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:
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:
- Initialize a variable, say len, to store the maximum length of alternating occurrences of a pair of characters.
- 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.
- Finally, print N – len as the required answer.
Below is the implementation of the above approach:
// 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 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 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# 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 |
<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> |
7
Time Complexity: O(N)
Auxiliary Space: O(1)