Make string S equal to T after replacing some prefix characters in S

Last Updated : 13 Sep, 2023

Given two strings S and T of equal length consisting of lowercase characters. In one operation, it is allowed to erase the 0^th character of the string S and move it to any place in the string S, the task is to find the minimum number of operations required to convert S equal to T.

Examples:

Input: S = abab, T = abba
Output: 2
Explanation: In 1st operation erase a and move to the end, S = baba, In 2nd operation erase b and move to 2nd position, S = abba. Hence in 2 operations, we can make string S equal to T.

Input: S = arc, T = cra
Output: 2
Explanation: In 1st move erase a and operation to the end, S = rca, In 2nd operation erase r and move to 1st position, S = cra. Hence in 2 operations, we can make string S equal to T.

Approach: This can be solved with the following idea:

Using binary search, look for the mid index and find the same index in the T string. If characters at both positions are equal, update the ans and right values. Repeat this step until the right becomes smaller than the left value.

Below are the steps involved in the implementation of the code:

First for the case where after sorting both strings, if they are not equal then it is not possible to make both strings equal, otherwise, it is always possible to make both strings equal in a maximum of N moves
Now, we can do a binary search on the answer taking the range as left = 1, right = N, where N is the length of string S.
Now for any mid = (left + right)/2, remove the first mid characters and check if the remaining suffix of length (N - mid) length exists as a subsequence in the string T.
If it exists, update right = mid, else left = mid + 1.

Below is the implementation of the above approach:

C++

// C++ code of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the minimum number
// of operation required.
int solve(string s, string t)
{
    int n = s.size();
    string ss = s, tt = t;
    sort(ss.begin(), ss.end()), sort(tt.begin(), tt.end());

    // Check for equality
    if (ss != tt) {
        return -1;
    }
    int ans = n;
    int left = 0, right = n - 1;
    reverse(t.begin(), t.end());
    while (left < right) {

        // Check for mid position
        int mid = (left + right) / 2;
        int cur = n - 1;
        for (auto ch : t) {
            if (cur == mid - 1)
                break;

            // If mid is equal to
            // current character
            if (ch == s[cur]) {
                --cur;
            }
        }

        if (cur == mid - 1) {
            right = mid;
            ans = right;
        }
        else {
            left = mid + 1;
        }
    }
    return ans;
}

// Driver code
int main()
{
    string s = "abab";
    string t = "abba";

    // Function call
    cout << solve(s, t);
    return 0;
}

Java

// Java code of the above approach
import java.util.*;

public class Main 
{

  // Function to find the minimum number
  // of operation required.
  public static int solve(String s, String t)
  {
    int n = s.length();
    String ss = s, tt = t;
    char[] sArray = ss.toCharArray();
    char[] tArray = tt.toCharArray();
    Arrays.sort(sArray);
    Arrays.sort(tArray);
    ss = new String(sArray);
    tt = new String(tArray);

    // Check for equality
    if (!ss.equals(tt)) {
      return -1;
    }
    int ans = n;
    int left = 0, right = n - 1;
    t = new StringBuilder(t).reverse().toString();
    while (left < right) {

      // Check for mid position
      int mid = (left + right) / 2;
      int cur = n - 1;
      for (char ch : t.toCharArray()) {
        if (cur == mid - 1)
          break;

        // If mid is equal to
        // current character
        if (ch == s.charAt(cur)) {
          --cur;
        }
      }

      if (cur == mid - 1) {
        right = mid;
        ans = right;
      }
      else {
        left = mid + 1;
      }
    }
    return ans;
  }

  // Driver code
  public static void main(String[] args)
  {
    String s = "abab";
    String t = "abba";

    // Function call
    System.out.println(solve(s, t));
  }
}

// This code is contributed by Prajwal Kandekar

Python3

# Python code of the above approach

# Function to find the minimum 
# number of operation required
def solve(s, t):
    n = len(s)
    sArray = sorted(s)
    tArray = sorted(t)
    ss = ''.join(sArray)
    tt = ''.join(tArray)
    
    # Check for equality
    if ss != tt:
        return -1

    ans = n
    left = 0
    right = n - 1
    t = t[::-1]

    while left < right:
        # Check for mid position
        mid = (left + right) // 2
        cur = n - 1
        for ch in t:
            if cur == mid - 1:
                break

            # If mid is equal to current character
            if ch == s[cur]:
                cur -= 1

        if cur == mid - 1:
            right = mid
            ans = right
        else:
            left = mid + 1

    return ans

s = "abab"
t = "abba"
# Function call
print(solve(s, t))

# This code is contributed by sankar.

// c# code of the above approach

using System;

public class GFG
{
    // Function to find the minimum number of operations required.
    public static int Solve(string s, string t)
    {
        int n = s.Length;
        string ss = s;
        string tt = t;
        char[] sArray = ss.ToCharArray();
        char[] tArray = tt.ToCharArray();
        Array.Sort(sArray);
        Array.Sort(tArray);
        ss = new string(sArray);
        tt = new string(tArray);

        // Check for equality
        if (ss != tt)
        {
            return -1;
        }

        int ans = n;
        int left = 0;
        int right = n - 1;
        t = new string(ReverseString(t.ToCharArray()));
        while (left < right)
        {
            // Check for mid position
            int mid = (left + right) / 2;
            int cur = n - 1;
            foreach (char ch in t)
            {
                if (cur == mid - 1)
                    break;

                // If mid is equal to current character
                if (ch == s[cur])
                {
                    --cur;
                }
            }

            if (cur == mid - 1)
            {
                right = mid;
                ans = right;
            }
            else
            {
                left = mid + 1;
            }
        }

        return ans;
    }

    // Helper function to reverse a string
    public static char[] ReverseString(char[] s)
    {
        int left = 0;
        int right = s.Length - 1;

        while (left < right)
        {
            char temp = s[left];
            s[left] = s[right];
            s[right] = temp;
            left++;
            right--;
        }

        return s;
    }

    // Driver code
    public static void Main(string[] args)
    {
        string s = "abab";
        string t = "abba";

        // Function call
        Console.WriteLine(Solve(s, t));
    }
}

JavaScript

// Function to find the minimum number
// of operation required.
function solve(s, t) {
    let n = s.length;
    let ss = s, tt = t;
    ss = ss.split('').sort().join('');
    tt = tt.split('').sort().join('');
    
    
    // Check for equality
    if (ss !== tt) {
        return -1;
    }
    let ans = n;
    let left = 0, right = n - 1;
    t = t.split('').reverse().join('');
    while (left < right) {
        
        // Check for mid position
        let mid = Math.floor((left + right) / 2);
        let cur = n - 1;
        for (let ch of t) {
            if (cur === mid - 1)
                break;
            
            // If mid is equal to
            // current character
            if (ch === s[cur]) {
                --cur;
            }
        }
        if (cur === mid - 1) {
            right = mid;
            ans = right;
        }
        else {
            left = mid + 1;
        }
    }
    
    // Return result
    return ans;
}

// Driver code
let s = "abab";
let t = "abba";

console.log(solve(s, t));

Output

Time Complexity: O(N)
Auxiliary Space: O(1)

Check if String S can be compressed to T by replacing some X characters with count X

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Make string S equal to T after replacing some prefix characters in S

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