Count characters to be shifted from the start or end of a string to obtain another string

Last Updated : 18 Jan, 2024

Given two strings A and B where string A is an anagram of string B. In one operation, remove the first or the last character of the string A and insert at any position in A. The task is to find the minimum number of such operations required to be performed to convert string A into string B.

Examples:

Input: A = "edacb", B = "abcde"
Output: 3
Explanation:
Perform the given operations as follows:
Step1: Take last character 'b' and insert it between 'a' and 'c' ( "edacb" becomes "edabc" )
Step2: Take first character 'e' insert it to last ( "edabc" becomes "dabce" )
Step3: Take first character 'd' and insert it between 'c' and 'e' ( "dabce" becomes "abcde" )
Therefore, the count of the operations is 3.

Input: A = "abcd", B = "abdc"
Output: 1
Explanation:
Perform the given operations as follows:
Step1: Take last character 'd' and insert it between 'b' and 'c' ( "abcd" becomes "abdc" )
Therefore, the count of the operations is 1.

Approach: The idea is to find the longest substring of string A which is also a subsequence of string B, and subtract this value from the given length of string to get the minimum count of operation required.

Below is the implementation of the above approach:

C++14

// C++14 program for the above approach
#include <bits/stdc++.h> 
using namespace std; 

// Function to find the minimum cost
// to convert string A to string B
int minCost(string A, string B)
{
    
    // Length of string
    int n = A.size();
 
    int i = 0;
 
    // Initialize maxlen as 0
    int maxlen = 0;
 
    // Traverse the string A
    while (i < n)
    {
        
        // Stores the length of
        // substrings of string A
        int length = 0;
 
        // Traversing string B for
        // each character of A
        for(int j = 0; j < n; ++j) 
        {
            
            // Shift i pointer towards
            // right and increment length,
            // if A[i] equals B[j]
            if (A[i] == B[j]) 
            {
                ++i;
                ++length;
 
                // If traverse till end
                if (i == n)
                break;
            }
        }
        
        // Update maxlen
        maxlen = max(maxlen,
                     length);
    }
    
    // Return minimum cost
    return n - maxlen;
}
 
// Driver Code
int main()
{
    
    // Given two strings A and B
    string A = "edacb";
    string B = "abcde";
 
    // Function call
    cout << minCost(A, B) << endl;
}

// This code is contributed by sanjoy_62

Java

// Java Program for the above approach

import java.util.*;
import java.lang.*;

class GFG {

    // Function to find the minimum cost
    // to convert string A to string B
    static int minCost(String A, String B)
    {
        // Length of string
        int n = A.length();

        int i = 0;

        // Initialize maxlen as 0
        int maxlen = 0;

        // Traverse the string A
        while (i < n) {

            // Stores the length of
            // substrings of string A
            int length = 0;

            // Traversing string B for
            // each character of A
            for (int j = 0; j < n; ++j) {

                // Shift i pointer towards
                // right and increment length,
                // if A[i] equals B[j]
                if (A.charAt(i) == B.charAt(j)) {

                    ++i;
                    ++length;

                    // If traverse till end
                    if (i == n)
                        break;
                }
            }

            // Update maxlen
            maxlen = Math.max(maxlen,
                              length);
        }

        // Return minimum cost
        return n - maxlen;
    }

    // Driver Code
    public static void main(String[] args)
    {
        // Given two strings A and B
        String A = "edacb";
        String B = "abcde";

        // Function call
        System.out.println(minCost(A, B));
    }
}

Python3

# Python3 Program for 
# the above approach
# Function to find the 
# minimum cost to convert 
# string A to string B
def minCost(A, B):
  
    # Length of string
    n = len(A);

    i = 0;

    # Initialize maxlen as 0
    maxlen = 0;

    # Traverse the string A
    while (i < n):

        # Stores the length of
        # substrings of string A
        length = 0;

        # Traversing string B for
        # each character of A
        for j in range(0, n):

            # Shift i pointer towards
            # right and increment length,
            # if A[i] equals B[j]
            if (A[i] == B[j]):

                i+= 1
                length+=1;

                # If traverse till end
                if (i == n):
                    break;

        # Update maxlen
        maxlen = max(maxlen, length);

    # Return minimum cost
    return n - maxlen;

# Driver Code
if __name__ == '__main__':
  
    # Given two strings A and B
    A = "edacb";
    B = "abcde";

    # Function call
    print(minCost(A, B));

# This code is contributed by Rajput-Ji

// C# program for the above approach  
using System;
using System.Collections.Generic; 

class GFG{
    
// Function to find the minimum cost
// to convert string A to string B
static int minCost(string A, string B)
{
    
    // Length of string
    int n = A.Length;

    int i = 0;

    // Initialize maxlen as 0
    int maxlen = 0;

    // Traverse the string A
    while (i < n)
    {
        
        // Stores the length of
        // substrings of string A
        int length = 0;

        // Traversing string B for
        // each character of A
        for(int j = 0; j < n; ++j)
        {
            
            // Shift i pointer towards
            // right and increment length,
            // if A[i] equals B[j]
            if (A[i] == B[j]) 
            {
                ++i;
                ++length;

                // If traverse till end
                if (i == n)
                    break;
            }
        }

        // Update maxlen
        maxlen = Math.Max(maxlen,
                          length);
    }

    // Return minimum cost
    return n - maxlen;
}

// Driver Code
public static void Main()
{
    
    // Given two strings A and B
    string A = "edacb";
    string B = "abcde";

    // Function call
    Console.WriteLine(minCost(A, B));
}
}

// This code is contributed by sanjoy_62

JavaScript

<script>

// Javascript program for the above approach

// Function to find the minimum cost
// to convert string A to string B
function minCost(A, B)
{
    
    // Length of string
    var n = A.length;
 
    var i = 0;
 
    // Initialize maxlen as 0
    var maxlen = 0;
 
    // Traverse the string A
    while (i < n)
    {
        
        // Stores the length of
        // substrings of string A
        var length = 0;
 
        // Traversing string B for
        // each character of A
        for(var j = 0; j < n; ++j) 
        {
            
            // Shift i pointer towards
            // right and increment length,
            // if A[i] equals B[j]
            if (A[i] == B[j]) 
            {
                ++i;
                ++length;
 
                // If traverse till end
                if (i == n)
                break;
            }
        }
        
        // Update maxlen
        maxlen = Math.max(maxlen,
                     length);
    }
    
    // Return minimum cost
    return n - maxlen;
}
 
// Driver Code
// Given two strings A and B
var A = "edacb";
var B = "abcde";

// Function call
document.write(minCost(A, B));

// This code is contributed by itsok.
</script>