Cost mínim per fer dues cadenes idèntiques
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#practiceLinkDiv { mostrar: cap !important; } Donades dues cadenes X i Y i dos valors costX i costY. Hem de trobar el cost mínim necessari per fer que les dues cadenes donades siguin idèntiques. Podem eliminar caràcters de les dues cadenes. El cost d'esborrar un caràcter de la cadena X és costX i de Y és costY. El cost d'eliminar tots els caràcters d'una cadena és el mateix.
Exemples:
Input : X = 'abcd' Y = 'acdb' costX = 10 costY = 20. Output: 30 For Making both strings identical we have to delete character 'b' from both the string hence cost will be = 10 + 20 = 30. Input : X = 'ef' Y = 'gh' costX = 10 costY = 20. Output: 60 For making both strings identical we have to delete 2-2 characters from both the strings hence cost will be = 10 + 10 + 20 + 20 = 60.Recommended Practice Cost mínim per fer dues cadenes idèntiques Prova-ho!
Aquest problema és una variació de la subseqüència comuna més llarga (LCS) . La idea és senzilla, primer trobarem la longitud de la subseqüència comuna més llarga de les cadenes X i Y. Ara restar len_LCS amb longituds de cadenes individuals ens dóna el nombre de caràcters que cal eliminar per fer-los idèntics.
// Cost of making two strings identical is SUM of following two // 1) Cost of removing extra characters (other than LCS) // from X[] // 2) Cost of removing extra characters (other than LCS) // from Y[] Minimum Cost to make strings identical = costX * (m - len_LCS) + costY * (n - len_LCS). m ==> Length of string X m ==> Length of string Y len_LCS ==> Length of LCS Of X and Y. costX ==> Cost of removing a character from X[] costY ==> Cost of removing a character from Y[] Note that cost of removing all characters from a string is same.
A continuació es mostra la implementació de la idea anterior.
C++ /* C++ code to find minimum cost to make two strings identical */ #include using namespace std ; /* Returns length of LCS for X[0..m-1] Y[0..n-1] */ int lcs ( char * X char * Y int m int n ) { int L [ m + 1 ][ n + 1 ]; /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for ( int i = 0 ; i <= m ; i ++ ) { for ( int j = 0 ; j <= n ; j ++ ) { if ( i == 0 || j == 0 ) L [ i ][ j ] = 0 ; else if ( X [ i -1 ] == Y [ j -1 ]) L [ i ][ j ] = L [ i -1 ][ j -1 ] + 1 ; else L [ i ][ j ] = max ( L [ i -1 ][ j ] L [ i ][ j -1 ]); } } /* L[m][n] contains length of LCS for X[0..n-1] and Y[0..m-1] */ return L [ m ][ n ]; } // Returns cost of making X[] and Y[] identical. costX is // cost of removing a character from X[] and costY is cost // of removing a character from Y[]/ int findMinCost ( char X [] char Y [] int costX int costY ) { // Find LCS of X[] and Y[] int m = strlen ( X ) n = strlen ( Y ); int len_LCS = lcs ( X Y m n ); // Cost of making two strings identical is SUM of // following two // 1) Cost of removing extra characters // from first string // 2) Cost of removing extra characters from // second string return costX * ( m - len_LCS ) + costY * ( n - len_LCS ); } /* Driver program to test above function */ int main () { char X [] = 'ef' ; char Y [] = 'gh' ; cout < < 'Minimum Cost to make two strings ' < < ' identical is = ' < < findMinCost ( X Y 10 20 ); return 0 ; }
