Minimum number of deletions and insertions to transform one string into another | Returns length of length common subsequence for str1 [ 0. . m - 1 ] , str2 [ 0. . n - 1 ] ; Following steps build L [ m + 1 ] [ n + 1 ] in bottom up fashion . Note that L [ i ] [ j ] contains length of LCS of str1 [ 0. . i - 1 ] and str2 [ 0. . j - 1 ] ; L [ m ] [ n ] contains length of LCS for X [ 0. . n - 1 ] and Y [ 0. . m - 1 ] ; function to find minimum number of deletions and insertions ; Driver code ; Function Call
function lcs ( str1 , str2 , m , n ) { let L = new Array ( m + 1 ) ; let i , j ; for ( i = 0 ; i <= m ; i ++ ) { L [ i ] = new Array ( n + 1 ) ; for ( j = 0 ; j <= n ; j ++ ) { L [ i ] [ j ] = 0 ; } } for ( i = 0 ; i <= m ; i ++ ) { for ( j = 0 ; j <= n ; j ++ ) { if ( i == 0 j == 0 ) L [ i ] [ j ] = 0 ; else if ( str1 [ i - 1 ] == str2 [ j - 1 ] ) L [ i ] [ j ] = L [ i - 1 ] [ j - 1 ] + 1 ; else L [ i ] [ j ] = Math . max ( L [ i - 1 ] [ j ] , L [ i ] [ j - 1 ] ) ; } } return L [ m ] [ n ] ; } function printMinDelAndInsert ( str1 , str2 ) { let m = str1 . length ; let n = str2 . length ; let len = lcs ( str1 , str2 , m , n ) ; document . write ( " " + " " ) ; document . write ( ( m - len ) + " " ) ; document . write ( " " + " " ) ; document . write ( ( n - len ) + " " ) ; } let str1 = " " ; let str2 = " " ; printMinDelAndInsert ( str1 , str2 ) ;