Find the minimum number of moves needed to move from one cell of matrix to another

Given a N X N matrix (M) filled with 1 , 0 , 2 , 3 . Find the minimum numbers of moves needed to move from source to destination (sink) . while traversing through blank cells only. You can traverse up, down, right and left.
A value of cell 1 means Source.
A value of cell 2 means Destination.
A value of cell 3 means Blank cell.
A value of cell 0 means Blank Wall.
Note : there is only single source and single destination.they may be more than one path from source to destination(sink).each move in matrix we consider as ‘1’

Examples:

Input : M[3][3] = {{ 0 , 3 , 2 },
                   { 3 , 3 , 0 },
                   { 1 , 3 , 0 }};
Output : 4 

Input : M[4][4] = {{ 3 , 3 , 1 , 0 },
                   { 3 , 0 , 3 , 3 },
                   { 2 , 3 , 0 , 3 },
                   { 0 , 3 , 3 , 3 }};
Output : 4

Asked in: Adobe Interview
.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to use Level graph ( Breadth First Traversal ). Consider each cell as a node and each boundary between any two adjacent cells be an edge . so total number of Node is N*N.

 1. Create an empty Graph having N*N node ( Vertex ).
 2. Push all node into graph.
 3. Note down source and sink vertices.
 4. Now Apply  level graph concept ( that we achieve using BFS  ) .
    In which we find level of every node from source vertex.
    After that we return  'Level[d]' ( d is destination ).
    (which is the minimum move from source to sink )

Below is implementation of above idea.

C++

// C++ program to find the minimum numbers 
// of moves needed to move from source to 
// destination . 
#include<bits/stdc++.h> 
using namespace std; 
#define N 4 
  
class Graph 
{ 
    int V ; 
    list < int > *adj; 
public : 
    Graph( int V ) 
    { 
        this->V = V ; 
        adj = new list<int>[V]; 
    } 
    void addEdge( int s , int d ) ; 
    int BFS ( int s , int d) ; 
}; 
  
// add edge to graph 
void Graph :: addEdge ( int s , int d ) 
{ 
    adj[s].push_back(d); 
    adj[d].push_back(s); 
} 
  
// Level  BFS function to find minimum path 
// from source to sink 
int Graph :: BFS(int s, int d) 
{ 
    // Base case 
    if (s == d) 
        return 0; 
  
    // make initial distance of all vertex -1 
    // from source 
    int *level = new int[V]; 
    for (int i = 0; i < V; i++) 
        level[i] = -1  ; 
  
    // Create a queue for BFS 
    list<int> queue; 
  
    // Mark the source node level[s] = '0' 
    level[s] = 0 ; 
    queue.push_back(s); 
  
    // it will be used to get all adjacent 
    // vertices of a vertex 
    list<int>::iterator i; 
  
    while (!queue.empty()) 
    { 
        // Dequeue a vertex from queue 
        s = queue.front(); 
        queue.pop_front(); 
  
        // Get all adjacent vertices of the 
        // dequeued vertex s. If a adjacent has 
        // not been visited ( level[i] < '0') , 
        // then update level[i] == parent_level[s] + 1 
        // and enqueue it 
        for (i = adj[s].begin(); i != adj[s].end(); ++i) 
        { 
            // Else, continue to do BFS 
            if (level[*i] < 0 || level[*i] > level[s] + 1 ) 
            { 
                level[*i] = level[s] + 1 ; 
                queue.push_back(*i); 
            } 
        } 
  
    } 
  
    // return minimum moves from source to sink 
    return level[d] ; 
} 
  
bool isSafe(int i, int j, int M[][N]) 
{ 
    if ((i < 0 || i >= N) || 
            (j < 0 || j >= N ) || M[i][j] == 0) 
        return false; 
    return true; 
} 
  
// Returns minimum numbers of  moves  from a source (a 
// cell with value 1) to a destination (a cell with 
// value 2) 
int MinimumPath(int M[][N]) 
{ 
    int s , d ; // source and destination 
    int V = N*N+2; 
    Graph g(V); 
  
