Given a Matrix of size N*N filled with 1‘s and 0‘s, the task is to find the maximum distance from a 0-cell to its nearest 1-cell. If the matrix is filled with only 0’s or only 1’s, return -1.
Note: Only horizontal and vertical movements are allowed in the matrix.
Examples:
Input:
mat[][] = {{0, 1, 0},
{0, 0, 1},
{0, 0, 0}}
Output: 3
Explanation:
Cell number (2, 0) is at the farthest
distance of 3 cells from both the
1-cells (0, 1) and (1, 2).
Input:
mat[][] = {{1, 0, 0},
{0, 0, 0},
{0, 0, 0}}
Output: 4
Explanation:
Cell number (2, 2) is at the farthest
distance of 4 cells from the only
1-cell (1, 1).
Approach 1: Naive Approach
For each 0-cell, compute its distance from every 1-cell and store the minimum. The maximum of all those minimal distances is the answer.
Below is the implementation of the above approach:
C++
// C++ Program to find the maximum // distance from a 0-cell to a 1-cell #include <bits/stdc++.h> using namespace std; int maxDistance(vector<vector<int> >& grid) { vector<pair<int, int> > one; int M = grid.size(); int N = grid[0].size(); int ans = -1; for (int i = 0; i < M; ++i) { for (int j = 0; j < N; ++j) { if (grid[i][j] == 1) one.emplace_back(i, j); } } // If the matrix consists of only 0's // or only 1's if (one.empty() || M * N == one.size()) return -1; for (int i = 0; i < M; ++i) { for (int j = 0; j < N; ++j) { if (grid[i][j] == 1) continue; // If it's a 0-cell int dist = INT_MAX; for (auto& p : one) { // calculate its distance // with every 1-cell int d = abs(p.first - i) + abs(p.second - j); // Compare and store the minimum dist = min(dist, d); if (dist <= ans) break; } // Compare and store the maximum ans = max(ans, dist); } } return ans; } // Driver code int main() { vector<vector<int> > arr = { { 0, 0, 1 }, { 0, 0, 0 }, { 0, 0, 0 } }; cout << maxDistance(arr) << endl; return 0; } |
Java
// Java Program to find the maximum // distance from a 0-cell to a 1-cell import java.util.*; class GFG{ static class pair { int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } static int maxDistance(int [][]grid) { Vector<pair> one = new Vector<pair>(); int M = grid.length; int N = grid[0].length; int ans = -1; for (int i = 0; i < M; ++i) { for (int j = 0; j < N; ++j) { if (grid[i][j] == 1) one.add(new pair(i, j)); } } // If the matrix consists of only 0's // or only 1's if (one.isEmpty() || M * N == one.size()) return -1; for (int i = 0; i < M; ++i) { for (int j = 0; j < N; ++j) { if (grid[i][j] == 1) continue; // If it's a 0-cell int dist = Integer.MAX_VALUE; for (pair p : one) { // calculate its distance // with every 1-cell int d = Math.abs(p.first - i) + Math.abs(p.second - j); // Compare and store the minimum dist = Math.min(dist, d); if (dist <= ans) break; } // Compare and store the maximum ans = Math.max(ans, dist); } } return ans; } // Driver code public static void main(String[] args) { int [][]arr = { { 0, 0, 1 }, { 0, 0, 0 }, { 0, 0, 0 } }; System.out.print(maxDistance(arr) +"\n"); } } // This code contributed by Princi Singh |
Python3
# Python3 program to find the maximum # distance from a 0-cell to a 1-cell def maxDistance(grid): one = [] M = len(grid) N = len(grid[0]) ans = -1 for i in range(M): for j in range(N): if (grid[i][j] == 1): one.append([i, j]) # If the matrix consists of only 0's # or only 1's if (one == [] or M * N == len(one)): return -1 for i in range(M): for j in range(N): if (grid[i][j] == 1): continue # If it's a 0-cell dist = float('inf') for p in one: # Calculate its distance # with every 1-cell d = abs(p[0] - i) + abs(p[1] - j) # Compare and store the minimum dist = min(dist, d) if (dist <= ans): break # Compare and store the maximum ans = max(ans, dist) return ans # Driver code arr = [ [ 0, 0, 1 ], [ 0, 0, 0 ], [ 0, 0, 0 ] ] print(maxDistance(arr)) # This code is contributed by rohitsingh07052 |
C#
// C# program to find the maximum // distance from a 0-cell to a 1-cell using System; using System.Collections.Generic; class GFG{ class pair { public int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } static int maxDistance(int [,]grid) { List<pair> one = new List<pair>(); int M = grid.GetLength(0); int N = grid.GetLength(1); int ans = -1; for(int i = 0; i < M; ++i) { for(int j = 0; j < N; ++j) { if (grid[i, j] == 1) one.Add(new pair(i, j)); } } // If the matrix consists of only 0's // or only 1's if (one.Count == 0 || M * N == one.Count) return -1; for(int i = 0; i < M; ++i) { for(int j = 0; j < N; ++j) { if (grid[i, j] == 1) continue; // If it's a 0-cell int dist = int.MaxValue; foreach (pair p in one) { // Calculate its distance // with every 1-cell int d = Math.Abs(p.first - i) + Math.Abs(p.second - j); // Compare and store the minimum dist = Math.Min(dist, d); if (dist <= ans) break; } // Compare and store the maximum ans = Math.Max(ans, dist); } } return ans; } // Driver code public static void Main(String[] args) { int [,]arr = { { 0, 0, 1 }, { 0, 0, 0 }, { 0, 0, 0 } }; Console.Write(maxDistance(arr) + "\n"); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // JavaScript Program to find the maximum // distance from a 0-cell to a 1-cell function maxDistance(grid) { let one = []; let M = grid.length; let N = grid[0].length; let ans = -1; for (let i = 0; i < M; ++i) { for (let j = 0; j < N; ++j) { if (grid[i][j] == 1) one.push([i, j]); } } // If the matrix consists of only 0's // or only 1's if (one.length==0 || M * N == one.length) return -1; for (let i = 0; i < M; ++i) { for (let j = 0; j < N; ++j) { if (grid[i][j] == 1) continue; // If it's a 0-cell let dist = Number.MAX_VALUE; for (let p of one) { // calculate its distance // with every 1-cell let d = Math.abs(p[0] - i) + Math.abs(p[1] - j); // Compare and store the minimum dist = Math.min(dist, d); if (dist <= ans) break; } // Compare and store the maximum ans = Math.max(ans, dist); } } return ans; } // Driver code let arr = [ [ 0, 0, 1 ], [ 0, 0, 0 ], [ 0, 0, 0 ] ]; document.write(maxDistance(arr),"</br>"); // This code is contributed by shinjanpatra. </script> |
Output:
4
Time complexity: O(M*N*P) where grid is of size M*N and P is the count of 1-cells.
Auxiliary Space: O(P)
Approach 2: Using BFS
Start from a 1-cell, and perform a Breadth First Search traversal, layer by layer. The maximum layer, up to which we can retrieve, is our answer.
Below is the implementation of the above approach:
C++
// C++ Program to find the maximum // distance from a 0-cell to a 1-cell #include <bits/stdc++.h> using namespace std; // Function to find the maximum distance int maxDistance(vector<vector<int> >& grid) { // Queue to store all 1-cells queue<pair<int, int> > q; // Grid dimensions int M = grid.size(); int N = grid[0].size(); int ans = -1; // Directions traversable from // a given a particular cell int dirs[4][2] = { { 0, 1 }, { 1, 0 }, { 0, -1 }, { -1, 0 } }; for (int i = 0; i < M; ++i) { for (int j = 0; j < N; ++j) { if (grid[i][j] == 1) q.emplace(i, j); } } // If the grid contains // only 0s or only 1s if (q.empty() || M * N == q.size()) return -1; while (q.size()) { int cnt = q.size(); while (cnt--) { // Access every 1-cell auto p = q.front(); q.pop(); // Traverse all possible // directions from the cells for (auto& dir : dirs) { int x = p.first + dir[0]; int y = p.second + dir[1]; // Check if the cell is // within the boundaries // or contains a 1 if (x < 0 || x >= M || y < 0 || y >= N || grid[x][y]) continue; q.emplace(x, y); grid[x][y] = 1; } } ++ans; } return ans; } // Driver code int main() { vector<vector<int> > arr = { { 0, 0, 1 }, { 0, 0, 0 }, { 0, 0, 1 } }; cout << maxDistance(arr) << endl; return 0; } |
