# 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:

 `// C++ Program to find the maximum` `// distance from a 0-cell to a 1-cell`   `#include ` `using` `namespace` `std;`   `int` `maxDistance(vector >& grid)` `{` `    ``vector > one;`   `    ``int` `M = grid.size();` `    ``int` `N = grid.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 > arr` `        ``= { { 0, 0, 1 },` `            ``{ 0, 0, 0 },` `            ``{ 0, 0, 0 } };`   `    ``cout << maxDistance(arr) << endl;` `    ``return` `0;` `}`

 `// 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 one = ``new` `Vector();` ` `  `    ``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`

 `// 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 one = ``new` `List();`   `    ``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`

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++ Program to find the maximum` `// distance from a 0-cell to a 1-cell`   `#include ` `using` `namespace` `std;`   `// Function to find the maximum distance` `int` `maxDistance(vector >& grid)` `{` `    ``// Queue to store all 1-cells` `    ``queue > q;`   `    ``// Grid dimensions` `    ``int` `M = grid.size();` `    ``int` `N = grid.size();` `    ``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.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;` `                ``int` `y = p.second + dir;`   `                ``// 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 > arr = { { 0, 0, 1 },` `                                 ``{ 0, 0, 0 },` `                                 ``{ 0, 0, 1 } };`   `    ``cout << maxDistance(arr) << endl;` `    ``return` `0;` `}`

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

 `// 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 q = ``new` `Queue();`   `  ``// 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;` `        ``int` `y = p.second + dir;`   `        ``// 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`

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++ Program to find the maximum` `// distance from a 0-cell to a 1-cell`   `#include ` `using` `namespace` `std;`   `// Function to find the maximum distance` `int` `maxDistance(vector >& grid)` `{` `    ``if` `(!grid.size())` `        ``return` `-1;` `    ``int` `N = grid.size();`   `    ``int` `INF = 1000000;`   `    ``// DP matrix` `    ``vector >` `    ``dp(N, vector<``int``>(N, 0));`   `    ``grid = grid == 1` `                     ``? 0` `                     ``: INF;`   `    ``// Set up top row and left column` `    ``for` `(``int` `i = 1; i < N; i++)` `        ``grid[i] = grid[i] == 1` `                         ``? 0` `                         ``: grid[i - 1] + 1;` `    ``for` `(``int` `i = 1; i < N; i++)` `        ``grid[i] = grid[i] == 1` `                         ``? 0` `                         ``: grid[i - 1] + 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 > arr = { { 0, 0, 1 },` `                                 ``{ 0, 0, 0 },` `                                 ``{ 0, 0, 0 } };`   `    ``cout << maxDistance(arr) << endl;` `    ``return` `0;` `}`

 `// 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`

Output
```4

```

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

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