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

Last Updated : 04 Jun, 2019

Given two strings $a$ and $b$ . Find the longest common prefix between them after performing zero or more operation on string $b$ . 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

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 <bits/stdc++.h> 
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<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; 
    } 
    cout<<(c)<<endl; 
} 
// Driver Code 
int main() 
{ 
string x="here", y =  "there"; 
   
LengthLCP(x, y); 
  
return 0; 
} 
//contributed by Arnab Kundu 

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<b ; i++) 
        { 
            // creating frequency array of 
            // characters of y 
            fr[y.charAt(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.charAt(i) - 97] > 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

<?php  
// PHP program to find the longest 
// common prefix between two strings  
// after performing swaps on the second string 
  
function LengthLCP($x, $y) 
{ 
      
  
    $fr = array_fill(0,26,NULL); 
      
    $a = strlen($x); // length of x 
    $b = strlen($y); // length of y 
      
    for ($i = 0 ;$i < $b ; $i++) 
    { 
        // creating frequency array of 
        // characters of y 
        $fr[ord($y[$i]) - 97] += 1; 
    } 
      
    // storing the length of  
    // longest common prefix 
    $c = 0; 
      
    for ($i = 0 ;$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[ord($x[$i]) - 97] > 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 
?> 

Output:

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP