Minimum Distance from a given Cell to all other Cells of a Matrix

Given two integers R and C, denoting the number of rows and columns in a matrix, and two integers X and Y, the task is to find the minimum distance from the given cell to all other cells of the matrix.

Examples:

Input: R = 5, C = 5, X = 2, Y = 2 
Output: 
2 2 2 2 2 
2 1 1 1 2 
2 1 0 1 2 
2 1 1 1 2 
2 2 2 2 2

Input: R = 5, C = 5, X = 1, Y = 1 
Output: 
1 1 1 2 3 
1 0 1 2 3 
1 1 1 2 3 
2 2 2 2 3 
3 3 3 3 3

Approach: 
Follow the steps below to solve the problem:



Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
int mat[1001][1001];
int r, c, x, y;
  
// Stores the accessible directions
int dx[] = { 0, -1, -1, -1, 0, 1, 1, 1 };
int dy[] = { 1, 1, 0, -1, -1, -1, 0, 1 };
  
// Function to find the minimum distance from a
// given cell to all other cells in the matrix
void FindMinimumDistance()
{
    // Stores the accessible cells
    // from current cell
    queue<pair<int, int> > q;
  
    // Insert pair (x, y)
    q.push({ x, y });
    mat[x][y] = 0;
  
    // Iterate untill queue is empty
    while (!q.empty()) {
  
        // Extract the pair
        x = q.front().first;
        y = q.front().second;
  
        // Pop them
        q.pop();
  
        for (int i = 0; i < 8; i++) {
            int a = x + dx[i];
            int b = y + dy[i];
  
            // Checking boundary condition
            if (a < 0 || a >= r || b >= c || b < 0)
                continue;
  
            // If the cell is not visited
            if (mat[a][b] == 0) {
  
                // Assign the minimum distance
                mat[a][b] = mat[x][y] + 1;
  
                // Insert the traversed neighbour
                // into the queue
                q.push({ a, b });
            }
        }
    }
}
  
// Driver Code
int main()
{
    r = 5, c = 5, x = 1, y = 1;
  
    int t = x;
    int l = y;
    mat[x][y] = 0;
  
    FindMinimumDistance();
  
    mat[t][l] = 0;
  
    // Print the required distances
    for (int i = 0; i < r; i++) {
        for (int j = 0; j < c; j++) {
            cout << mat[i][j] << " ";
        }
        cout << endl;
    }
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to implement
// the above approach
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 [][]mat = new int[1001][1001];
static int r, c, x, y;
  
// Stores the accessible directions
static int dx[] = { 0, -1, -1, -1, 0, 1, 1, 1 };
static int dy[] = { 1, 1, 0, -1, -1, -1, 0, 1 };
  
// Function to find the minimum distance from a
// given cell to all other cells in the matrix
static void FindMinimumDistance()
{
      
    // Stores the accessible cells
    // from current cell
    Queue<pair> q = new LinkedList<>();
  
    // Insert pair (x, y)
    q.add(new pair(x, y));
    mat[x][y] = 0;
  
    // Iterate untill queue is empty
    while (!q.isEmpty())
    {
          
        // Extract the pair
        x = q.peek().first;
        y = q.peek().second;
  
        // Pop them
        q.remove();
  
        for(int i = 0; i < 8; i++)
        {
            int a = x + dx[i];
            int b = y + dy[i];
  
            // Checking boundary condition
            if (a < 0 || a >= r ||
               b >= c || b < 0)
                continue;
  
            // If the cell is not visited
            if (mat[a][b] == 0
            {
                  
                // Assign the minimum distance
                mat[a][b] = mat[x][y] + 1;
  
                // Insert the traversed neighbour
                // into the queue
                q.add(new pair(a, b));
            }
        }
    }
}
  
// Driver Code
public static void main(String[] args)
{
    r = 5; c = 5; x = 1; y = 1;
  
    int t = x;
    int l = y;
    mat[x][y] = 0;
  
    FindMinimumDistance();
  
    mat[t][l] = 0;
  
