Minimum steps required to reach the end of a matrix | Set 2

Given a 2d-matrix mat[][] consisting of positive integers, the task is to find the minimum number of steps required to reach the end of the matrix. If we are at cell (i, j) we can go to cells (i, j + arr[i][j]) or (i + arr[i][j], j). We cannot go out of bounds. If no path exists then print -1.

Examples:

Input: mat[][] = {
{2, 1, 2},
{1, 1, 1},
{1, 1, 1}}
Output: 2
The path will be {0, 0} -> {0, 2} -> {2, 2}
Thus, we are reaching there in two steps.

Input: mat[][] = {
{1, 1, 1},
{1, 1, 1},
{1, 1, 1}}
Output: 4



Approach: We have already discussed a dynamic programming based approach for this problem in this article. This problem can also be solved using breadth first search (BFS).

The algorithm is as follows:

  • Push (0, 0) in a queue.
  • Traverse (0, 0) i.e. push all the cells it can visit in queue.
  • Repeat the above steps, i.e. traverse all the elements in the queue individually again if they have not been visited/traversed before.
  • Repeat till we don’t reach the cell (N-1, N-1).
  • The depth of this traversal will give the minimum steps required to reach the end.

Remember to mark a cell visited after it has been traversed. For this, we will use a 2D boolean array.

Why BFS works?

  • This whole scenario can be considered equivalent to a directed-graph where each cell is connected to at-most two more cells({i, j+arr[i][j]} and {i+arr[i][j], j}).
  • The graph is un-weighted. BFS can find shortest path in such scenarios.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
#define n 3
using namespace std;
  
// Function to return the minimum steps
// required to reach the end of the matrix
int minSteps(int arr[][n])
{
    // Array to determine whether
    // a cell has been visited before
    bool v[n][n] = { 0 };
  
    // Queue for bfs
    queue<pair<int, int> > q;
  
    // Initializing queue
    q.push({ 0, 0 });
  
    // To store the depth of search
    int depth = 0;
  
    // BFS algorithm
    while (q.size() != 0) {
  
        // Current queue size
        int x = q.size();
        while (x--) {
  
            // Top-most element of queue
            pair<int, int> y = q.front();
  
            // To store index of cell
            // for simplicity
            int i = y.first, j = y.second;
            q.pop();
  
            // Base case
            if (v[i][j])
                continue;
  
            // If we reach (n-1, n-1)
            if (i == n - 1 && j == n - 1)
                return depth;
  
            // Marking the cell visited
            v[i][j] = 1;
  
            // Pushing the adjacent cells in the
            // queue that can be visited
            // from the current cell
            if (i + arr[i][j] < n)
                q.push({ i + arr[i][j], j });
            if (j + arr[i][j] < n)
                q.push({ i, j + arr[i][j] });
        }
        depth++;
    }
  
    return -1;
}
  
// Driver code
int main()
{
    int arr[n][n] = { { 1, 1, 1 },
                      { 1, 1, 1 },
                      { 1, 1, 1 } };
  
    cout << minSteps(arr);
  
    return 0;
}

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Python3

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# Python 3 implementation of the approach
n = 3
  
# Function to return the minimum steps
# required to reach the end of the matrix
def minSteps(arr):
      
    # Array to determine whether
    # a cell has been visited before
    v = [[0 for i in range(n)] for j in range(n)]
  
    # Queue for bfs
    q = [[0,0]]
  
    # To store the depth of search
    depth = 0
  
    # BFS algorithm
    while (len(q) != 0):
          
        # Current queue size
        x = len(q)
        while (x > 0):
              
            # Top-most element of queue
            y = q[0]
  
            # To store index of cell
            # for simplicity
            i = y[0]
            j = y[1]
            q.remove(q[0])
  
            x -= 1
  
            # Base case
            if (v[i][j]):
                continue
  
            # If we reach (n-1, n-1)
            if (i == n - 1 and j == n - 1):
                return depth
  
            # Marking the cell visited
            v[i][j] = 1
  
            # Pushing the adjacent cells in the
            # queue that can be visited
            # from the current cell
            if (i + arr[i][j] < n):
                q.append([i + arr[i][j], j])
            if (j + arr[i][j] < n):
                q.append([i, j + arr[i][j]])
  
        depth += 1
  
    return -1
  
# Driver code
if __name__ == '__main__':
    arr = [[1, 1, 1],
            [1, 1, 1],
            [1, 1, 1]]
  
    print(minSteps(arr))
  
# This code is contributed by
# Surendra_Gangwar

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Output:

4

Time Complexity: O(n2)



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