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

• Difficulty Level : Medium
• Last Updated : 14 Jun, 2021

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:

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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:

• Assign the distance of the initial cells as 0.
• Initialize a Queue and insert the pair {X, Y} into the Queue.
• Iterate until the Queue is not empty, and for every unvisited cell, assign the current distance and insert the index {i, j} into the Queue using the BFS technique.
• Print the distance of each cell at the end.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``using` `namespace` `std;` `int` `mat;``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 > 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;``    ``}``}`

## Java

 `// 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 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`

## Python3

 `# 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`

## C#

 `// 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 q = ``new` `Queue();` `    ``// 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`

## Javascript

 ``
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)

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