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

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

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

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++ 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 Code int 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 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

 # Pyton3 implementation of the above approach    INF = 99999 size = 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 Code 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 ]]    x, y = 2, 3    print(findMinimumSteps(mat, x, y, size))    # This code is contributed by Mohit kumar 29

 // 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 }

Output:
2

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