Given a matrix of size N consisting of 0‘s and 1‘s, the task is to find the minimum time required to fill the entire matrix with 1‘s. Every 1 at an instant in the matrix, can convert all 0‘s to 1 in its eight adjacent cells,i.e. a 1 present at (i,j) can convert all 0‘s to 1 at positions (i, j-1), (i, j+1), (i-1, j), (i+1, j), (i-1, j-1), (i-1, j+1), (i+1, j-1), (i+1, j+1).
Examples:
Input: N = 3, mtrx[][] = {{1,0,0},{0,0,1},{0,0,0}}
Output: 2
Explanation:
Initially the matrix appears to be
1, 0, 0
0, 0, 1
0, 0, 0
After the first instant of time, the new matrix is
1, 1, 1
1, 1, 1
0, 1, 1
After the 2nd instant the remaining 0 is converted to 1.Input: N = 5,
mtrx[][] = {{0,0,0,0,0},
{1,0,0,0,0},
{0,0,0,0,0},
{0,0,1,0,0},
{0,0,0,1,0}}
Output: 3
Approach:
In order to solve this problem, we are using the BFS approach. Traverse through the matrix and store the indices of the matrix with 1 initially in a queue. Loop until the queue gets empty and converts the adjacent valid cells of all queue elements to 1. The number of levels of BFS traversal that has led to the conversion of at least one 0 to 1 is the answer.
The below code is the implementation of the above approach:
C++
// C++ program to calculate the number of steps// in fill all entire matrix with '1'#include<bits/stdc++.h>using namespace std;// Function to return total number // of steps required to fill// entire matrix with '1'int numberOfSteps(int n,vector<vector<int>> mtrx){ // Queue to store indices with 1 queue<pair<int,int>> q; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (mtrx[i][j] == 1) { q.push({i,j}); } } } // Initialise step variable as zero int step = 0 ; // BFS approach while (!q.empty()) { int qsize = q.size(); // Visit all indices with 1 and // fill its neighbouring cells by 1 while (qsize--) { pair<int,int> p = q.front(); q.pop(); int i = p.first; int j = p.second; // Convert the neighbour from '0' to '1' if((j > 0) && mtrx[i][j-1] == 0) { mtrx[i][j-1] = 1; q.push({i,j-1}); } // Convert the neighbour from '0' to '1' if((i < n-1) && mtrx[i+1][j] == 0) { mtrx[i+1][j] = 1; q.push({i+1,j}); } // Convert the neighbour from '0' to '1' if((j < n-1) && mtrx[i][j+1] == 0) { mtrx[i][j+1] = 1; q.push({i,j + 1}); } // Convert the neighbour from '0' to '1' if((i > 0) && mtrx[i-1][j] == 0) { mtrx[i-1][j] = 1; q.push({i-1,j}); } // Convert the neighbour from '0' to '1' if((i > 0) && (j > 0) && mtrx[i-1][j-1] == 0) { mtrx[i-1][j-1] = 1; q.push({i-1,j-1}); } // Convert the neighbour from '0' to '1' if((i > 0) && (j < (n-1)) && mtrx[i-1][j+1] == 0) { mtrx[i-1][j+1] = 1; q.push({i-1,j+1}); } // Convert the neighbour from '0' to '1' if((i < (n-1)) && (j < (n-1)) && mtrx[i+1][j+1] == 0) { mtrx[i+1][j+1] = 1; q.push({i+1,j + 1}); } // Convert the neighbour from '0' to '1' if((i < (n-1)) && (j > 0) && mtrx[i+1][j-1] == 0) { mtrx[i+1][j-1] = 1; q.push({i+1,j-1}); } } //count steps step++; } return step-1;}// Driver codeint main(){ int n = 5 ; //dimension of matrix NXN vector<vector<int>> mtrx = {{0,0,0,0,0}, {1,0,0,0,0}, {0,0,0,0,0}, {0,0,1,0,0}, {0,0,0,1,0}}; // Print number of steps cout << numberOfSteps(n, mtrx); return 0;} |
