LeetCode //C - 1224. Maximum Equal Frequency

1224. Maximum Equal Frequency

Given an array nums of positive integers, return the longest possible length of an array prefix of nums, such that it is possible to remove exactly one element from this prefix so that every number that has appeared in it will have the same number of occurrences.

If after removing one element there are no remaining elements, it's still considered that every appeared number has the same number of ocurrences (0).

Example 1:

Input: nums = 2,2,1,1,5,3,3,5

Output: 7

Explanation: For the subarray 2,2,1,1,5,3,3 of length 7, if we remove nums4 = 5, we will get 2,2,1,1,3,3, so that each number will appear exactly twice.

Example 2:

Input: nums = 1,1,1,2,2,2,3,3,3,4,4,4,5

Output: 13

Constraints:

• 2 < = n u m s . l e n g t h < = 10 5 2 <= nums.length <= 10^5 2<=nums.length<=105
• 1 < = n u m s i < = 10 5 1 <= numsi <= 10^5 1<=numsi<=105

From: LeetCode

Link: 1224. Maximum Equal Frequency

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Solution:

Ideas:

We can solve this in one pass by tracking two things: each value's frequency, and how many values currently have each frequency. The key is checking when the current prefix can be fixed by deleting exactly one number.

Code:

int maxEqualFreq(int* nums, int numsSize) {
    int count[100001] = {0};   // count[x] = frequency of number x
    int freq[100002] = {0};    // freq[f] = how many numbers appear f times

    int ans = 0;
    int maxFreq = 0;

    for (int i = 0; i < numsSize; i++) {
        int x = nums[i];

        if (count[x] > 0) {
            freq[count[x]]--;
        }

        count[x]++;
        freq[count[x]]++;

        if (count[x] > maxFreq) {
            maxFreq = count[x];
        }

        int len = i + 1;

        /*
            Valid cases:

            1. maxFreq == 1
               Every number appears once.
               Remove any one element.

            2. One number appears maxFreq times,
               all others appear maxFreq - 1 times.
               Remove one occurrence from that number.

            3. One number appears once,
               all others appear maxFreq times.
               Remove that single-occurrence number.
        */
        if (maxFreq == 1 ||
            freq[maxFreq] == 1 &&
            freq[maxFreq] * maxFreq + freq[maxFreq - 1] * (maxFreq - 1) == len ||
            freq[1] == 1 &&
            freq[maxFreq] * maxFreq + 1 == len) {
            ans = len;
        }
    }

    return ans;
}