Minimum deletions required to make frequency of each letter unique in a String in Java

Minimum deletions to make frequency of each character unique

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This is a popular interview question asked in Tier-1 companies.

Given a string, find the minimum number of characters to be deleted in a string so that the frequency of each character in the string is unique.

Question 1:

String s = 'aaaabbbcccdddd'

Frequency of characters : { a=4, b=3, c=3, d=4 }

a nd d both have same frequency 4, similarly b and c have same frequency 3. so frequency is repeated and we need to make it unique.

we need to delete three character from a, one characters from b to make their frequency different.

s = 'abbcccdddd'

Frequency of characters : { a=1, b=2, c=3, d=4 }

Detailed Description:

we need to make the frequency unique,

Either we need to delete one character from a or d so that we break the same frequency.

so we delete a, now string is "aaabbbcccdddd"

Now the frequency of a is 3, b is 3, c is 3 and d is 4

the frequency of a, b and c is repeating, we need to delete one character from a,

s = 'aabbbcccdddd'

Now the frequency of a is 2, b is 3, c is 3 and d is 4

the frequency of b and c is repeating, we need to delete one character from b, but this will make b count to 2 which will be same as a count, so we will delete 2 characters from b so the frequency will be 1.

s = 'aabcccdddd'

Now the frequency of a is 2, b is 1, c is 3 and d is 4, so is unique now.

Answer is 4 because we delete 4 characters to make the frequency unique.

Java Program to find minimum deletions in a String to make the frequency of each character unique

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package javabypatel;

import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
import java.util.PriorityQueue;

public class MinimumDeletionUsingPriorityQueue {
    public static void main(String[] args) {
        System.out.println(countFrequency("aaaabbbcccdddd"));
    }

    private static int countFrequency(String str) {

        if (str == null || str.length() < 2) {
            return 0;
        }

        //Taking a map to find unique occurrence of each character in a string.
        Map<Character, Integer> frequencyCount = new HashMap<>();
        for (char ch: str.toCharArray()) {
            int count = frequencyCount.getOrDefault(ch, 0);
            frequencyCount.put(ch, count + 1);
        }

        //Taking a PriorityQueue for processing the count of each occurrence of a character
        //Putting the values in reverse order so that higher counts are at head
        PriorityQueue<Integer> priorityQueue = new PriorityQueue<>(Collections.reverseOrder());

        //dumping the frequency of each characters in a Priority queue.
        for (Map.Entry<Character, Integer> entry : frequencyCount.entrySet()) {
            priorityQueue.add(entry.getValue());
        }

        int deletionCount = 0;

        while (!priorityQueue.isEmpty()) {
            int frequency = priorityQueue.poll();

            //If there is no element left after poll, It means we are done, return deletionCount.
            if (priorityQueue.size() == 0 ) {
                return deletionCount;
            }

            //Here we are comparing the polled count with the count at the head. since our PriorityQueue is in reverse order
            //Counts that are same will be grouped together. So idea is to check polled and peek and if they are same it means we
            //need to delete one occurrence of a character to make it unique, but the catch is after deletion, the count which we encountered
            //may also be the occurrence count of other character present in a queue, so we have to repeat this process.
            //number of time we delete the occurrence, we add that in our deletionCount.
            if (frequency == priorityQueue.peek()) {
                //it means we have to delete one character to make the frequency unique
                priorityQueue.add(frequency-1);

                //Number of time we delete the character to make frequency unique, we have to increase the deletionCount.
                deletionCount ++;
            }
        }
        return deletionCount;
    }
}

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Enjoy !!!! 

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