Java
// Java program to calculate the// number of steps in fill all // entire matrix with '1'import java.util.*;class GFG{ static class pair{ int first, second; public pair(int first, int second) { this.first = first; this.second = second; } }// Function to return total number // of steps required to fill// entire matrix with '1'static int numberOfSteps(int n, int[][] mtrx){ // Queue to store indices with 1 Queue<pair> q = new LinkedList<>(); for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { if (mtrx[i][j] == 1) { q.add(new pair(i, j)); } } } // Initialise step variable as zero int step = 0; // BFS approach while (!q.isEmpty()) { int qsize = q.size(); // Visit all indices with 1 and // fill its neighbouring cells by 1 while (qsize-- > 0) { pair p = q.peek(); q.remove(); int i = p.first; int j = p.second; // Convert the neighbour from '0' to '1' if ((j > 0) && mtrx[i][j - 1] == 0) { mtrx[i][j - 1] = 1; q.add(new pair(i, j - 1)); } // Convert the neighbour from '0' to '1' if ((i < n - 1) && mtrx[i + 1][j] == 0) { mtrx[i + 1][j] = 1; q.add(new pair(i + 1, j)); } // Convert the neighbour from '0' to '1' if ((j < n - 1) && mtrx[i][j + 1] == 0) { mtrx[i][j + 1] = 1; q.add(new pair(i, j + 1)); } // Convert the neighbour from '0' to '1' if ((i > 0) && mtrx[i - 1][j] == 0) { mtrx[i - 1][j] = 1; q.add(new pair(i - 1, j)); } // Convert the neighbour from '0' to '1' if ((i > 0) && (j > 0) && mtrx[i - 1][j - 1] == 0) { mtrx[i - 1][j - 1] = 1; q.add(new pair(i - 1, j - 1)); } // Convert the neighbour from '0' to '1' if ((i > 0) && (j < (n - 1)) && mtrx[i - 1][j + 1] == 0) { mtrx[i - 1][j + 1] = 1; q.add(new pair(i - 1, j + 1)); } // Convert the neighbour from '0' to '1' if ((i < (n - 1)) && (j < (n - 1)) && mtrx[i + 1][j + 1] == 0) { mtrx[i + 1][j + 1] = 1; q.add(new pair(i + 1, j + 1)); } // Convert the neighbour from '0' to '1' if ((i < (n - 1)) && (j > 0) && mtrx[i + 1][j - 1] == 0) { mtrx[i + 1][j - 1] = 1; q.add(new pair(i + 1, j - 1)); } } // Count steps step++; } return step - 1;}// Driver codepublic static void main(String[] args){ // Dimension of matrix NXN int n = 5; int [][]mtrx = { { 0, 0, 0, 0, 0 }, { 1, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0 }, { 0, 0, 1, 0, 0 }, { 0, 0, 0, 1, 0 } }; // Print number of steps System.out.print(numberOfSteps(n, mtrx));}}// This code is contributed by Amit Katiyar |
Python3
# Python 3 program to calculate the number of steps# in fill all entire matrix with '1'# Function to return total number # of steps required to fill# entire matrix with '1'def numberOfSteps(n,mtrx): # Queue to store indices with 1 q = [] for i in range(n): for j in range(n): if (mtrx[i][j] == 1): q.append([i,j]) # Initialise step variable as zero step = 0 # BFS approach while (len(q)): qsize = len(q) # Visit all indices with 1 and # fill its neighbouring cells by 1 