Java
// Java program to find the maximum // distance from a 0-cell to a 1-cell 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 find the maximum distance static int maxDistance(int [][]grid) { // Queue to store all 1-cells Queue<pair> q = new LinkedList<pair>(); // Grid dimensions int M = grid.length; int N = grid[0].length; int ans = -1; // Directions traversable from // a given a particular cell int dirs[][] = { { 0, 1 }, { 1, 0 }, { 0, -1 }, { -1, 0 } }; for(int i = 0; i < M; ++i) { for(int j = 0; j < N; ++j) { if (grid[i][j] == 1) q.add(new pair(i, j)); } } // If the grid contains // only 0s or only 1s if (q.isEmpty() || M * N == q.size()) return -1; while (q.size() > 0) { int cnt = q.size(); while (cnt-->0) { // Access every 1-cell pair p = q.peek(); q.remove(); // Traverse all possible // directions from the cells for(int []dir : dirs) { int x = p.first + dir[0]; int y = p.second + dir[1]; // Check if the cell is // within the boundaries // or contains a 1 if (x < 0 || x >= M || y < 0 || y >= N || grid[x][y] > 0) continue; q.add(new pair(x, y)); grid[x][y] = 1; } } ++ans; } return ans; } // Driver code public static void main(String[] args) { int [][]arr = { { 0, 0, 1 }, { 0, 0, 0 }, { 0, 0, 1 } }; System.out.print(maxDistance(arr) + "\n"); } } // This code is contributed by Amit Katiyar |
Python3
# Python3 program to find the maximum # distance from a 0-cell to a 1-cell # Function to find the maximum distance def maxDistance(grid): # Queue to store all 1-cells q = [] # Grid dimensions M = len(grid) N = len(grid[0]) ans = -1 # Directions traversable from # a given a particular cell dirs = [ [ 0, 1 ], [ 1, 0 ], [ 0, -1 ], [ -1, 0 ] ] for i in range(M): for j in range(N): if (grid[i][j] == 1): q.append([i, j]) # If the grid contains # only 0s or only 1s if (len(q) == 0 or M * N == len(q)): return -1 while (len(q) > 0): cnt = len(q) while (cnt > 0): # Access every 1-cell p = q[0] q.pop() # Traverse all possible # directions from the cells for Dir in dirs: x = p[0] + Dir[0] y = p[1] + Dir[1] # Check if the cell is # within the boundaries # or contains a 1 if (x < 0 or x >= M or y < 0 or y >= N or grid[x][y]): continue q.append([x, y]) grid[x][y] = 1 cnt -= 1 ans += 2 return ans # Driver code arr = [ [ 0, 0, 1 ], [ 0, 0, 0 ], [ 0, 0, 1 ] ] print(maxDistance(arr)) # This code is contributed by divyeshrabadiya07 |
C#
// C# program to find // the maximum distance // from a 0-cell to a 1-cell using System; using System.Collections.Generic; class GFG{ static int index = 0; class pair { public int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } // Function to find the // maximum distance static int maxDistance(int [,]grid) { // Queue to store all 1-cells Queue<pair> q = new Queue<pair>(); // Grid dimensions int M = grid.GetLength(0); int N = grid.GetLength(1); int ans = -1; // Directions traversable from // a given a particular cell int [,]dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; for(int i = 0; i < M; ++i) { for(int j = 0; j < N; ++j) { if (grid[i, j] == 1) q.Enqueue(new pair(i, j)); } } // If the grid contains // only 0s or only 1s if (q.Count==0 || M * N == q.Count) return -1; while (q.Count > 0) { int cnt = q.Count; while (cnt-- > 0) { // Access every 1-cell pair p = q.Peek(); q.Dequeue(); // Traverse all possible // directions from the cells for(int i = 0; i < dirs.GetLength(0);) { int []dir = GetRow(dirs, i++); int x = p.first + dir[0]; int y = p.second + dir[1]; // Check if the cell is // within the boundaries // or contains a 1 if (x < 0 || x >= M || y < 0 || y >= N || grid[x, y] > 0) continue; q.Enqueue(new pair(x, y)); grid[x, y] = 1; } } ++ans; } return ans; } public static int[] GetRow(int[,] matrix, int row) { var rowLength = matrix.GetLength(1); var rowVector = new int[rowLength]; for (var i = 0; i < rowLength; i++) rowVector[i] = matrix[row, i]; return rowVector; } // Driver code public static void Main(String[] args) { int [,]arr = {{0, 0, 1}, {0, 0, 0}, {0, 0, 1}}; Console.Write(maxDistance(arr) + "\n"); } } // This code is contributed by shikhasingrajput |
Javascript
<script> // JavaScript program to find the maximum // distance from a 0-cell to a 1-cell // Function to find the maximum distance function maxDistance(grid){ // Queue to store all 1-cells let q = [] // Grid dimensions let M = grid.length let N = grid[0].length let ans = -1 // Directions traversable from // a given a particular cell let dirs = [ [ 0, 1 ], [ 1, 0 ], [ 0, -1 ], [ -1, 0 ] ] for(let i = 0; i < M; i++) { for(let j = 0; j < N; j++) { if (grid[i][j] == 1) q.push([i, j]) } } // If the grid contains // only 0s or only 1s if (q.length == 0 || M * N == q.length) return -1 while (q.length > 0){ let cnt = q.length while (cnt > 0){ // Access every 1-cell let p = q[0] q.pop() // Traverse all possible // directions from the cells for(let Dir of dirs){ let x = p[0] + Dir[0] let y = p[1] + Dir[1] // Check if the cell is // within the boundaries // or contains a 1 if (x < 0 || x >= M || y < 0 || y >= N || grid[x][y]) continue q.push([x, y]) grid[x][y] = 1 } cnt -= 1 } ans += 2 } return ans } // Driver code let arr = [ [ 0, 0, 1 ], [ 0, 0, 0 ], [ 0, 0, 1 ] ] console.log(maxDistance(arr)) // This code is contributed by shinjanpatra </script> |
Output:
3
Time complexity: O(M*N)
Auxiliary Space: O(M*N)
Approach 3: Using Dynamic Programming
- Keep updating the matrix of the maximum distances that have been traveled.
- Traverse from the top left-hand cell (0, 0) of the matrix to the bottom right. Let grid[i][j] represent the maximum distance from the nearest 1-cell on the left or above(or of course on itself).
- Do a second pass from bottom right to top left, updating the grid array, defining the cell grid[i][j] as the minimum of grid[i][j], grid[i+1][j], and grid[i][j+1].
- Keep track of the maximum value during the bottom right to top left traversal and return the value at the end. In case the value is 0, i.e the grid is filled with only 0’s or only 1’s, return -1.
Below is the implementation of the above approach:
C++
// C++ Program to find the maximum // distance from a 0-cell to a 1-cell #include <bits/stdc++.h> using namespace std; // Function to find the maximum distance int maxDistance(vector<vector<int> >& grid) { if (!grid.size()) return -1; int N = grid.size(); int INF = 1000000; // DP matrix vector<vector<int> > dp(N, vector<int>(N, 0)); grid[0][0] = grid[0][0] == 1 ? 0 : INF; // Set up top row and left column for (int i = 1; i < N; i++) grid[0][i] = grid[0][i] == 1 ? 0 : grid[0][i - 1] + 1; for (int i = 1; i < N; i++) grid[i][0] = grid[i][0] == 1 ? 0 : grid[i - 1][0] + 1; // Pass one: top left to bottom right for (int i = 1; i < N; i++) { for (int j = 1; j < N; j++) { grid[i][j] = grid[i][j] == 1 ? 