# Nearest 1 in a binary matrix

Given a binary matrix of order m*n, the task is to find the distance of nearest 1 for each 0 in the matrix and print final distance matrix. From any cell (i,j), we can move only in four directions up, down, left and right.

__Note :__ Distance from one cell to immediate another cell is always incremented by 1.

**Examples:**

Input : m = 3, n = 4 mat[m][n] = {{0, 0, 0, 1}, {0, 0, 1, 1}, {0, 1, 1, 0}} Output: 3 2 1 0 2 1 0 0 1 0 0 1

A **simple solution** for this problem is to for each 0 in the matrix recursively check the nearest 1 in the matrix.

An **efficient solution** solution for this problem is to use BFS. Here is the algorithm to solve this problem :

- Take distance matrix dist[m][n] and initialize it with INT_MAX.
- Now traverse the matrix and make_pair(i,j) of indices of cell (i, j) having value ‘1’ and push this pair into queue and update dist[i][j] = 0 because distance of ‘1’ from itself will be always 0.
- Now pop elements from queue one by one until it gets empty and call
**BFS**on it. - Here we need to find the distance of nearest one and we are calling
**BFS**for the cells having ‘1’, so whenever we take adjacent of popped element from queue, we try to minimize the distance by putting condition if (dist[i][j]+1) < dist[ADCi][ADCj]. Then we update the distance of adjacent element in the distance matrix and push this adjacent in the queue to complete the**BFS**traversal and filling the complete distance matrix. - After completing the
**BFS**traversal each cell of distance matrix will contain the distance of nearest ‘1’.

`// C++ program to find the minimum distance from a ` `// "1" in binary matrix. ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` `const` `int` `MAX = 1000; ` ` ` `// distance matrix which stores the distance of ` `// nearest '1' ` `int` `dist[MAX][MAX]; ` ` ` `// Function to find the nearest '1' ` `void` `nearestOne(` `int` `mat[][MAX], ` `int` `m, ` `int` `n) ` `{ ` ` ` `// two array when respective values of newx and ` ` ` `// newy are added to (i,j) it gives up, down, ` ` ` `// left or right adjacent of (i,j) cell ` ` ` `int` `newx[] = {-1, 0, 1, 0}; ` ` ` `int` `newy[] = {0, -1, 0, 1}; ` ` ` ` ` `// queue of pairs to store nodes for bfs ` ` ` `queue< pair<` `int` `,` `int` `> > q; ` ` ` ` ` `// traverse matrix and make pair of indices of ` ` ` `// cell (i,j) having value '1' and push them ` ` ` `// in queue ` ` ` `for` `(` `int` `i=0; i<m; i++) ` ` ` `{ ` ` ` `for` `(` `int` `j=0; j<n; j++) ` ` ` `{ ` ` ` `dist[i][j] = INT_MAX; ` ` ` ` ` `if` `(mat[i][j] == 1) ` ` ` `{ ` ` ` `// distance of '1' from itself is always 0 ` ` ` `dist[i][j] = 0; ` ` ` ` ` `// make pair and push it in queue ` ` ` `q.push(make_pair(i, j)); ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` ` ` `// now do bfs traversal ` ` ` `// pop element from queue one by one until it gets empty ` ` ` `// pair element to hold the currently popped element ` ` ` `pair<` `int` `,` `int` `> poped; ` ` ` `while` `(!q.empty()) ` ` ` `{ ` ` ` `poped = q.front(); ` ` ` `q.pop(); ` ` ` ` ` `// coordinate of currently popped node ` ` ` `int` `x = poped.first; ` ` ` `int` `y = poped.second; ` ` ` ` ` `// now check for all adjancent of popped element ` ` ` `for` `(` `int` `i=0; i<4; i++) ` ` ` `{ ` ` ` `int` `adjx = x + newx[i]; ` ` ` `int` `adjy = y + newy[i]; ` ` ` ` ` `// if new coordinates are within boundary and ` ` ` `// we can minimize the distance of adjacent ` ` ` `// then update the distance of adjacent in ` ` ` `// distance matrix and push this adjacent ` ` ` `// element in queue for further bfs ` ` ` `if` `(adjx>=0 && adjx<m && adjy>=0 && adjy<n && ` ` ` `dist[adjx][adjy] > dist[x][y] + 1) ` ` ` `{ ` ` ` `// update distance ` ` ` `dist[adjx][adjy] = dist[x][y] + 1; ` ` ` `q.push(make_pair(adjx,adjy)); ` ` ` `} ` ` ` `} ` ` ` `} ` `} ` ` ` `// Driver program to run the case ` `int` `main() ` `{ ` ` ` `int` `m = 3, n = 4; ` ` ` `int` `mat[][MAX] = {{0, 0, 0, 1}, ` ` ` `{0, 0, 1, 1}, ` ` ` `{0, 1, 1, 0} ` ` ` `}; ` ` ` ` ` `// Fills values in dist[][] ` ` ` `nearestOne(mat, m, n); ` ` ` ` ` `// print distance matrix ` ` ` `for` `(` `int` `i=0; i<m; i++) ` ` ` `{ ` ` ` `for` `(` `int` `j=0; j<n; j++) ` ` ` `cout << dist[i][j] << ` `" "` `; ` ` ` `cout << endl; ` ` ` `} ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

Output:

3 2 1 0 2 1 0 0 1 0 0 1

This article is contributed by **Shashank Mishra ( Gullu )**. 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.

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