Maximum Equal Frequency LeetCode solution in Java

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import java.util.NavigableMap;
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import java.util.TreeMap;
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class Solution {
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  public int maxEqualFreq(int[] nums) {
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    Map<Integer, Integer> F = new HashMap<>(); // Frequencies
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    NavigableMap<Integer, Integer> V = new TreeMap<>(); // Values for frequencies
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    int max = 0;
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    for (int i = 0; i < nums.length; i++) {
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      increaseCount(F, nums[i]);
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      int frequency = F.get(nums[i]);
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      decreaseCount(V, frequency - 1);
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      increaseCount(V, frequency);
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      if (isPossibleToRemove(V)) max = i;
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    }
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    return max + 1;
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  }
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  public boolean isPossibleToRemove(NavigableMap<Integer, Integer> frequenciesMap) {
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    if (frequenciesMap.size() > 2) return false; // more than 2 different frequencies
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    Map.Entry<Integer, Integer> first = frequenciesMap.firstEntry();
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    Map.Entry<Integer, Integer> last = frequenciesMap.lastEntry();
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    if (frequenciesMap.size() == 1)
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      return first.getKey() == 1 || first.getValue() == 1; // should be [a,a,a,a] or [a,b,c,d]
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    int firstReduced = removeElement(first);
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    int lastReduced = removeElement(last);
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    if (firstReduced > 0 && lastReduced > 0 && first.getKey() != lastReduced) return false;
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    return true;
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  }
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  // Try to remove element which contributes to this frequency:
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  // if there's only 1 occurence of such frequency, the frequency itself will become smaller by 1
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  // if there are more than 1 occurences of such frequency, removing 1 element will not change it
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  public int removeElement(Map.Entry<Integer, Integer> frequencyValue) {
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    if (frequencyValue.getValue() == 1) return frequencyValue.getKey() - 1;
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    return frequencyValue.getKey();
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  }
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  public void decreaseCount(Map<Integer, Integer> map, int element) {
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    if (!map.containsKey(element)) return;
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    map.put(element, map.get(element) - 1);
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    if (map.get(element) == 0) map.remove(element);
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  }
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  public void increaseCount(Map<Integer, Integer> map, int element) {
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    if (!map.containsKey(element)) map.put(element, 0);
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    map.put(element, map.get(element) + 1);
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  }
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}