Minimum number of removals required such that no subsequence of length 2 occurs more than once

Given a string S consisting of N lowercase characters, the task is to modify the given string such that no subsequence of length two repeats in the string by removing minimum number of characters.

Examples:

Input: S = “abcaadbcd”
Output: abcd
Explanation: Removing the characters at indices {2, 3, 4, 5, 6, 7} modifies the string to “abcd”, that contains every subsequence of length 2 exactly once.

Input: S = “cadbcc”
Output: cadbc

Approach: The given problem can be solved based on the observation that the final string can contain only unique characters with the exception that the first character can occur 2 times in the string, one at the start and the other at the end(if possible). Follow the steps below to solve the problem:

• Initialize an empty string, say ans to store the resultant final string.
• Initialize a boolean array C[] of size 26 to check if the character is present in the final string or not.
• Initialize a variable, say pos as 0 to store the index of the last character added to the string, ans.
• Traverse the given string S and if the current character is not present in ans, then append it to ans, mark it as visited in the array C[], and update the value of pos to i.
• Iterate over the range [pos + 1, N – 1] using the variable i and if S[i] is equal to S[0], then append it to the final string ans and break out of the loop.
• After completing the above steps, print the string ans as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to remove the minimum count
// of characters from the string such
// that no subsequence of length 2 repeats
void RemoveCharacters(string s)
{
    // Initialize the final string
    string ans = "";

    // Stores if any character occurs
    // in the final string or not
    bool c[26];

    for (int i = 0; i < 26; i++)
        c[i] = 0;

    // Store the index of the last
    // character added in the string
    int pos = 0;

    // Traverse the string
    for (int i = 0; i < s.size(); i++) {

        // Add all the unique
        // characters
        if (c[s[i] - 'a'] == 0) {
            c[s[i] - 'a'] = 1;
            pos = i;
            ans += s[i];
        }
    }

    // Check if S[0] appears in the
    // range [pos+1, N-1]
    for (int i = pos + 1;
         i < (int)s.size(); i++) {

        // If the characters are the
        // same
        if (s[i] == s[0]) {
            ans += s[i];
            break;
        }
    }

    // Print the resultant string
    cout << ans;
}

// Driver Code
int main()
{
    string S = "abcaadbcd";
    RemoveCharacters(S);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;

class GFG{

// Function to remove the minimum count
// of characters from the string such
// that no subsequence of length 2 repeats
static void RemoveCharacters(String s)
{

    // Initialize the final string
    String ans = "";

    // Stores if any character occurs
    // in the final string or not
    int []c = new int[26];

    for (int i = 0; i < 26; i++)
        c[i] = 0;

    // Store the index of the last
    // character added in the string
    int pos = 0;

    // Traverse the string
    for (int i = 0; i < s.length(); i++) {

        // Add all the unique
        // characters
        if (c[(int)s.charAt(i) - 97] == 0) {
            c[(int)s.charAt(i) - 97] = 1;
            pos = i;
            ans += s.charAt(i);
        }
    }

    // Check if S[0] appears in the
    // range [pos+1, N-1]
    for (int i = pos + 1;
         i < s.length(); i++) {

        // If the characters are the
        // same
        if (s.charAt(i) == s.charAt(0)) {
            ans += s.charAt(i);
            break;
        }
    }

    // Print the resultant string
    System.out.println(ans);
}

// Driver code
public static void main(String[] args)
{
    String S = "abcaadbcd";
    RemoveCharacters(S);
}
}

// This code is contributed by code_hunt.

Python3

# Python 3 program for the above approach

# Function to remove the minimum count
# of characters from the string such
# that no subsequence of length 2 repeats
def RemoveCharacters(s):

    # Initialize the final string
    ans = ""

    # Stores if any character occurs
    # in the final string or not
    c = [0 for i in range(26)]

    # Store the index of the last
    # character added in the string
    pos = 0

    # Traverse the string
    for i in range(len(s)):

        # Add all the unique
        # characters
        if (c[ord(s[i]) - 97] == 0):
            c[ord(s[i]) - 97] = 1
            pos = i
            ans += s[i]

    # Check if S[0] appears in the
    # range [pos+1, N-1]
    for i in range(pos + 1, len(s), 1):

        # If the characters are the
        # same
        if (s[i] == s[0]):
            ans += s[i]
            break

    # Print the resultant string
    print(ans)

# Driver Code
if __name__ == '__main__':
    S = "abcaadbcd"
    RemoveCharacters(S)

    # This code is contributed by ipg2016107.

C#

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

class GFG{

// Function to remove the minimum count
// of characters from the string such
// that no subsequence of length 2 repeats
static void RemoveCharacters(string s)
{

    // Initialize the final string
    string ans = "";

    // Stores if any character occurs
    // in the final string or not
    int []c = new int[26];

    for (int i = 0; i < 26; i++)
        c[i] = 0;

    // Store the index of the last
    // character added in the string
    int pos = 0;

    // Traverse the string
    for (int i = 0; i < s.Length; i++) {

        // Add all the unique
        // characters
        if (c[(int)s[i] - 97] == 0) {
            c[(int)s[i] - 97] = 1;
            pos = i;
            ans += s[i];
        }
    }

    // Check if S[0] appears in the
    // range [pos+1, N-1]
    for (int i = pos + 1;
         i < s.Length; i++) {

        // If the characters are the
        // same
        if (s[i] == s[0]) {
            ans += s[i];
            break;
        }
    }

    // Print the resultant string
    Console.Write(ans);
}

// Driver Code
public static void Main()
{
    string S = "abcaadbcd";
    RemoveCharacters(S);
}
}

// This code is contributed by SURENDRA_GANGWAR.

Javascript

<script>

// javascript program for the above approach

// Function to remove the minimum count
// of characters from the string such
// that no subsequence of length 2 repeats
function RemoveCharacters(s)
{
    // Initialize the final string
    var ans = "";

    // Stores if any character occurs
    // in the final string or not
    var c = new Array(26);
    var i;
    for (i = 0; i < 26; i++)
        c[i] = 0;

    // Store the index of the last
    // character added in the string
    var pos = 0;

    // Traverse the string
    for (i = 0; i < s.length; i++) {

        // Add all the unique
        // characters
        if (c[s[i].charCodeAt() - 97] == 0) {
            c[s[i].charCodeAt() - 97] = 1;
            pos = i;
            ans += s[i];
        }
    }

    // Check if S[0] appears in the
    // range [pos+1, N-1]
    for (i = pos + 1;
         i < s.length; i++) {

        // If the characters are the
        // same
        if (s[i] == s[0]) {
            ans += s[i];
            break;
        }
    }

    // Print the resultant string
    document.write(ans);
}

// Driver Code
    var S = "abcaadbcd";
    RemoveCharacters(S);

// This code is contributed by bgangwar59.
</script>

Output:

abcd

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