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// 229. Majority Element II
#include <limits.h>
#include <stdio.h>
#include <stdlib.h>
/*
* given an integer array of size 'n', find all elements that appear more than n
* / 3 times (floored)
*/
int *majorityElement(int *nums, int nums_size, int *return_size) {
int one = INT_MIN, two = INT_MIN, cnt1 = 0, cnt2 = 0, n = nums_size;
for (int i = 0; i < n; i++) {
if (nums[i] == one)
cnt1++;
else if (nums[i] == two)
cnt2++;
else {
if (!cnt1) {
one = nums[i];
cnt1 = 1;
} else if (!cnt2) {
two = nums[i];
cnt2 = 1;
} else {
cnt1--;
cnt2--;
}
}
}
cnt1 = 0;
cnt2 = 0;
for (int i = 0; i < n; i++) {
if (nums[i] == one)
cnt1++;
else if (nums[i] == two)
cnt2++;
}
int *ans;
if (cnt1 > n / 3 && cnt2 > n / 3) {
ans = malloc(2 * sizeof(int));
ans[0] = one;
ans[1] = two;
*return_size = 2;
return ans;
} else if (cnt1 > n / 3) {
ans = malloc(sizeof(int));
ans[0] = one;
*return_size = 2;
return ans;
} else if (cnt1 > n / 3) {
ans = malloc(sizeof(int));
ans[0] = two;
*return_size = 1;
return ans;
} else {
*return_size = 0;
return ans;
}
}
int main() {
int n1[] = {3, 2, 3}, n2[] = {1}, n3[] = {1, 2};
int *me1 = majorityElement(n1, 3, NULL);
int *me2 = majorityElement(n2, 1, NULL);
int *me3 = majorityElement(n3, 2, NULL);
printf("%d\n", me1[0]); // expect: 3
printf("%d\n", me2[0]); // expect: 1
printf("%d", me3[0]); // expect: 1
printf("%d\n", me3[1]); // expect: 2
}
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