matrix”);

    int s = in.nextInt();

    int[][] a = new int[s][s];

    int q=0;

    System.out.println(“Enter the matrix”);

    for(int i=0;i<s;i++){

    for(int j=0;j<s;j++){

    a[i][j] = in.nextInt();

    }

    }

    //Algorithm marking zero positions

    for(int i=0;i<s;i++){

    for(int j=0;j<s;j++){

    if(a[i][j]==0){

    a[i][j]=9999;

    }

    }

    }

    //Algorithm setting rows and columns as zeros

    for(int i=0;i<s;i++){

    for(int j=0;j<s;j++){

    if(a[i][j]==9999){

    for(q=0;q<s;q++){

    a[i][q]=0;

    a[q][j]=0;

    }

    }

    }

    }

    //printing

    for(int i=0;i<s;i++){

    for(int j=0;j<s;j++){

    System.out.print(a[i][j]);

    }

    System.out.println("");

    }

    }

    }

  • Thomas Wolf

    I think you are not considering the case, that the first row or first column is not 0 and therefore should contain the values at the end that it contained at the beginning.
    In your solution the current values of row and column 0 would be overwritten.

  • maniac87

    public class SetMatrixZeros {

    public static int[][] setMatrixZeros(int[][] array) {

    if(array.length == 0) {
    return array;
    }

    int numColumns = array[0].length;
    int numrows = array.length;

    //we just need to keep track of what row and column sho //uld be 0
    Set rows = new HashSet();
    Set columns = new HashSet();

    for(int i = 0; i < numrows ; i++) {
    for(int j = 0 ; j < numColumns ; j++) {
    if(array[i][j] == 0) {
    rows.add(i);
    columns.add(j);
    }
    }
    }

    //now set rows and columns to 0s
    for(int row : rows) {
    for(int i = 0 ; i < numColumns ; i++) {
    array[row][i] = 0;
    }
    }

    for(int column : columns) {
    for(int i = 0 ; i < numrows ; i++) {
    array[i][column] = 0;
    }
    }
    return array;
    }
    }

  • Will

    Nice solution but in this case we need to “..Do it in place.”

  • Here is my intuitive solution accepted by LeetCode judge :

    public class Solution {
    public void setZeroes(int[][] matrix) {
    int n = matrix.length;
    int m = matrix[0].length;
    boolean[] historyRow = new boolean[n];
    boolean[] historyCol = new boolean[m];
    for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
    if (matrix[i][j] == 0) {
    if (!historyRow[i])
    historyRow[i] = true;
    if (!historyCol[j])
    historyCol[j] = true;
    }
    }
    }
    for (int i = 0; i < historyRow.length; i++) {
    if (historyRow[i])
    for (int k = 0; k < m; k++)
    matrix[i][k] = 0;
    }
    for (int j = 0; j < historyCol.length; j++) {
    if (historyCol[j])
    for (int k = 0; k < n; k++)
    matrix[k][j] = 0;
    }
    }
    }

  • Kaushik Ghosh

    readable, clean, efficient code. O(1) in space which is cool.

  • yigal

    Your solution takes O(n*m(n+m)) if the matrix is full of zeros, which is O(n^3) if n==m. Wang’s solution is wiser and takes O(n*m).

  • e

    fr
    r
    r
    r

  • David Prun

    works .thanks

  • hector

    O(n*m)

  • hector

    I think this is a more elegant solution and O(n2)

    public void clearColumnRows(int [][] m, int i, int j)
    {
    if(j >= m[0].length)
    {
    j = 0;
    i += 1;
    }
    if(i >= m.length)
    {
    return;
    }
    if(m[i][j] != 0)
    {
    clearColumnRows(m, i, j + 1);
    }
    else {
    clearColumnRows(m, i, j + 1);
    for(int x = 0; x < m.length; x++)
    {
    m[x][j] = 0;
    }
    for(int x = 0; x < m[0].length; x++)
    {
    m[i][x] = 0;
    }
    }
    }