Минимальные затраты на создание одинаковых двух строк
Опубликовано: 20 Января, 2022
Даны две строки X и Y и два значения costX и costY. Нам нужно найти минимальную стоимость, необходимую для того, чтобы данные две строки были идентичными. Мы можем удалять символы из обеих строк. Стоимость удаления символа из строки X - costX, а из Y - costY. Стоимость удаления всех символов из строки такая же.
Примеры :
Ввод: X = "abcd", Y = "acdb", costX = 10, costY = 20. Выход: 30 Чтобы сделать обе строки идентичными, мы должны удалить символ 'b' из обеих строк, следовательно, стоимость будет быть = 10 + 20 = 30. Ввод: X = "ef", Y = "gh", costX = 10, costY = 20. Выход: 60 Чтобы сделать обе строки идентичными, мы должны удалить 2-2 символов из обеих строк, поэтому стоимость будет = 10 + 10 + 20 + 20 = 60.
Рекомендуется: сначала решите эту проблему на «ПРАКТИКЕ», прежде чем переходить к решению.
Эта проблема является разновидностью самой длинной общей подпоследовательности (LCS). Идея проста: сначала мы находим длину самой длинной общей подпоследовательности строк X и Y. Теперь вычитание len_LCS из длин отдельных строк дает нам количество символов, которые нужно удалить, чтобы сделать их идентичными.
// Стоимость совмещения двух строк равна СУММУ следующих двух // 1) Стоимость удаления лишних символов (кроме LCS) // из X [] // 2) Стоимость удаления лишних символов (кроме LCS) // из Y [] Минимальные затраты на создание одинаковых строк = costX * (m - len_LCS) + costY * (n - len_LCS). m ==> Длина строки X m ==> Длина строки Y len_LCS ==> Длина LCS X и Y. costX ==> Стоимость удаления персонажа из X [] costY ==> Стоимость удаления символа из Y [] Обратите внимание, что стоимость удаления всех символов из строки такой же.
Below is the implementation of above idea.
C++
/* C++ code to find minimum cost to make two strings identical */ #include<bits/stdc++.h> 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 elif 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 /* 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++) { РЕКОМЕНДУЕМЫЕ СТАТЬИ |