# Minimum steps required to convert X to Y where a binary matrix represents the possible conversions

• Difficulty Level : Medium
• Last Updated : 27 May, 2021

Given a binary matrix of size NxN where 1 denotes that the number i can be converted to j, and 0 denotes it cannot be converted to. Also given are two numbers X(<N)and Y(<N), the task is to find the minimum number of steps required to convert the number X to Y. If there is no such way possible, print -1.
Examples:

Input:
{{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 1, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}
{ 0, 0, 0, 1, 0, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
{ 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}}

X = 2, Y = 3
Output: 2
Convert 2 -> 4 -> 3, which is the minimum way possible.

Input:
{{ 0, 0, 0, 0}
{ 0, 0, 0, 1}
{ 0, 0, 0, 0}
{ 0, 1, 0, 0}}

X = 1, Y = 2
Output: -1

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Approach: This problem is a variant of Floyd-warshall algorithm where there is an edge of weight 1 between i and j i.e. mat[i][j]==1, else they don’t have an edge and we can assign edges as infinity as we do in Floyd-Warshall. Find the solution matrix and return dp[i][j] if it is not infinite. Return -1 if it is infinite which means there is no path possible between them.
Below is the implementation of the above approach:

## C++

 // C++ implementation of the above approach#include using namespace std;#define INF 99999#define size 10 int findMinimumSteps(int mat[size][size], int x, int y, int n){    // dist[][] will be the output matrix that    // will finally have the shortest    // distances between every pair of numbers    int dist[n][n], i, j, k;     // Initially same as mat    for (i = 0; i < n; i++) {        for (j = 0; j < n; j++) {            if (mat[i][j] == 0)                dist[i][j] = INF;            else                dist[i][j] = 1;             if (i == j)                dist[i][j] = 1;        }    }     // Add all numbers one by one to the set    // of intermediate numbers. Before start of    // an iteration, we have shortest distances    // between all pairs of numbers such that the    // shortest distances consider only the numbers    // in set {0, 1, 2, .. k-1} as intermediate numbers.    // After the end of an iteration, vertex no. k is    // added to the set of intermediate numbers and    // the set becomes {0, 1, 2, .. k}    for (k = 0; k < n; k++) {         // Pick all numbers as source one by one        for (i = 0; i < n; i++) {             // Pick all numbers as destination for the            // above picked source            for (j = 0; j < n; j++) {                 // If number k is on the shortest path from                // i to j, then update the value of dist[i][j]                if (dist[i][k] + dist[k][j] < dist[i][j])                    dist[i][j] = dist[i][k] + dist[k][j];            }        }    }     // If no path    if (dist[x][y] < INF)        return dist[x][y];    else        return -1;} // Driver Codeint main(){     int mat[size][size] = { { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                            { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                            { 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 },                            { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 },                            { 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 },                            { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                            { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                            { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                            { 0, 1, 0, 0, 0, 0, 0, 0, 0, 0 },                            { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 } };     int x = 2, y = 3;     cout << findMinimumSteps(mat, x, y, size);}

## Java

 // Java implementation of the above approach class GFG{         static int INF=99999;         static int findMinimumSteps(int mat[][], int x, int y, int n)    {        // dist[][] will be the output matrix that        // will finally have the shortest        // distances between every pair of numbers        int i, j, k;        int [][] dist= new int[n][n];             // Initially same as mat        for (i = 0; i < n; i++) {            for (j = 0; j < n; j++) {                if (mat[i][j] == 0)                    dist[i][j] = INF;                else                    dist[i][j] = 1;                     if (i == j)                    dist[i][j] = 1;            }        }             // Add all numbers one by one to the set        // of intermediate numbers. Before start of        // an iteration, we have shortest distances        // between all pairs of numbers such that the        // shortest distances consider only the numbers        // in set {0, 1, 2, .. k-1} as intermediate numbers.        // After the end of an iteration, vertex no. k is        // added to the set of intermediate numbers and        // the set becomes {0, 1, 2, .. k}        for (k = 0; k < n; k++) {                 // Pick all numbers as source one by one            for (i = 0; i < n; i++) {                     // Pick all numbers as destination for the                // above picked source                for (j = 0; j < n; j++) {                         // If number k is on the shortest path from                    // i to j, then update the value of dist[i][j]                    if (dist[i][k] + dist[k][j] < dist[i][j])                        dist[i][j] = dist[i][k] + dist[k][j];                }            }        }             // If no path        if (dist[x][y] < INF)            return dist[x][y];        else            return -1;    }         // Driver Code    public static void main(String []args)    {             int [][] mat =  { { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 },                        { 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 1, 0, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 } };             int x = 2, y = 3;        int size=mat.length;                 System.out.println( findMinimumSteps(mat, x, y, size));    } }  // This code is contributed by ihritik

