Minimale kosten om twee snaren identiek te maken
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Uitvoer
#practiceLinkDiv {weergave: geen! belangrijk; } Gegeven twee strings X en Y en twee waarden costX en costY. We moeten de minimale kosten vinden die nodig zijn om de gegeven twee strings identiek te maken. We kunnen tekens uit beide strings verwijderen. De kosten voor het verwijderen van een teken uit string X zijn costX en uit Y zijn costY. De kosten voor het verwijderen van alle tekens uit een string zijn hetzelfde.
Voorbeelden:
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 Minimale kosten om twee snaren identiek te maken Probeer het!
Dit probleem is een variatie op de langste gemeenschappelijke deelreeks (LCS) . Het idee is eenvoudig: we vinden eerst de lengte van de langste gemeenschappelijke deelreeks van strings X en Y. Als we nu len_LCS aftrekken met de lengtes van individuele strings, krijgen we het aantal tekens dat moet worden verwijderd om ze identiek te maken.
// 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.
Hieronder ziet u de implementatie van bovenstaand idee.
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>
Uitvoer
Minimum Cost to make two strings identical is = 60
Tijdcomplexiteit: O(m*n)
Hulpruimte: O(m*n)
Dit artikel is beoordeeld door team geeksforgeeks.