Java // Java code to find minimum cost to // make two strings identical import java.io.* ; class GFG { // Returns length of LCS for X[0..m-1] Y[0..n-1] static int lcs ( String X String Y int m int n ) { int L [][]= new int [ m + 1 ][ n + 1 ] ; /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for ( int i = 0 ; i <= m ; i ++ ) { for ( int j = 0 ; j <= n ; j ++ ) { if ( i == 0 || j == 0 ) L [ i ][ j ] = 0 ; else if ( X . charAt ( i - 1 ) == Y . charAt ( 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 ] ); } } // L[m][n] contains length of LCS // for X[0..n-1] and Y[0..m-1] return L [ m ][ n ] ; } // Returns cost of making X[] and Y[] identical. // costX is cost of removing a character from X[] // and costY is cost of removing a character from Y[]/ static int findMinCost ( String X String Y int costX int costY ) { // Find LCS of X[] and Y[] int m = X . length (); int n = Y . length (); int len_LCS ; len_LCS = lcs ( X Y m n ); // Cost of making two strings identical // is SUM of following two // 1) Cost of removing extra characters // from first string // 2) Cost of removing extra characters // from second string return costX * ( m - len_LCS ) + costY * ( n - len_LCS ); } // Driver code public static void main ( String [] args ) { String X = 'ef' ; String Y = 'gh' ; System . out . println ( 'Minimum Cost to make two strings ' + ' identical is = ' + findMinCost ( X Y 10 20 )); } } // This code is contributed by vt_m
Python3 # Python code to find minimum cost # to make two strings identical # Returns length of LCS for # X[0..m-1] Y[0..n-1] def lcs ( X Y m n ): L = [[ 0 for i in range ( n + 1 )] for i in range ( m + 1 )] # Following steps build # L[m+1][n+1] in bottom # up fashion. Note that # L[i][j] contains length # of LCS of X[0..i-1] and Y[0..j-1] for i in range ( m + 1 ): for j in range ( n + 1 ): if i == 0 or j == 0 : L [ i ][ j ] = 0 else if X [ i - 1 ] == Y [ j - 1 ]: L [ i ][ j ] = L [ i - 1 ][ j - 1 ] + 1 else : L [ i ][ j ] = max ( L [ i - 1 ][ j ] L [ i ][ j - 1 ]) # L[m][n] contains length of # LCS for X[0..n-1] and Y[0..m-1] return L [ m ][ n ] # Returns cost of making X[] # and Y[] identical. costX is # cost of removing a character # from X[] and costY is cost # of removing a character from Y[] def findMinCost ( X Y costX costY ): # Find LCS of X[] and Y[] m = len ( X ) n = len ( Y ) len_LCS = lcs ( X Y m n ) # Cost of making two strings # identical is SUM of following two # 1) Cost of removing extra # characters from first string # 2) Cost of removing extra # characters from second string return ( costX * ( m - len_LCS ) + costY * ( n - len_LCS )) # Driver Code X = 'ef' Y = 'gh' print ( 'Minimum Cost to make two strings ' end = '' ) print ( 'identical is = ' findMinCost ( X Y 10 20 )) # This code is contributed # by sahilshelangia
C# // C# code to find minimum cost to // make two strings identical using System ; class GFG { // Returns length of LCS for X[0..m-1] Y[0..n-1] static int lcs ( String X String Y int m int n ) { int [] L = new int [ m + 1 n + 1 ]; /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for ( int i = 0 ; i <= m ; i ++ ) { for ( int j = 0 ; j <= n ; j ++ ) { if ( i == 0 || j == 0 ) L [ i j ] = 0 ; else if ( X [ i - 1 ] == Y [ 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 ]); } } // L[m][n] contains length of LCS // for X[0..n-1] and Y[0..m-1] return L [ m n ]; } // Returns cost of making X[] and Y[] identical. // costX is cost of removing a character from X[] // and costY is cost of removing a character from Y[] static int findMinCost ( String X String Y int costX int costY ) { // Find LCS of X[] and Y[] int m = X . Length ; int n = Y . Length ; int len_LCS ; len_LCS = lcs ( X Y m n ); // Cost of making two strings identical // is SUM of following two // 1) Cost of removing extra characters // from first string // 2) Cost of removing extra characters // from second string return costX * ( m - len_LCS ) + costY * ( n - len_LCS ); } // Driver code public static void Main () { String X = 'ef' ; String Y = 'gh' ; Console . Write ( 'Minimum Cost to make two strings ' + ' identical is = ' + findMinCost ( X Y 10 20 )); } } // This code is contributed by nitin mittal.
