/* 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 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 

# 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# 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[ij] = 0;    else if (X[i - 1] == Y[j - 1])  L[ij] = L[i - 1j - 1] + 1;    else  L[ij] = Math.Max(L[i - 1j] L[ij - 1]);  }  }    // L[m][n] contains length of LCS   // for X[0..n-1] and Y[0..m-1]   return L[mn];  }    // 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 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; ?> 

<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(XYcostXcostY)  {  // 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>