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Minimum removals required such that given string consists only of a pair of alternating characters

  • Difficulty Level : Easy
  • Last Updated : 11 Jun, 2021

Given a string S, the task is to find the minimum removal of characters required such that the string S consists only of two alternating characters.

Examples:

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Input: S = “adebbeeaebd”
Output: 7
Explanation: Removing all occurrences of ‘b’ and ‘e’ modifies the string to “adad”, which consist of alternating occurrences of ‘a’ and ‘d’.



Input: S = “abccd”
Output: 3
Explanation: Removing all occurrences of ‘c’ and ‘d’ modifies the string to “ab”, which consist of alternating ‘a’ and ‘b’.

 

Approach: The problem can be solved by generating all the possible 262 pairs of English letters and find the pair with a maximum length of alternating occurrences, say len, in the string S. Then, print the number of characters required to be removed to achieve that by subtracting len from N.

Follow the steps below to solve the given problem:

  1. Initialize a variable, say len, to store the maximum length of alternating occurrences of a pair of characters.
  2. Iterate over every possible pair of English alphabets and for each pair, perform the following operations:
    • Iterate over the characters of the string S and find the length, say newlen, of alternating occurrences of two characters from the string S.
    • Check if len is smaller than newlen or not. If found to be true, then update len = newlen.
  3. Finally, print N – len as the required answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum
// length of alternating occurrences
// of a pair of characters in a string s
int findLength(string s, char i, char j)
{
    // Stores the next character
    // for alternating sequence
    char required = i;
 
    // Stores the length of alternating
    // occurrences of a pair of characters
    int length = 0;
 
    // Traverse the given string
    for (char curr : s) {
 
        // If current character is same
        // as the required character
        if (curr == required) {
 
            // Increase length by 1
            length += 1;
 
            // Reset required character
            if (required == i)
                required = j;
            else
                required = i;
        }
    }
 
    // Return the length
    return length;
}
 
// Function to find minimum characters
// required to be deleted from S to
// obtain an alternating sequence
int minimumDeletions(string S)
{
 
    // Stores maximum length
    // of alternating sequence
    // of two characters
    int len = 0;
 
    // Stores length of the string
    int n = S.length();
 
    // Generate every pair
    // of English alphabets
    for (char i = 'a'; i <= 'z'; i++) {
        for (char j = i + 1; j <= 'z'; j++) {
 
            // Function call to find length
            // of alternating sequence for
            // current pair of characters
            int newLen = findLength(S, i, j);
 
            // Update len to store the maximum
            // of len and newLen in len
            len = max(len, newLen);
        }
    }
 
    // Return n - len as the final result
    return n - len;
}
 
// Driver Code
int main()
{
    // Given Input
    string S = "adebbeeaebd";
 
    // Function call to find minimum
    // characters required to be removed
    // from S to make it an alternating
    // sequence of a pair of characters
    cout << minimumDeletions(S);
    return 0;
}

Java




// Java program for the above approach
class GFG{
     
// Function to find the maximum
// length of alternating occurrences
// of a pair of characters in a string s
static int findLength(String s, char i, char j)
{
     
    // Stores the next character
    // for alternating sequence
    char required = i;
 
    // Stores the length of alternating
    // occurrences of a pair of characters
    int length = 0;
 
    // Traverse the given string
    for(int k = 0; k < s.length(); k++)
    {
        char curr = s.charAt(k);
         
        // If current character is same
        // as the required character
        if (curr == required)
        {
             
            // Increase length by 1
            length += 1;
 
            // Reset required character
            if (required == i)
                required = j;
            else
                required = i;
        }
    }
 
    // Return the length
    return length;
}
 
// Function to find minimum characters
// required to be deleted from S to
// obtain an alternating sequence
static int minimumDeletions(String S)
{
     
    // Stores maximum length
    // of alternating sequence
    // of two characters
    int len = 0;
 
    // Stores length of the string
    int n = S.length();
 
    // Generate every pair
    // of English alphabets
    for(int i = 0; i < 26; i++)
    {
        for(int j = i + 1; j < 26; j++)
        {
             
            // Function call to find length
            // of alternating sequence for
            // current pair of characters
            int newLen = findLength(S, (char)(i + 97),
                                       (char)(j + 97));
 
            // Update len to store the maximum
            // of len and newLen in len
            len = Math.max(len, newLen);
        }
    }
 
    // Return n - len as the final result
    return n - len;
}
 
// Driver Code
public static void main (String[] args)
{
     
    // Given Input
    String S = "adebbeeaebd";
 
    // Function call to find minimum
    // characters required to be removed
    // from S to make it an alternating
    // sequence of a pair of characters
    System.out.print(minimumDeletions(S));
}
}
 
// This code is contributed by AnkThon

Python3




# Python3 program for the above approach
 
# Function to find the maximum
# length of alternating occurrences
# of a pair of characters in a string s
def findLength(s, i, j):
     
    # Stores the next character
    # for alternating sequence
    required = i
 
    # Stores the length of alternating
    # occurrences of a pair of characters
    length = 0
 
    # Traverse the given string
    for curr in s:
         
        # If current character is same
        # as the required character
        if (curr == required):
             
            # Increase length by 1
            length += 1
 
            # Reset required character
            if (required == i):
                required = j
            else:
                required = i
 
    # Return the length
    return length
 
# Function to find minimum characters
# required to be deleted from S to
# obtain an alternating sequence
def minimumDeletions(S):
     
    # Stores maximum length
    # of alternating sequence
    # of two characters
    len1 = 0
 
    # Stores length of the string
    n = len(S)
 
    # Generate every pair
    # of English alphabets
    for i in range(0, 26, 1):
        for j in range(i + 1, 26, 1):
             
            # Function call to find length
            # of alternating sequence for
            # current pair of characters
            newLen = findLength(S, chr(i + 97),
                                   chr(j + 97))
 
            # Update len to store the maximum
            # of len and newLen in len
            len1 = max(len1, newLen)
 
    # Return n - len as the final result
    return n - len1
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    S = "adebbeeaebd"
 
    # Function call to find minimum
    # characters required to be removed
    # from S to make it an alternating
    # sequence of a pair of characters
    print(minimumDeletions(S))
 
# This code is contributed by ipg2016107

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find the maximum
// length of alternating occurrences
// of a pair of characters in a string s
static int findLength(string s, char i, char j)
{
     
    // Stores the next character
    // for alternating sequence
    char required = i;
 
    // Stores the length of alternating
    // occurrences of a pair of characters
    int length = 0;
 
    // Traverse the given string
    for(int k = 0; k < s.Length; k++)
    {
        char curr = s[k];
         
        // If current character is same
        // as the required character
        if (curr == required)
        {
             
            // Increase length by 1
            length += 1;
 
            // Reset required character
            if (required == i)
                required = j;
            else
                required = i;
        }
    }
 
    // Return the length
    return length;
}
 
// Function to find minimum characters
// required to be deleted from S to
// obtain an alternating sequence
static int minimumDeletions(string S)
{
     
    // Stores maximum length
    // of alternating sequence
    // of two characters
    int len = 0;
 
    // Stores length of the string
    int n = S.Length;
 
    // Generate every pair
    // of English alphabets
    for(int i = 0; i < 26; i++)
    {
        for(int j = i + 1; j < 26; j++)
        {
             
            // Function call to find length
            // of alternating sequence for
            // current pair of characters
            int newLen = findLength(S, (char)(i + 97),
                                       (char)(j + 97));
 
            // Update len to store the maximum
            // of len and newLen in len
            len = Math.Max(len, newLen);
        }
    }
 
    // Return n - len as the final result
    return n - len;
}
 
// Driver Code
public static void Main(string[] args)
{
     
    // Given Input
    string S = "adebbeeaebd";
 
    // Function call to find minimum
    // characters required to be removed
    // from S to make it an alternating
    // sequence of a pair of characters
    Console.WriteLine(minimumDeletions(S));
}
}
 
// This code is contributed by avijitmondal1998

Javascript




<script>
 
// JavaScript program for the above approach
 
     
// Function to find the maximum
// length of alternating occurrences
// of a pair of characters in a string s
function findLength( s,  i,  j)
{
     
    // Stores the next character
    // for alternating sequence
    var required = i;
 
    // Stores the length of alternating
    // occurrences of a pair of characters
    let length = 0;
 
    // Traverse the given string
    for(let k = 0; k < s.length; k++)
    {
        var curr = s.charAt(k);
         
        // If current character is same
        // as the required character
        if (curr == required)
        {
             
            // Increase length by 1
            length += 1;
 
            // Reset required character
            if (required == i)
                required = j;
            else
                required = i;
        }
    }
 
    // Return the length
    return length;
}
 
// Function to find minimum characters
// required to be deleted from S to
// obtain an alternating sequence
function minimumDeletions( S)
{
     
    // Stores maximum length
    // of alternating sequence
    // of two characters
    let len = 0;
 
    // Stores length of the string
    let n = S.length;
 
    // Generate every pair
    // of English alphabets
    for(let i = 0; i < 26; i++)
    {
        for(let j = i + 1; j < 26; j++)
        {
             
            // Function call to find length
            // of alternating sequence for
            // current pair of characters
            let newLen =
            findLength(S, String.fromCharCode(i + 97),
                          String.fromCharCode(j + 97));
 
            // Update len to store the maximum
            // of len and newLen in len
            len = Math.max(len, newLen);
        }
    }
 
    // Return n - len as the final result
    return n - len;
}
 
// Driver Code
 
// Given Input
let S = "adebbeeaebd";
     
// Function call to find minimum
// characters required to be removed
// from S to make it an alternating
// sequence of a pair of characters
document.write(minimumDeletions(S));
 
     
     
</script>
Output: 
7

 

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

 




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