Last updated on January 19th, 2025 at 05:49 pm

Here, we see a Set Matrix Zeroes LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.

List of all LeetCode Solution

Topics

Array

Companies

Amazon, Microsoft

Level of Question

Medium

Set Matrix Zeroes LeetCode Solution

1. Problem Statement

Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0‘s.

You must do it in place.

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]]

2. Coding Pattern Used in Solution

The coding pattern used in all the provided implementations is Matrix Manipulation. This pattern involves traversing a 2D matrix, identifying specific conditions, and modifying the matrix based on those conditions.

3. Code Implementation in Different Languages

3.1 Set Matrix Zeroes C++

class Solution {
public:
    void setZeroes(vector<vector<int>>& matrix) {
        int m=matrix.size();
        int n=matrix[0].size();
        vector<pair<int,int>> cor;
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(matrix[i][j]==0)
                {
                   cor.push_back({i,j});
                }
            }
        }
        for(int i=0;i<cor.size();i++)
        {
            int x=cor[i].first;
            int y=cor[i].second;
            int row=0;
            int col=0;
            while(row<m)
            {
                matrix[row][y]=0;
                row++;
            }
            while(col<n)
            {
                matrix[x][col]=0;
                col++;
            }
        }       
    }
};

3.2 Set Matrix Zeroes Java

class Solution {
    public void setZeroes(int[][] matrix) {
    int m = matrix.length, n = matrix[0].length, k = 0;
    while (k < n && matrix[0][k] != 0) ++k;
    for (int i = 1; i < m; ++i)
        for (int j = 0; 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 = n - 1; j >= 0; --j)
            if (matrix[0][j] == 0 || matrix[i][0] == 0)
                matrix[i][j] = 0;
    if (k < n) Arrays.fill(matrix[0], 0);        
    }
}

3.3 Set Matrix Zeroes JavaScript

var setZeroes = function(matrix) {
  let isCol = false, r = matrix.length, c = matrix[0].length;
  for (let i = 0; i < r; i++) {
    if (!matrix[i][0]) isCol = true;
    for (let j = 1; j < c; j++) {
      if (!matrix[i][j]) {
        matrix[i][0] = 0;
        matrix[0][j] = 0;
      };
    }
  }
  for (let i = 1; i < r; i++) {
    for (let j = 1; j < c; j++) {
      if (!matrix[i][0] || !matrix[0][j]) matrix[i][j] = 0;
    }
  }
  if (!matrix[0][0]) {
    for (let j = 0; j < c; j++) matrix[0][j] = 0;
  }
  if (isCol) {
    for (let i = 0; i < r; i++) {
      matrix[i][0] = 0;
    }
  }   
};

3.4 Set Matrix Zeroes Python

class Solution(object):
    def setZeroes(self, matrix):
        row = set()
        column = set()
        
        for i in range(len(matrix)):
            for j in range(len(matrix[0])):
                if matrix[i][j] == 0:
                    row.add(i)
                    column.add(j)          
        for i in row:
            for j in range(len(matrix[0])):
                matrix[i][j] = 0
        for i in column:
            for j in range(len(matrix)):
                matrix[j][i] = 0 

4. Time and Space Complexity

	Time Complexity	Space Complexity
C++	O(m * n)	O(k)
Java	O(m * n)	O(1)
JavaScript	O(m * n)	O(1)
Python	O(m * n)	O(m + n)

where k is the number of zeroes.

• The C++ implementation is less space-efficient due to the use of an auxiliary vector.
• The Java and JavaScript implementations are the most space-efficient, as they use the matrix itself for marking.
• The Python implementation is straightforward but uses additional space for sets.