## Python3

 # Pyton3 implementation of the above approach INF = 99999size = 10 def findMinimumSteps(mat, x, y, n):     # dist[][] will be the output matrix    # that will finally have the shortest    # distances between every pair of numbers    dist = [[0 for i in range(n)]               for i in range(n)]    i, j, k = 0, 0, 0     # Initially same as mat    for i in range(n):        for j in range(n):            if (mat[i][j] == 0):                dist[i][j] = INF            else:                dist[i][j] = 1             if (i == j):                dist[i][j] = 1             # Add all numbers one by one to the set    # of intermediate numbers. Before start    # of an iteration, we have shortest distances    # between all pairs of numbers such that the    # shortest distances consider only the numbers    # in set {0, 1, 2, .. k-1} as intermediate    # numbers. After the end of an iteration, vertex    # no. k is added to the set of intermediate    # numbers and the set becomes {0, 1, 2, .. k}    for k in range(n):         # Pick all numbers as source one by one        for i in range(n):             # Pick all numbers as destination            # for the above picked source            for j in range(n):                 # If number k is on the shortest path from                # i to j, then update the value of dist[i][j]                if (dist[i][k] + dist[k][j] < dist[i][j]):                    dist[i][j] = dist[i][k] + dist[k][j]     # If no path    if (dist[x][y] < INF):        return dist[x][y]    else:        return -1 # Driver Codemat = [[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ],       [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ],       [ 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 ],       [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 ],       [ 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 ],       [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ],       [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ],       [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ],       [ 0, 1, 0, 0, 0, 0, 0, 0, 0, 0 ],       [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 ]] x, y = 2, 3 print(findMinimumSteps(mat, x, y, size)) # This code is contributed by Mohit kumar 29

## C#

 // C# implementation of the above approach using System;class GFG{         static int INF=99999;         static int findMinimumSteps(int [,]mat, int x, int y, int n)    {        // dist[][] will be the output matrix that        // will finally have the shortest        // distances between every pair of numbers        int i, j, k;        int [,] dist= new int[n,n];             // Initially same as mat        for (i = 0; i < n; i++) {            for (j = 0; j < n; j++) {                if (mat[i,j] == 0)                    dist[i,j] = INF;                else                    dist[i,j] = 1;                     if (i == j)                    dist[i,j] = 1;            }        }             // Add all numbers one by one to the set        // of intermediate numbers. Before start of        // an iteration, we have shortest distances        // between all pairs of numbers such that the        // shortest distances consider only the numbers        // in set {0, 1, 2, .. k-1} as intermediate numbers.        // After the end of an iteration, vertex no. k is        // added to the set of intermediate numbers and        // the set becomes {0, 1, 2, .. k}        for (k = 0; k < n; k++) {                 // Pick all numbers as source one by one            for (i = 0; i < n; i++) {                     // Pick all numbers as destination for the                // above picked source                for (j = 0; j < n; j++) {                         // If number k is on the shortest path from                    // i to j, then update the value of dist[i][j]                    if (dist[i,k] + dist[k,j] < dist[i,j])                        dist[i,j] = dist[i,k] + dist[k,j];                }            }        }             // If no path        if (dist[x,y] < INF)            return dist[x,y];        else            return -1;    }         // Driver Code    public static void Main()    {             int [,] mat = { { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 },                        { 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 1, 0, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 } };             int x = 2, y = 3;        int size = mat.GetLength(0) ;                 Console.WriteLine( findMinimumSteps(mat, x, y, size));    }    // This code is contributed by Ryuga}

## Javascript


Output:
2

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