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# Minimum steps to reach any of the boundary edges of a matrix | Set-2

• Difficulty Level : Easy
• Last Updated : 26 Feb, 2020

Given an N X M matrix, where ai, j = 1 denotes the cell is not empty, ai, j = 0 denotes the cell is empty and ai, j = 2, denotes that you are standing at that cell. You can move vertically up or down and horizontally left or right to any cell which is empty. The task is to find the minimum number of steps to reach any boundary edge of the matrix. Print -1 if not possible to reach any of the boundary edges.

Note: There will be only one cell with value 2 in the entire matrix.

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Examples:

```Input: matrix[] = {1, 1, 1, 0, 1}
{1, 0, 2, 0, 1}
{0, 0, 1, 0, 1}
{1, 0, 1, 1, 0}
Output: 2
Move to the right and then move
upwards to reach the nearest boundary
edge.

Input: matrix[] = {1, 1, 1, 1, 1}
{1, 0, 2, 0, 1}
{1, 0, 1, 0, 1}
{1, 1, 1, 1, 1}
Output: -1
```

Approach: The problem has been solved in Set-1 using Dynamic Programming. In this article, we will be discussing a method using BFS. Given below are the steps to solve the above problem.

• Find the index of the ‘2’ in the matrix.
• Check if this index is a boundary edge or not, if it is, then no moves are required.
• Insert the index x and index y of ‘2’ in the queue with moves as 0.
• Use a 2-D vis array to mark the visiting positions in the matrix.
• Iterate till the queue is empty or we reach any boundary edge.
• Get the front element(x, y, val = moves) in the queue and mark vis[x][y] as visited. Do all the possible moves(right, left, up and down) possible.
• Re-insert val+1 and their indexes of all the valid moves to the queue.
• If the x and y become the boundary edges any time return val.
• If all the moves are made, and the queue is empty, then it is not possible, hence return -1.

Below is the implementation of the above approach:

## CPP

 `// C++ program to find Minimum steps``// to reach any of the boundary``// edges of a matrix``#include ``using` `namespace` `std;``#define r 4``#define c 5`` ` `// Function to check validity``bool` `check(``int` `i, ``int` `j, ``int` `n, ``int` `m, ``int` `mat[r])``{``    ``if` `(i >= 0 && i < n && j >= 0 && j < m) {``        ``if` `(mat[i][j] == 0)``            ``return` `true``;``    ``}``    ``return` `false``;``}`` ` `// Function to find out minimum steps``int` `findMinSteps(``int` `mat[r], ``int` `n, ``int` `m)``{`` ` `    ``int` `indx, indy;``    ``indx = indy = -1;`` ` `    ``// Find index of only 2 in matrix``    ``for` `(``int` `i = 0; i < n; i++) {``        ``for` `(``int` `j = 0; j < m; j++) {``            ``if` `(mat[i][j] == 2) {``                ``indx = i, indy = j;``                ``break``;``            ``}``        ``}``        ``if` `(indx != -1)``            ``break``;``    ``}`` ` `    ``// Push elements in the queue``    ``queue > > q;`` ` `    ``// Push the position 2 with moves as 0``    ``q.push(make_pair(0, make_pair(indx, indy)));`` ` `    ``// If already at boundary edge``    ``if` `(check(indx, indy, n, m, mat))``        ``return` `0;`` ` `    ``// Marks the visit``    ``bool` `vis[r];``    ``memset``(vis, 0, ``sizeof` `vis);`` ` `    ``// Iterate in the queue``    ``while` `(!q.empty()) {``        ``// Get the front of the queue``        ``auto` `it = q.front();`` ` `        ``// Pop the first element from the queue``        ``q.pop();`` ` `        ``// Get the position``        ``int` `x = it.second.first;``        ``int` `y = it.second.second;`` ` `        ``// Moves``        ``int` `val = it.first;`` ` `        ``// If a boundary edge``        ``if` `(x == 0 || x == (n - 1) || y == 0 || y == (m - 1)) {``            ``return` `val;``        ``}`` ` `        ``// Marks the visited array``        ``vis[x][y] = 1;`` ` `        ``// If a move is possible``        ``if` `(check(x - 1, y, n, m, mat)) {`` ` `            ``// If not visited previously``            ``if` `(!vis[x - 1][y])``                ``q.push(make_pair(val + 1, make_pair(x - 1, y)));``        ``}`` ` `        ``// If a move is possible``        ``if` `(check(x + 1, y, n, m, mat)) {`` ` `            ``// If not visited previously``            ``if` `(!vis[x + 1][y])``                ``q.push(make_pair(val + 1, make_pair(x + 1, y)));``        ``}`` ` `        ``// If a move is possible``        ``if` `(check(x, y + 1, n, m, mat)) {`` ` `            ``// If not visited previously``            ``if` `(!vis[x][y + 1])``                ``q.push(make_pair(val + 1, make_pair(x, y + 1)));``        ``}`` ` `        ``// If a move is possible``        ``if` `(check(x, y - 1, n, m, mat)) {`` ` `            ``// If not visited previously``            ``if` `(!vis[x][y - 1])``                ``q.push(make_pair(val + 1, make_pair(x, y - 1)));``        ``}``    ``}`` ` `    ``return` `-1;``}`` ` `// Driver Code``int` `main()``{``    ``int` `mat[r] = { { 1, 1, 1, 0, 1 },``                      ``{ 1, 0, 2, 0, 1 },``                      ``{ 0, 0, 1, 0, 1 },``                      ``{ 1, 0, 1, 1, 0 } };`` ` `    ``cout << findMinSteps(mat, r, c);``}`