    // Print the required distances
    for(int i = 0; i < r; i++)
    {
        for(int j = 0; j < c; j++)
        {
            System.out.print(mat[i][j] + " ");
        }
        System.out.println();
    }
}
}
  
// This code is contributed by Amit Katiyar
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to implement
# the above approach
mat = [[0 for x in range(1001)] 
          for y in range(1001)]
  
# Stores the accessible directions
dx = [ 0, -1, -1, -1, 0, 1, 1, 1 ]
dy = [ 1, 1, 0, -1, -1, -1, 0, 1 ]
  
# Function to find the minimum distance
# from a given cell to all other cells
# in the matrix
def FindMinimumDistance():
      
    global x, y, r, c
  
    # Stores the accessible cells
    # from current cell
    q = []
  
    # Insert pair (x, y)
    q.append([x, y])
    mat[x][y] = 0
  
    # Iterate untill queue is empty
    while(len(q) != 0):
  
        # Extract the pair
        x = q[0][0]
        y = q[0][1]
  
        # Pop them
        q.pop(0)
  
        for i in range(8):
            a = x + dx[i]
            b = y + dy[i]
  
            # Checking boundary condition
            if(a < 0 or a >= r or 
              b >= c or b < 0):
                continue
  
            # If the cell is not visited
            if(mat[a][b] == 0):
  
                # Assign the minimum distance
                mat[a][b] = mat[x][y] + 1
  
                # Insert the traversed neighbour
                # into the queue
                q.append([a, b])
  
# Driver Code
r = 5
c = 5
x = 1
y = 1
t = x
l = y
  
mat[x][y] = 0
  
FindMinimumDistance()
mat[t][l] = 0
  
# Print the required distances 
for i in range(r):
    for j in range(c):
        print(mat[i][j], end = " ")
          
    print()
  
# This code is contributed by Shivam Singh
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to implement
// the above approach
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 [,]mat = new int[1001, 1001];
static int r, c, x, y;
  
// Stores the accessible directions
static int []dx = { 0, -1, -1, -1, 0, 1, 1, 1 };
static int []dy = { 1, 1, 0, -1, -1, -1, 0, 1 };
  
// Function to find the minimum distance from a
// given cell to all other cells in the matrix
static void FindMinimumDistance()
{
      
    // Stores the accessible cells
    // from current cell
    Queue<pair> q = new Queue<pair>();
  
    // Insert pair (x, y)
    q.Enqueue(new pair(x, y));
    mat[x, y] = 0;
  
    // Iterate untill queue is empty
    while (q.Count != 0)
    {
          
        // Extract the pair
        x = q.Peek().first;
        y = q.Peek().second;
  
        // Pop them
        q.Dequeue();
  
        for(int i = 0; i < 8; i++)
        {
            int a = x + dx[i];
            int b = y + dy[i];
  
            // Checking boundary condition
            if (a < 0 || a >= r ||
                b >= c || b < 0)
                continue;
  
            // If the cell is not visited
            if (mat[a, b] == 0) 
            {
                  
                // Assign the minimum distance
                mat[a, b] = mat[x, y] + 1;
  
                // Insert the traversed neighbour
                // into the queue
                q.Enqueue(new pair(a, b));
            }
        }
    }
}
  
// Driver Code
public static void Main(String[] args)
{
    r = 5; c = 5; x = 1; y = 1;
  
    int t = x;
    int l = y;
    mat[x, y] = 0;
  
    FindMinimumDistance();
  
    mat[t, l] = 0;
  
    // Print the required distances
    for(int i = 0; i < r; i++)
    {
        for(int j = 0; j < c; j++)
        {
            Console.Write(mat[i, j] + " ");
        }
        Console.WriteLine();
    }
}
}
  
// This code is contributed by shikhasingrajput
chevron_right

Output: 
1 1 1 2 3 
1 0 1 2 3 
1 1 1 2 3 
2 2 2 2 3 
3 3 3 3 3

Time Complexity: O(R * C) 
Auxiliary Space: O(R * C)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




Recommended Posts:


Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :