Maximum of all distances to the nearest 1 cell from any 0 cell in a Binary matrix

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: 



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// 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 ans 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;
}
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// 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 ans 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
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// 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 ans 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
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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:  

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// 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;
}
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// 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
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// 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
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Output: 
3








 

Time complexity: O(M*N) 
Auxiliary Space: O(M*N)

Approach 3: Using Dynamic Programming  

Below is the implementation of the above approach:  

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// 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;
}
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// 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
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Output
4



Time complexity: O(M*N) 
Auxiliary Space: O(1)




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