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LeetCode 1224: Maximum Equal Frequency Solution

Master LeetCode problem 1224 (Maximum Equal Frequency), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1224. Maximum Equal Frequency

Hard
ArrayHash Table

Problem Explanation

Explanation:

The problem can be solved by iterating through the array and keeping track of the frequencies of each element and their corresponding counts. We need to find the longest prefix such that after removing one element, all remaining elements have the same frequency. This can be achieved by observing the pattern of counts and frequencies for different prefixes.

  1. Iterate through the array and for each element:

    • Update the count of that element in a hashmap.
    • Update the count of frequency in another hashmap.

  2. Check the conditions for a valid prefix:

    • If all elements have the same frequency except one with frequency 1.
    • If all elements except one have frequency one less than the others.
    • If there is only one element in the prefix.

  3. Keep track of the maximum length encountered that satisfies the conditions.

  4. Return the maximum length encountered.

Time Complexity: O(N), where N is the number of elements in the array. Space Complexity: O(N), for storing frequencies and counts.

:

Solution Code

class Solution {
    public int maxEqualFreq(int[] nums) {
        int n = nums.length;
        int maxLen = 0;
        Map<Integer, Integer> countMap = new HashMap<>();
        Map<Integer, Integer> freqMap = new HashMap<>();
        
        for (int i = 0; i < n; i++) {
            int num = nums[i];
            countMap.put(num, countMap.getOrDefault(num, 0) + 1);
            int count = countMap.get(num);
            freqMap.put(count, freqMap.getOrDefault(count, 0) + 1);
            freqMap.put(count - 1, freqMap.getOrDefault(count - 1, 0) - 1);
            
            if (freqMap.get(count) * count == i || (freqMap.get(count - 1) * (count - 1) + 1) == i || count == 1) {
                maxLen = Math.max(maxLen, i + 1);
            }
        }
        
        return maxLen;
    }
}

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