Related Articles

# 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 ```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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 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

 `// 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 ``=` `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#

 `// 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`

My Personal Notes arrow_drop_up