Count Elements With Maximum Frequency

class Solution {
public:
    int retrieveMaxFrequencyTotal(vector<int>& elements) {
        unordered_map<int, int> elemFreq;
        int highestFreq = 0;
        int frequencySum = 0;

        for (int element : elements) {
            elemFreq[element]++;
            int currentFreq = elemFreq[element];

            if (currentFreq > highestFreq) {
                highestFreq = currentFreq;
                frequencySum = currentFreq;
            } else if (currentFreq == highestFreq) {
                frequencySum += currentFreq;
            }
        }
        return frequencySum;
    }
};

The provided C++ solution focuses on determining the sum of the maximum frequencies of elements in an array. Here’s a step-by-step breakdown:

1. Define a function retrieveMaxFrequencyTotal that accepts a vector of integers elements.
2. Utilize an unordered_map named elemFreq to track the frequency of each element.
3. Initialize two integer variables: highestFreq to track the highest frequency encountered, and frequencySum to store the sum of frequencies at which the highest frequency occurs.
4. Loop through each element in elements:
   • Increment the frequency of the current element in the map.
   • If this frequency is greater than highestFreq, update highestFreq with this new frequency and reset frequencySum to this frequency since it starts counting a new highest frequency.
   • If the frequency of the element matches highestFreq, add this frequency to frequencySum.
5. Return frequencySum which holds the sum of all occurrences of elements having the maximum frequency.

This solution efficiently computes the required sum by using a hashmap to keep count of occurrences, and adjusting the sum and highest frequency on-the-fly as the elements are processed.

public class Solution {
    public int findMostFrequentElements(int[] elements) {
        Map<Integer, Integer> countMap = new HashMap<>();
        int highestFreq = 0;
        int sumOfMaxFreq = 0;

        for (int element : elements) {
            countMap.put(element, countMap.getOrDefault(element, 0) + 1);
            int currentFreq = countMap.get(element);

            if (currentFreq > highestFreq) {
                highestFreq = currentFreq;
                sumOfMaxFreq = currentFreq;
            } else if (currentFreq == highestFreq) {
                sumOfMaxFreq += currentFreq;
            }
        }
        return sumOfMaxFreq;
    }
}

Solution Summary

The problem requires counting the sum of the frequencies of the elements that appear most frequently in an array. The solution is implemented in Java.

• Begin by creating a HashMap called countMap to store each element and its frequency from the input array.
• Initialize two integers, highestFreq to track the highest frequency of any element, and sumOfMaxFreq to keep the sum of the frequencies of the most frequently occurring elements.
• Iterate through the array of elements:
  • Update the frequency of each element in the countMap. If the element isn't already in the map, add it with an initial frequency of one; otherwise, increment its current frequency.
  • Check the frequency of the current element:
    • If this frequency is greater than highestFreq, update highestFreq to this new frequency. Also, reset sumOfMaxFreq to this frequency as the current element is now the most frequent.
    • If this frequency matches highestFreq, increment sumOfMaxFreq by this frequency.
• Finally, return sumOfMaxFreq, which holds the cumulative frequency of elements that are most frequent in the array.

class Solution:
    def highestFrequencySum(self, elements):
        element_counts = {}
        highest_freq = 0
        sum_highest_freq = 0

        for element in elements:
            element_counts[element] = element_counts.get(element, 0) + 1
            count = element_counts[element]

            if count > highest_freq:
                highest_freq = count
                sum_highest_freq = count
            elif count == highest_freq:
                sum_highest_freq += count

        return sum_highest_freq

Count Elements With Maximum Frequency solution is implemented in Python and effectively determines the sum of elements that appear with the highest frequency in a given list called elements.

The solution involves a few key steps:

1. Initialize a dictionary element_counts to store the frequency of each element in the list.
2. Initialize variables highest_freq to zero, which will store the maximum frequency found, and sum_highest_freq to zero, which will store the cumulative sum of the elements that appear with this frequency.
3. Iterate over each element in the provided list, updating its count in the element_counts dictionary.
4. For each element update, check if the current element's count surpasses the highest_freq. If it does, update highest_freq with this new count and reset sum_highest_freq to the element value multiplied by its count.
5. If the current element’s count equals highest_freq, add the element's count to sum_highest_freq.
6. Ultimately, return sum_highest_freq, which will be the sum of the counts of all elements appearing the most frequently in the list.

By following the above steps, you will be able to compute the sum of the frequencies of the elements that are most frequent in the elements list.