Minimum deletions from string to reduce it to string with at most 2 unique characters

Given a string $S$ containing lowercase English alphabets. The task is to find the minimum number of characters needed to be removed so that the remaining string contains at most 2 unique characters.

Note: The final string can have duplicate characters. The task is only to reduce the string with minimum deletions such that there can be a maximum of 2 unique characters in the resultant string.

Examples:

Input: S = “geeksforgeeks”
Output: 7
After removing 7 characters the final string will be “geegee”

Input: S = “helloworld”
Output: 5

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: First count the occurrences of each character within the given string then just select two characters with maximum occurrence i.e. the two most frequently occurring characters in the string. And the result will be:

String length – (Occurrence of 1st most frequent character + Occurrence of 2nd most frequent character)

Below is the implementation of the above approach:

Java

// Java implementation of the above approach 
  
public class GFG{ 
  
// Function to find the minimum deletions 
static int check(String s) 
{ 
      
    int i,j; 
    // Array to store the occurrences  
    // of each characters 
    int fr[] = new int[26] ; 
  
    // Length of the string 
    int n = s.length() ; 
    for(i = 0; i < n; i++) 
    { 
          
        // ASCII of the character 
        char x = s.charAt(i) ; 
          
        // Increasing the frequency for this character 
        fr[x-'a'] += 1 ; 
      
    } 
  
  
    int minimum = Integer.MAX_VALUE; 
  
    for(i = 0 ; i < 26; i++) 
    { 
        for( j = i + 1;j < 26; j++) 
        { 
  
            // Choosing two character 
            int z = fr[i] + fr[j] ; 
  
            // Finding the minimum deletion 
            minimum = Math.min(minimum, n-z) ; 
        } 
    } 
  
    return minimum ; 
} 
  
    /* Driver program to test above functions */
     public static void main(String []args){ 
              
        String s ="geeksforgeeks" ; 
        System.out.println(check(s)) ; 
  
        } 
    // This code is contributed by ANKITRAI1 
} 

Python3

# Python3 implementation of the above approach 
  
# Function to find the minimum deletions 
def check(s): 
  
    # Array to store the occurrences  
    # of each characters 
    fr =[0]*26
  
    # Length of the string 
    n = len(s) 
    for i in range(n): 
  
        # ASCII of the character 
        x = ord(s[i]) 
  
        # Increasing the frequency for this character 
        fr[x-97] += 1
  
    minimum = 99999999999
  
    for i in range(26): 
        for j in range(i + 1, 26): 
  
            # Choosing two character 
            z = fr[i] + fr[j] 
  
            # Finding the minimum deletion 
            minimum = min(minimum, n-z) 
  
    return minimum 
  
# Driver code 
s ="geeksforgeeks"
print(check(s)) 

C#

// C# implementation of the above approach 
   
using System; 
public class GFG{ 
   
// Function to find the minimum deletions 
static int check(string s) 
{ 
       
    int i,j; 
    // Array to store the occurrences  
    // of each characters 
    int[] fr = new int[26] ; 
   
    // Length of the string 
    int n = s.Length ; 
    for(i = 0; i < n; i++) 
    { 
           
        // ASCII of the character 
        char x = s[i] ; 
           
        // Increasing the frequency for this character 
        fr[x-'a'] += 1 ; 
       
    } 
   
   
    int minimum = int.MaxValue; 
   
    for(i = 0 ; i < 26; i++) 
    { 
        for( j = i + 1;j < 26; j++) 
        { 
   
            // Choosing two character 
            int z = fr[i] + fr[j] ; 
   
            // Finding the minimum deletion 
            minimum = Math.Min(minimum, n-z) ; 
        } 
    } 
   
    return minimum ; 
} 
   
    /* Driver program to test above functions */
     public static void Main(){ 
               
        string s ="geeksforgeeks" ; 
        Console.Write(check(s)) ; 
   
        } 
} 