    // create graph with n*n node 
    // each cell consider as node 
    int k = 1 ; // Number of current vertex 
    for (int i =0 ; i < N ; i++) 
    { 
        for (int j = 0 ; j < N; j++) 
        { 
            if (M[i][j] != 0) 
            { 
                // connect all 4 adjacent cell to 
                // current cell 
                if ( isSafe ( i , j+1 , M ) ) 
                    g.addEdge ( k , k+1 ); 
                if ( isSafe ( i , j-1 , M ) ) 
                    g.addEdge ( k , k-1 ); 
                if (j< N-1 && isSafe ( i+1 , j , M ) ) 
                    g.addEdge ( k , k+N ); 
                if ( i > 0 && isSafe ( i-1 , j , M ) ) 
                    g.addEdge ( k , k-N ); 
            } 
  
            // source index 
            if( M[i][j] == 1 ) 
                s = k ; 
  
            // destination index 
            if (M[i][j] == 2) 
                d = k; 
            k++; 
        } 
    } 
  
    // find minimum moves 
    return g.BFS (s, d) ; 
} 
  
// driver program to check above function 
int main() 
{ 
    int M[N][N] = {{ 3 , 3 , 1 , 0 }, 
        { 3 , 0 , 3 , 3 }, 
        { 2 , 3 , 0 , 3 }, 
        { 0 , 3 , 3 , 3 } 
    }; 
  
    cout << MinimumPath(M) << endl; 
  
    return 0; 
} 

Python3

# Python3 program to find the minimum numbers  
# of moves needed to move from source to  
# destination .  
  
class Graph: 
    def __init__(self, V): 
        self.V = V  
        self.adj = [[] for i in range(V)] 
  
    # add edge to graph  
    def addEdge (self, s , d ): 
        self.adj[s].append(d)  
        self.adj[d].append(s) 
      
    # Level BFS function to find minimum  
    # path from source to sink  
    def BFS(self, s, d): 
          
        # Base case  
        if (s == d):  
            return 0
      
        # make initial distance of all  
        # vertex -1 from source  
        level = [-1] * self.V 
      
        # Create a queue for BFS  
        queue = [] 
      
        # Mark the source node level[s] = '0'  
        level[s] = 0
        queue.append(s)  
      
        # it will be used to get all adjacent  
        # vertices of a vertex  
      
        while (len(queue) != 0): 
              
            # Dequeue a vertex from queue  
            s = queue.pop()  
      
            # Get all adjacent vertices of the  
            # dequeued vertex s. If a adjacent has  
            # not been visited ( level[i] < '0') ,  
            # then update level[i] == parent_level[s] + 1  
            # and enqueue it  
            i = 0
            while i < len(self.adj[s]): 
                  
                # Else, continue to do BFS  
                if (level[self.adj[s][i]] < 0 or 
                    level[self.adj[s][i]] > level[s] + 1 ): 
                    level[self.adj[s][i]] = level[s] + 1
                    queue.append(self.adj[s][i]) 
                i += 1
      
        # return minimum moves from source 
        # to sink  
        return level[d] 
  
def isSafe(i, j, M): 
    global N 
    if ((i < 0 or i >= N) or
        (j < 0 or j >= N ) or M[i][j] == 0):  
        return False
    return True
  
# Returns minimum numbers of moves from a  
# source (a cell with value 1) to a destination  
# (a cell with value 2)  
def MinimumPath(M): 
    global N 
    s , d = None, None # source and destination  
    V = N * N + 2
    g = Graph(V)  
  
    # create graph with n*n node  
    # each cell consider as node  
    k = 1 # Number of current vertex 
    for i in range(N): 
        for j in range(N): 
            if (M[i][j] != 0): 
                  
                # connect all 4 adjacent cell to  
                # current cell  
                if (isSafe (i , j + 1 , M)):  
                    g.addEdge (k , k + 1) 
                if (isSafe (i , j - 1 , M)): 
                    g.addEdge (k , k - 1)  
                if (j < N - 1 and isSafe (i + 1 , j , M)):  
                    g.addEdge (k , k + N)  
                if (i > 0 and isSafe (i - 1 , j , M)):  
                    g.addEdge (k , k - N) 
  
            # source index  
            if(M[i][j] == 1): 
                s = k  
  
            # destination index  
            if (M[i][j] == 2):  
                d = k  
            k += 1
  
    # find minimum moves  
    return g.BFS (s, d) 
  
# Driver Code 
N = 4
M = [[3 , 3 , 1 , 0 ], [3 , 0 , 3 , 3 ],  
     [2 , 3 , 0 , 3 ], [0 , 3 , 3 , 3]] 
  
print(MinimumPath(M)) 
  
# This code is contributed by PranchalK 

Output:

This article is contributed by Nishant Singh . 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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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Python3

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