## Python3

 `# Python3 program to find Minimum steps``# to reach any of the boundary``# edges of a matrix``from` `collections ``import` `deque``r ``=` `4``c ``=` `5`` ` `# Function to check validity``def` `check(i, j, n, m, mat):`` ` `    ``if` `(i >``=` `0` `and` `i < n ``and` `j >``=` `0` `and` `j < m):``        ``if` `(mat[i][j] ``=``=` `0``):``            ``return` `True`` ` `    ``return` `False`` ` `# Function to find out minimum steps``def` `findMinSteps(mat, n, m):`` ` `    ``indx ``=` `indy ``=` `-``1``;`` ` `    ``# Find index of only 2 in matrix``    ``for` `i ``in` `range``(n):``        ``for` `j ``in` `range``(m):``            ``if` `(mat[i][j] ``=``=` `2``):``                ``indx ``=` `i``                ``indy ``=` `j``                ``break`` ` `        ``if` `(indx !``=` `-``1``):``            ``break`` ` `    ``# Push elements in the queue``    ``q ``=` `deque()`` ` `    ``# Push the position 2 with moves as 0``    ``q.append([``0``, indx, indy])`` ` `    ``# If already at boundary edge``    ``if` `(check(indx, indy, n, m, mat)):``        ``return` `0`` ` `    ``# Marks the visit``    ``vis``=``[ [``0` `for` `i ``in` `range``(r)]``for` `i ``in` `range``(r)]`` ` `    ``# Iterate in the queue``    ``while` `len``(q) > ``0``:``         ` `        ``# Get the front of the queue``        ``it ``=` `q.popleft()`` ` `        ``#Pop the first element from the queue`` ` `        ``# Get the position``        ``x ``=` `it[``1``]``        ``y ``=` `it[``2``]`` ` `        ``# Moves``        ``val ``=` `it[``0``]`` ` `        ``# If a boundary edge``        ``if` `(x ``=``=` `0` `or` `x ``=``=` `(n ``-` `1``) ``or` `y ``=``=` `0` `or` `y ``=``=` `(m ``-` `1``)):``            ``return` `val`` ` ` ` `        ``# Marks the visited array``        ``vis[x][y] ``=` `1`` ` `        ``# If a move is possible``        ``if` `(check(x ``-` `1``, y, n, m, mat)):`` ` `            ``# If not visited previously``            ``if` `(``not` `vis[x ``-` `1``][y]):``                ``q.append([val ``+` `1``, x ``-` `1``, y])`` ` `        ``# If a move is possible``        ``if` `(check(x ``+` `1``, y, n, m, mat)):`` ` `            ``# If not visited previously``            ``if` `(``not` `vis[x ``+` `1``][y]):``                ``q.append([val ``+` `1``, x ``+` `1``, y])`` ` `        ``# If a move is possible``        ``if` `(check(x, y ``+` `1``, n, m, mat)):`` ` `            ``# If not visited previously``            ``if` `(``not` `vis[x][y ``+` `1``]):``                ``q.append([val ``+` `1``, x, y ``+` `1``])`` ` `        ``# If a move is possible``        ``if` `(check(x, y ``-` `1``, n, m, mat)):`` ` `            ``# If not visited previously``            ``if` `(``not` `vis[x][y ``-` `1``]):``                ``q.append([val ``+` `1``, x, y ``-` `1``])`` ` `    ``return` `-``1`` ` `# Driver Code``mat ``=` `[[``1``, ``1``, ``1``, ``0``, ``1` `],``       ``[``1``, ``0``, ``2``, ``0``, ``1` `],``       ``[``0``, ``0``, ``1``, ``0``, ``1` `],``       ``[``1``, ``0``, ``1``, ``1``, ``0` `] ];`` ` `print``(findMinSteps(mat, r, c))`` ` `# This code is contributed by mohit kumar 29`
Output:
```2
```

Time Complexity: O(N^2)
Auxiliary Space: O(N^2)

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