Minimum steps required to convert X to Y where a binary matrix represents the possible conversions | Java implementation of the above approach ; dist [ ] [ ] will be the output matrix that will finally have the shortest distances between every pair of numbers ; Initially same as mat ; 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 } ; Pick all numbers as source one by one ; Pick all numbers as destination for the above picked source ; If number k is on the shortest path from i to j , then update the value of dist [ i ] [ j ] ; If no path ; Driver Code
class GFG { static int INF = 99999 ; static int findMinimumSteps ( int mat [ ] [ ] , int x , int y , int n ) { int i , j , k ; int [ ] [ ] dist = new int [ n ] [ n ] ; 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 ; } } for ( k = 0 ; k < n ; k ++ ) { for ( i = 0 ; i < n ; i ++ ) { for ( j = 0 ; j < n ; j ++ ) { if ( dist [ i ] [ k ] + dist [ k ] [ j ] < dist [ i ] [ j ] ) dist [ i ] [ j ] = dist [ i ] [ k ] + dist [ k ] [ j ] ; } } } if ( dist [ x ] [ y ] < INF ) return dist [ x ] [ y ] ; else return - 1 ; } 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 ) ) ; } }