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 convert the adjacent valid cells of all queue elements to 1. The number of levels of BFS traversal that has led to conversion of at least one 0 to 1 is the answer.
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}); } } } // Intialise 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 code int 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; } |
Python 3
# 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]) # Intialise 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 code if __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 |
3
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