A Space Optimized Solution of LCS | Java Code for A Space Optimized Solution of LCS ; Returns length of LCS for X [ 0. . m - 1 ] , Y [ 0. . n - 1 ] ; Find lengths of two strings ; Binary index , used to index current row and previous row . ; Compute current binary index ; Last filled entry contains length of LCS for X [ 0. . n - 1 ] and Y [ 0. . m - 1 ] ; Driver Code
class GFG { public static int lcs ( String X , String Y ) { int m = X . length ( ) , n = Y . length ( ) ; int L [ ] [ ] = new int [ 2 ] [ n + 1 ] ; int bi = 0 ; for ( int i = 0 ; i <= m ; i ++ ) { bi = i & 1 ; for ( int j = 0 ; j <= n ; j ++ ) { if ( i == 0 j == 0 ) L [ bi ] [ j ] = 0 ; else if ( X . charAt ( i - 1 ) == Y . charAt ( j - 1 ) ) L [ bi ] [ j ] = L [ 1 - bi ] [ j - 1 ] + 1 ; else L [ bi ] [ j ] = Math . max ( L [ 1 - bi ] [ j ] , L [ bi ] [ j - 1 ] ) ; } } return L [ bi ] [ n ] ; } public static void main ( String [ ] args ) { String X = " AGGTAB " ; String Y = " GXTXAYB " ; System . out . println ( " Length ▁ of ▁ LCS ▁ is ▁ " + lcs ( X , Y ) ) ; } }