0 : min(grid[i - 1][j], grid[i][j - 1]) + 1; } } // Check if there was no "One" Cell if (grid[N - 1][N - 1] >= INF) return -1; // Set up top row and left column int maxi = grid[N - 1][N - 1]; for (int i = N - 2; i >= 0; i--) { grid[N - 1][i] = min(grid[N - 1][i], grid[N - 1][i + 1] + 1); maxi = max(grid[N - 1][i], maxi); } for (int i = N - 2; i >= 0; i--) { grid[i][N - 1] = min(grid[i][N - 1], grid[i + 1][N - 1] + 1); maxi = max(grid[i][N - 1], maxi); } // Past two: bottom right to top left for (int i = N - 2; i >= 0; i--) { for (int j = N - 2; j >= 0; j--) { grid[i][j] = min(grid[i][j], min(grid[i + 1][j] + 1, grid[i][j + 1] + 1)); maxi = max(grid[i][j], maxi); } } return !maxi ? -1 : maxi; } // Driver code int main() { vector<vector<int> > arr = { { 0, 0, 1 }, { 0, 0, 0 }, { 0, 0, 0 } }; cout << maxDistance(arr) << endl; return 0; } |
Java
// Java program to find the maximum // distance from a 0-cell to a 1-cell import java.util.*; import java.lang.*; class GFG{ // Function to find the maximum distance static int maxDistance(int[][] grid) { if (grid.length == 0) return -1; int N = grid.length; int INF = 1000000; grid[0][0] = grid[0][0] == 1 ? 0 : INF; // Set up top row and left column for(int i = 1; i < N; i++) grid[0][i] = grid[0][i] == 1 ? 0 : grid[0][i - 1] + 1; for(int i = 1; i < N; i++) grid[i][0] = grid[i][0] == 1 ? 0 : grid[i - 1][0] + 1; // Pass one: top left to bottom right for(int i = 1; i < N; i++) { for(int j = 1; j < N; j++) { grid[i][j] = grid[i][j] == 1 ? 0 : Math.min(grid[i - 1][j], grid[i][j - 1]) + 1; } } // Check if there was no "One" Cell if (grid[N - 1][N - 1] >= INF) return -1; // Set up top row and left column int maxi = grid[N - 1][N - 1]; for(int i = N - 2; i >= 0; i--) { grid[N - 1][i] = Math.min( grid[N - 1][i], grid[N - 1][i + 1] + 1); maxi = Math.max(grid[N - 1][i], maxi); } for(int i = N - 2; i >= 0; i--) { grid[i][N - 1] = Math.min( grid[i][N - 1], grid[i + 1][N - 1] + 1); maxi = Math.max(grid[i][N - 1], maxi); } // Past two: bottom right to top left for(int i = N - 2; i >= 0; i--) { for(int j = N - 2; j >= 0; j--) { grid[i][j] = Math.min( grid[i][j], Math.min(grid[i + 1][j] + 1, grid[i][j + 1] + 1)); maxi = Math.max(grid[i][j], maxi); } } return maxi == 0 ? -1 : maxi; } // Driver code public static void main(String[] args) { int[][] arr = { { 0, 0, 1 }, { 0, 0, 0 }, { 0, 0, 0 } }; System.out.println(maxDistance(arr)); } } // This code is contributed by offbeat |
Python3
# Python3 program to find the maximum # distance from a 0-cell to a 1-cell # Function to find the maximum distance def maxDistance(grid): if (len(grid) == 0): return -1 N = len(grid) INF = 1000000 if grid[0][0] == 1: grid[0][0] = 0 else: grid[0][0] = INF # Set up top row and left column for i in range(1, N): if grid[0][i] == 1: grid[0][i] = 0 else: grid[0][i] = grid[0][i - 1] + 1 for i in range(1, N): if grid[i][0] == 1: grid[i][0] = 0 else: grid[i][0] = grid[i - 1][0] + 1 # Pass one: top left to bottom right for i in range(1, N): for j in range(1, N): if grid[i][j] == 1: grid[i][j] = 0 else: grid[i][j] = min(grid[i - 1][j], grid[i][j - 1] + 1) # Check if there was no "One" Cell if (grid[N - 1][N - 1] >= INF): return -1 # Set up top row and left column maxi = grid[N - 1][N - 1] for i in range(N - 2, -1, -1): grid[N - 1][i] = min(grid[N - 1][i], grid[N - 1][i + 1] + 1) maxi = max(grid[N - 1][i], maxi) for i in range(N - 2, -1, -1): grid[i][N - 1] = min(grid[i][N - 1], grid[i + 1][N - 1] + 1) maxi = max(grid[i][N - 1], maxi + 1) # Past two: bottom right to top left for i in range(N - 2, -1, -1): for j in range(N - 2, -1, -1): grid[i][j] = min(grid[i][j], min(grid[i + 1][j] + 1, grid[i][j + 1] + 1)) maxi = max(grid[i][j], maxi) if maxi == 0: return -1 else: return maxi # Driver code arr = [ [ 0, 0, 1 ], [ 0, 0, 0 ], [ 0, 0, 0 ] ] print(maxDistance(arr)) # This code is contributed by divyesh072019 |
C#
