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// 1224. Maximum Equal Frequency
#include "leetcode.h"
/*
* 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 ever number has appear in it
* will have the same number occurrences. if after removing one element there
* are no remaining elements, it's still considered that every appear number has
* the same number of occurrences.
*/
typedef struct {
int key;
int cnt;
} data;
int maxEqualFreq(int *nums, int numsSize) {
int n = numsSize, m = n, type = 0, ans = 0, max_freq = 1;
data **num_freq = calloc(m, sizeof(data *));
int *cnt = calloc(n + 1, sizeof(int));
for (int i = 0; i < n; i++) {
int val = nums[i], curr = val;
while (1) {
if (!num_freq[curr % m]) {
num_freq[curr % m] = malloc(sizeof(data));
num_freq[curr % m]->key = val;
num_freq[curr % m]->cnt = 1;
cnt[1]++;
type++;
break;
} else if (num_freq[curr % m]->key == val) {
num_freq[curr % m]->cnt++;
cnt[num_freq[curr % m]->cnt]++;
cnt[num_freq[curr % m]->cnt - 1]--;
if (num_freq[curr % m]->cnt > max_freq)
max_freq = num_freq[curr % m]->cnt;
break;
} else
curr++;
}
if (type == i + 1)
ans = fmax(ans, i + 1);
else if (cnt[1] == 1 && max_freq * cnt[max_freq] == i)
ans = fmax(ans, i + 1);
else if ((max_freq - 1) * (cnt[max_freq - 1] + 1) == i)
ans = fmax(ans, i + 1);
}
for (int i = 0; i < m; i++)
if (num_freq[i])
free(num_freq[i]);
free(num_freq), free(cnt);
return ans;
}
int main() {
int n1[] = {2, 2, 1, 1, 5, 3, 3, 5},
n2[] = {1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5};
printf("%d\n", maxEqualFreq(n1, ARRAY_SIZE(n1))); // expect: 7
printf("%d\n", maxEqualFreq(n2, ARRAY_SIZE(n2))); // expect: 13
}
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