In the Leetcode Set Matrix Zeroes problem solution 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]]
Constraints:
- m == matrix.length
- n == matrix[0].length
- 1 <= m, n <= 200
- -231 <= matrix[i][j] <= 231 – 1
Solution in C Programming
void setZeroes(int** matrix, int matrixSize, int* matrixColSize){
bool row=false;
bool column=false;
for(int i=0;i<matrixSize;i++){
if(matrix[i][0]==0){
column=true;
break;
}
}
for(int j=0;j<*matrixColSize;j++){
if(matrix[0][j]==0){
row=true;
break;
}
}
for(int i=1;i<matrixSize;i++){
for(int j=1;j<*matrixColSize;j++){
if(matrix[i][j]==0){
matrix[0][j]=0;
matrix[i][0]=0;
}
}
}
for(int i=1; i<matrixSize; i++)
for(int j=1;j<*matrixColSize; j++)
{
if(matrix[0][j] == 0 || matrix[i][0] == 0)
{
matrix[i][j] = 0;
}
}
if(row){
for(int i=0;i<*matrixColSize;i++)
{
matrix[0][i]=0;
}
}
if(column){
for(int i=0;i<matrixSize;i++)
{
matrix[i][0]=0;
}
}
}Solution in C++ Programming
class Solution {
public:
void setZeroes(vector<vector<int>>& matrix) {
set<pair<int, int>> zeroCoords;
for(int i = 0; i < matrix.size(); i++){
for(int j = 0; j < matrix.at(i).size(); j++){
if(matrix.at(i).at(j) == 0) zeroCoords.insert({i, j});
}
}
for(int i = 0; i < matrix.size(); i++){
for(int j = 0; j < matrix.at(i).size(); j++){
if(matrix.at(i).at(j) == 0 && zeroCoords.find({i, j}) != zeroCoords.end()){
fillRows(matrix, i);
fillColumns(matrix, j);
zeroCoords.erase({i, j});
}
}
}
}
void fillRows(vector<vector<int>> &matrix, int row){
for(int j = 0; j < matrix.at(0).size(); j++)
matrix.at(row).at(j) = 0;
}
void fillColumns(vector<vector<int>> &matrix, int col){
for(int i = 0; i < matrix.size(); i++)
matrix.at(i).at(col) = 0;
}
};Solution in Java Programming
class Solution {
public void setZeroes(int[][] matrix) {
int r = matrix.length;
int c = matrix[0].length;
int col0 = 1;
for (int i = 0; i < r; i++) {
if (matrix[i][0] == 0) col0 = 0;
for (int j = 1; j < c; j++)
if (matrix[i][j] == 0)
matrix[i][0] = matrix[0][j] = 0;
}
for (int i = r - 1; i >= 0; i--) {
for (int j = c - 1; j >= 1; j--)
if (matrix[i][0] == 0 || matrix[0][j] == 0)
matrix[i][j] = 0;
if (col0 == 0) matrix[i][0] = 0;
}
}
}Solution in Python Programming
class Solution:
def setZeroes(self, matrix):
"""
Do not return anything, modify matrix in-place instead.
"""
row = []
col = []
for r in range(0,len(matrix)):
for c in range(0,len(matrix[0])):
if matrix[r][c] == 0:
row.append(r)
col.append(c)
for a in row:
for b in range(0,len(matrix[0])):
matrix[a][b] = 0
for i in col:
for j in range(0,len(matrix)):
matrix[j][i] = 0Solution in C# Programming
public class Solution {
public void SetZeroes(int[][] matrix) {
var r = new Dictionary<int,int>();
var c = new Dictionary<int,int>();
for(int i=0;i<matrix.Length;i++){
for(int j=0;j<matrix[0].Length;j++){
if(matrix[i][j]==0){
if(!r.ContainsKey(i)){
r.Add(i,0);
}
if(!c.ContainsKey(j)){
c.Add(j,0);
}
}
}
}
foreach(var i in r.Keys){
RMakeZero(i,matrix);
}
foreach(var i in c.Keys){
CMakeZero(i,matrix);
}
}
public static void RMakeZero(int row,int[][] matrix){
for(int i=0;i<matrix[0].Length;i++){
matrix[row][i] =0;
}
}
public static void CMakeZero(int col,int[][] matrix){
for(int i=0;i<matrix.Length;i++){
matrix[i][col] =0;
}
}
}Also read,