// 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 }