while(qsize): p = q[0] q.remove(q[0]) i = p[0] j = p[1] # Convert the neighbour from '0' to '1' if((j > 0) and mtrx[i][j - 1] == 0): mtrx[i][j - 1] = 1 q.append([i, j - 1]) # Convert the neighbour from '0' to '1' if((i < n-1) and mtrx[i + 1][j] == 0): mtrx[i + 1][j] = 1 q.append([i + 1, j]) # Convert the neighbour from '0' to '1' if((j < n-1) and mtrx[i][j + 1] == 0): mtrx[i][j + 1] = 1 q.append([i, j + 1]) # Convert the neighbour from '0' to '1' if((i > 0) and mtrx[i - 1][j] == 0): mtrx[i - 1][j] = 1 q.append([i - 1, j]) # Convert the neighbour from '0' to '1' if((i > 0) and (j > 0) and mtrx[i - 1][j - 1] == 0): mtrx[i - 1][j - 1] = 1 q.append([i - 1, j - 1]) # Convert the neighbour from '0' to '1' if((i > 0) and (j < (n-1)) and mtrx[i - 1][j + 1] == 0): mtrx[i - 1][j + 1] = 1 q.append([i - 1, j + 1]) # Convert the neighbour from '0' to '1' if((i < (n - 1)) and (j < (n - 1)) and mtrx[i + 1][j + 1] == 0): mtrx[i + 1][j + 1] = 1 q.append([i + 1, j + 1]) # Convert the neighbour from '0' to '1' if((i < (n - 1)) and (j > 0) and mtrx[i + 1][j - 1] == 0): mtrx[i + 1][j - 1] = 1 q.append([i + 1,j - 1]) qsize -= 1 #count steps step += 1 return step-1# Driver codeif __name__ == '__main__': #dimension of matrix NXN n = 5 mtrx = [[ 0, 0, 0, 0, 0 ], [ 1, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0 ], [ 0, 0, 1, 0, 0 ], [ 0, 0, 0, 1, 0 ]] # Print number of steps print(numberOfSteps(n, mtrx)) # This code is contributed by Samarth |
C#
// C# program to calculate the// number of steps in fill all // entire matrix with '1'using System;using System.Collections.Generic;class GFG{ public class pair{ public int first, second; public pair(int first, int second) { this.first = first; this.second = second; } }// Function to return total number // of steps required to fill// entire matrix with '1'static int numberOfSteps(int n, int[,] mtrx){ // Queue to store indices with 1 Queue<pair> q = new Queue<pair>(); for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { if (mtrx[i, j] == 1) { q.Enqueue(new pair(i, j)); } } } // Initialise step variable as zero int step = 0; // BFS approach while (q.Count != 0) { int qsize = q.Count; // Visit all indices // with 1 and fill // its neighbouring // cells by 1 while (qsize-- > 0) { pair p = q.Peek(); q.Dequeue(); int i = p.first; int j = p.second; // Convert the neighbour // from '0' to '1' if ((j > 0) && mtrx[i, j - 1] == 0) { mtrx[i, j - 1] = 1; q.Enqueue(new pair(i, j - 1)); } // Convert the neighbour // from '0' to '1' if ((i < n - 1) && mtrx[i + 1, j] == 0) { mtrx[i + 1, j] = 1; q.Enqueue(new pair(i + 1, j)); } // Convert the neighbour // from '0' to '1' if ((j < n - 1) && mtrx[i, j + 1] == 0) { mtrx[i, j + 1] = 1; q.Enqueue(new pair(i, j + 1)); } // Convert the neighbour // from '0' to '1' if ((i > 0) && mtrx[i - 1, j] == 0) { mtrx[i - 1, j] = 1; q.Enqueue(new pair(i - 1, j)); } // Convert the neighbour // from '0' to '1' if ((i > 0) && (j > 0) && mtrx[i - 1, j - 1] == 0) { mtrx[i - 1, j - 1] = 1; q.Enqueue(new