PHP /* PHP code to find minimum cost to make two strings identical */ /* Returns length of LCS for X[0..m-1] Y[0..n-1] */ function lcs ( $X $Y $m $n ) { $L = array_fill ( 0 ( $m + 1 ) array_fill ( 0 ( $n + 1 ) NULL )); /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for ( $i = 0 ; $i <= $m ; $i ++ ) { for ( $j = 0 ; $j <= $n ; $j ++ ) { if ( $i == 0 || $j == 0 ) $L [ $i ][ $j ] = 0 ; else if ( $X [ $i - 1 ] == $Y [ $j - 1 ]) $L [ $i ][ $j ] = $L [ $i - 1 ][ $j - 1 ] + 1 ; else $L [ $i ][ $j ] = max ( $L [ $i - 1 ][ $j ] $L [ $i ][ $j - 1 ]); } } /* L[m][n] contains length of LCS for X[0..n-1] and Y[0..m-1] */ return $L [ $m ][ $n ]; } // Returns cost of making X[] and Y[] identical. costX is // cost of removing a character from X[] and costY is cost // of removing a character from Y[]/ function findMinCost ( & $X & $Y $costX $costY ) { // Find LCS of X[] and Y[] $m = strlen ( $X ); $n = strlen ( $Y ); $len_LCS = lcs ( $X $Y $m $n ); // Cost of making two strings identical is SUM of // following two // 1) Cost of removing extra characters // from first string // 2) Cost of removing extra characters from // second string return $costX * ( $m - $len_LCS ) + $costY * ( $n - $len_LCS ); } /* Driver program to test above function */ $X = 'ef' ; $Y = 'gh' ; echo 'Minimum Cost to make two strings ' . ' identical is = ' . findMinCost ( $X $Y 10 20 ); return 0 ; ?>
JavaScript < script > // Javascript code to find minimum cost to // make two strings identical // Returns length of LCS for X[0..m-1] Y[0..n-1] function lcs ( X Y m n ) { let L = new Array ( m + 1 ); for ( let i = 0 ; i < m + 1 ; i ++ ) { L [ i ] = new Array ( n + 1 ); } for ( let i = 0 ; i < m + 1 ; i ++ ) { for ( let j = 0 ; j < n + 1 ; j ++ ) { L [ i ][ j ] = 0 ; } } /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for ( let i = 0 ; i <= m ; i ++ ) { for ( let j = 0 ; j <= n ; j ++ ) { if ( i == 0 || j == 0 ) L [ i ][ j ] = 0 ; else if ( X [ i - 1 ] == Y [ 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 ]); } } // L[m][n] contains length of LCS // for X[0..n-1] and Y[0..m-1] return L [ m ][ n ]; } // Returns cost of making X[] and Y[] identical. // costX is cost of removing a character from X[] // and costY is cost of removing a character from Y[]/ function findMinCost ( X Y costX costY ) { // Find LCS of X[] and Y[] let m = X . length ; let n = Y . length ; let len_LCS ; len_LCS = lcs ( X Y m n ); // Cost of making two strings identical // is SUM of following two // 1) Cost of removing extra characters // from first string // 2) Cost of removing extra characters // from second string return costX * ( m - len_LCS ) + costY * ( n - len_LCS ); } // Driver code let X = 'ef' ; let Y = 'gh' ; document . write ( 'Minimum Cost to make two strings ' + ' identical is = ' + findMinCost ( X Y 10 20 )); // This code is contributed by avanitrachhadiya2155 < /script>
Sortida
Minimum Cost to make two strings identical is = 60
Complexitat temporal: O(m*n)
Espai auxiliar: O(m*n)
Aquest article és revisat per l'equip de geeksforgeeks.