// C# program to find the maximum // distance from a 0-cell to a 1-cell using System; class GFG { // Function to find the maximum distance static int maxDistance(int[, ] grid) { if (grid.GetLength(0) == 0) return -1; int N = grid.GetLength(0); int INF = 1000000; grid[0, 0] = grid[0, 0] == 1 ? 0 : INF; // Set up top row and left column for (int i = 1; i < N; i++) grid[0, i] = grid[0, i] == 1 ? 0 : grid[0, i - 1] + 1; for (int i = 1; i < N; i++) grid[i, 0] = grid[i, 0] == 1 ? 0 : grid[i - 1, 0] + 1; // Pass one: top left to bottom right for (int i = 1; i < N; i++) { for (int j = 1; j < N; j++) { grid[i, j] = grid[i, j] == 1 ? 0 : Math.Min(grid[i - 1, j], grid[i, j - 1]) + 1; } } // Check if there was no "One" Cell if (grid[N - 1, N - 1] >= INF) return -1; // Set up top row and left column int maxi = grid[N - 1, N - 1]; for (int i = N - 2; i >= 0; i--) { grid[N - 1, i] = Math.Min( grid[N - 1, i], grid[N - 1, i + 1] + 1); maxi = Math.Max(grid[N - 1, i], maxi); } for (int i = N - 2; i >= 0; i--) { grid[i, N - 1] = Math.Min( grid[i, N - 1], grid[i + 1, N - 1] + 1); maxi = Math.Max(grid[i, N - 1], maxi); } // Past two: bottom right to top left for (int i = N - 2; i >= 0; i--) { for (int j = N - 2; j >= 0; j--) { grid[i, j] = Math.Min( grid[i, j], Math.Min(grid[i + 1, j] + 1, grid[i, j + 1] + 1)); maxi = Math.Max(grid[i, j], maxi); } } return maxi == 0 ? -1 : maxi; } // Driver code public static void Main() { int[, ] arr = { { 0, 0, 1 }, { 0, 0, 0 }, { 0, 0, 0 } }; Console.WriteLine(maxDistance(arr)); } } // This code is contributed by subhammahato348 |
Javascript
<script> // JavaScript Program to find the maximum // distance from a 0-cell to a 1-cell // Function to find the maximum distance function maxDistance(grid) { if (!grid.length) return -1 let N = grid.length let INF = 1000000 // DP matrix let dp = new Array(N) for(let i=0;i<N;i++){ dp[i] = new Array(N).fill(0) } grid[0][0] = grid[0][0] == 1 ? 0 : INF // Set up top row and left column for (let i = 1; i < N; i++) grid[0][i] = grid[0][i] == 1 ? 0 : grid[0][i - 1] + 1; for (let i = 1; i < N; i++) grid[i][0] = grid[i][0] == 1 ? 0 : grid[i - 1][0] + 1; // Pass one: top left to bottom right for (let i = 1; i <N; i++) { for (let j = 1; j < N; j++) { grid[i][j] = grid[i][j] == 1 ? 0 : Math.min(grid[i - 1][j], grid[i][j - 1]) + 1; } } // Check if there was no "One" Cell if (grid[N - 1][N - 1] >= INF) return -1; // Set up top row and left column let maxi = grid[N - 1][N - 1]; for (let i = N - 2; i >= 0; i--) { grid[N - 1][i] = Math.min(grid[N - 1][i], grid[N - 1][i + 1] + 1); maxi = Math.max(grid[N - 1][i], maxi); } for (let i = N - 2; i >= 0; i--) { grid[i][N - 1] = Math.min(grid[i][N - 1], grid[i + 1][N - 1] + 1); maxi = Math.max(grid[i][N - 1], maxi); } // Past two: bottom right to top left for (let i = N - 2; i >= 0; i--) { for (let j = N - 2; j >= 0; j--) { grid[i][j] = Math.min(grid[i][j], Math.min(grid[i + 1][j] + 1, grid[i][j + 1] + 1)); maxi = Math.max(grid[i][j], maxi); } } return !maxi ? -1 : maxi; } // Driver code let arr = [ [ 0, 0, 1 ], [ 0, 0, 0 ], [ 0, 0, 0 ] ] document.write(maxDistance(arr),"</br>") // This code is contributed by shinjanpatra </script> |
Output
4
Time complexity: O(M*N)
Auxiliary Space: O(N2)
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