pair(i - 1, j - 1)); } // Convert the neighbour // from '0' to '1' if ((i > 0) && (j < (n - 1)) && mtrx[i - 1, j + 1] == 0) { mtrx[i - 1, j + 1] = 1; q.Enqueue(new pair(i - 1, j + 1)); } // Convert the neighbour // from '0' to '1' if ((i < (n - 1)) && (j < (n - 1)) && mtrx[i + 1, j + 1] == 0) { mtrx[i + 1, j + 1] = 1; q.Enqueue(new pair(i + 1, j + 1)); } // Convert the neighbour // from '0' to '1' if ((i < (n - 1)) && (j > 0) && mtrx[i + 1, j - 1] == 0) { mtrx[i + 1, j - 1] = 1; q.Enqueue(new pair(i + 1, j - 1)); } } // Count steps step++; } return step - 1;}// Driver codepublic static void Main(String[] args){ // Dimension of matrix NXN int n = 5; int [,]mtrx = {{0, 0, 0, 0, 0}, {1, 0, 0, 0, 0}, {0, 0, 0, 0, 0}, {0, 0, 1, 0, 0}, {0, 0, 0, 1, 0}}; // Print number of steps Console.Write(numberOfSteps(n, mtrx));}}// This code is contributed by Rajput-Ji |
Javascript
<script>// JavaScript program to calculate the number of steps// in fill all entire matrix with '1'// Function to return total number // of steps required to fill// entire matrix with '1'function numberOfSteps(n,mtrx){ // Queue to store indices with 1 let q = [] for(let i = 0; i < n; i++) { for(let j = 0; j < n; j++) { if (mtrx[i][j] == 1) q.push([i, j]) } } // Initialise step variable as zero let step = 0 // BFS approach while (q.length){ let qsize = q.length // Visit all indices with 1 and // fill its neighbouring cells by 1 while(qsize){ let p = q.shift() let i = p[0] let j = p[1] // Convert the neighbour from '0' to '1' if((j > 0) && mtrx[i][j - 1] == 0){ mtrx[i][j - 1] = 1 q.push([i, j - 1]) } // Convert the neighbour from '0' to '1' if((i < n-1) && mtrx[i + 1][j] == 0){ mtrx[i + 1][j] = 1 q.push([i + 1, j]) } // Convert the neighbour from '0' to '1' if((j < n-1) && mtrx[i][j + 1] == 0){ mtrx[i][j + 1] = 1 q.push([i, j + 1]) } // Convert the neighbour from '0' to '1' if((i > 0) && mtrx[i - 1][j] == 0){ mtrx[i - 1][j] = 1 q.push([i - 1, j]) } // Convert the neighbour from '0' to '1' if((i > 0) && (j > 0) && mtrx[i - 1][j - 1] == 0){ mtrx[i - 1][j - 1] = 1 q.push([i - 1, j - 1]) } // Convert the neighbour from '0' to '1' if((i > 0) && (j < (n-1)) && mtrx[i - 1][j + 1] == 0){ mtrx[i - 1][j + 1] = 1 q.push([i - 1, j + 1]) } // Convert the neighbour from '0' to '1' if((i < (n - 1)) && (j < (n - 1)) && mtrx[i + 1][j + 1] == 0){ mtrx[i + 1][j + 1] = 1 q.push([i + 1, j + 1]) } // Convert the neighbour from '0' to '1' if((i < (n - 1)) && (j > 0) && mtrx[i + 1][j - 1] == 0){ mtrx[i + 1][j - 1] = 1 q.push([i + 1,j - 1]) } qsize -= 1 } // count steps step += 1 } return step-1}// Driver code // dimension of matrix NXNlet n = 5let mtrx = [[ 0, 0, 0, 0, 0 ], [ 1, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0 ], [ 0, 0, 1, 0, 0 ], [ 0, 0, 0, 1, 0 ]] // Print number of stepsdocument.write(numberOfSteps(n, mtrx),"</br>") // This code is contributed by shinjanpatra</script> |
Output:
3
Time Complexity: O(n*n)
Auxiliary Space: O(n*n)
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