Minimum deletions from string to reduce it to string with at most 2 unique characters
Given a string 
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
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:
C++
// C++ implementation of the above approach #include<bits/stdc++.h> using namespace std; // Function to find the minimum deletions int check(string s) { int i, j; // Array to store the occurrences // of each characters int fr[26] = {0} ; // Length of the string int n = s.size() ; 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_MAX; 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 = min(minimum, n - z) ; } } return minimum ; } /* Driver code */int main() { string s ="geeksforgeeks" ; cout << check(s) ; } // This code is contributed by ihritik |
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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 } |
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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)) |
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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)) ; } } |
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Output:
7
Time Complexity: O(N), where N is the length of the given String.
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