# Find the longest common prefix between two strings after performing swaps on second string

Given two strings and . Find the longest common prefix between them after performing zero or more operations on string . In each operation, you can swap any two letters.
Examples

Input : a = "here", b = "there"
Output : 4
The 2nd string can be made "heret" by just
swapping characters and thus the longest
prefix is of length 4.

Input : a = "you", b = "me"
Output : 0

Given that we are only allowed to performs swaps in the string and the length of prefix should be maximized. So the idea is to traverse string and check if the frequency of current character in string is same or less of that in string . If yes then move forward in string an otherwise break and print the length of the part of string an up to which a character is matched in string .
Below is the implementation of the above approach:

## C++

 // C++ program to find the longest // common prefix between two strings // after performing swaps on the second string #include using namespace std;      void LengthLCP(string x, string y) {             int fr[26]={0};            int a = x.length(); // length of x     int b = y.length(); // length of y            for (int i=0 ;i 0){             c += 1;             fr[x[i] - 97] -= 1;         }         else             break;     }     cout<<(c)<

## Java

 // Java program to find the longest // common prefix between two strings // after performing swaps on the second string   public class GFG {       static void LengthLCP(String x, String y)     {                     int fr[]=new int [26];                    int a = x.length(); // length of x         int b = y.length(); // length of y                    for (int i=0 ;i 0){                 c += 1;                 fr[x.charAt(i) - 97] -= 1;             }             else                 break;         }         System.out.println((c)) ;     }         public static void main(String args[])     {         String x="here", y =  "there";                    LengthLCP(x, y);         }     // This code is contributed by ANKITRAI1 }

## Python3

 # Python program to find the longest # common prefix between two strings # after performing swaps on the second string   def LengthLCP(x, y):     fr = [0] * 26           a = len(x) # length of x     b = len(y) # length of y           for i in range(b):         # creating frequency array of         # characters of y         fr[ord(y[i]) - 97] += 1           # storing the length of     # longest common prefix     c = 0           for i in range(a):         # checking if the frequency of the character at         # position i in x in b is greater than zero or not         # if zero we increase the prefix count by 1         if (fr[ord(x[i]) - 97] > 0):             c += 1             fr[ord(x[i]) - 97] -= 1         else:             break     print(c)   # Driver Code   x, y = "here", "there"   LengthLCP(x, y)

## C#

 // C# program to find the longest // common prefix between two strings // after performing swaps on the // second string using System;   class GFG {   static void LengthLCP(String x, String y) {     int []fr = new int [26];           int a = x.Length; // length of x     int b = y.Length; // length of y           for (int i = 0 ; i < b; i++)     {         // creating frequency array         // of characters of y         fr[y[i] - 97] += 1;     }           // storing the length of     // longest common prefix     int c = 0;           for (int i = 0 ; i < a; i++)     {         // checking if the frequency of         // the character at position i         // in x in b is greater than zero         // or not if zero we increase the         // prefix count by 1         if (fr[x[i] - 97] > 0)         {             c += 1;             fr[x[i] - 97] -= 1;         }         else             break;     }     Console.Write((c)) ; }   // Driver Code public static void Main() {     String x = "here", y = "there";           LengthLCP(x, y); } }   // This code is contributed by 29AjayKumar

## PHP

 0)         {             \$c += 1;             \$fr[ord(\$x[\$i]) - 97] -= 1;         }         else             break;     }     echo \$c; }   // Driver Code \$x="here"; \$y = "there";   LengthLCP(\$x, \$y);   return 0;   // This code is contributed by ChitraNayal ?>

## Javascript



Output:

4

Time Complexity: O(length(x) + length(y))

Space Complexity: O(26)

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