**Submajority Vote - Amazon Top Interview Questions**

### Problem Statement :

You are given a list of integers nums where each number represents a vote to a candidate. Return the ids of the candidates that have greater than \lfloor \frac{n}{3}\rfloor⌊ 3 n ⌋ votes, in ascending order. Bonus: solve in \mathcal{O}(1)O(1) space. Constraints n ≤ 100,000 where n is the length of nums. Example 1 Input nums = [2, 1, 5, 5, 5, 5, 6, 6, 6, 6, 6] Output [5, 6] Explanation Both 5 and 6 have 40% of the votes. Example 2 Input nums = [1, 1, 1, 1, 2, 3] Output [1]

### Solution :

` ````
Solution in C++ :
vector<int> solve(vector<int>& nums) {
int a = -1;
int b = -1;
if (nums.size() < 2) {
return nums;
}
int count = 1;
int s = nums.size();
sort(nums.begin(), nums.end());
int i = 0;
for (i = 1; i < nums.size(); i++) {
if (nums[i] == nums[i - 1]) {
count++;
} else {
if (count > (s / 3)) {
if (a == -1) {
a = nums[i - 1];
} else
b = nums[i - 1];
}
count = 1;
}
}
if (count > (s / 3)) {
if (a == -1) {
a = nums[i - 1];
} else
b = nums[i - 1];
}
if (a == -1) {
return {};
} else if (b == -1) {
return {a};
} else
return {a, b};
}
```

` ````
Solution in Python :
class Solution:
def solve(self, A):
a = b = count_a = count_b = 0
for x in A:
if count_a == 0:
a, count_a = x, 1
elif x == a:
count_a += 1
elif count_b == 0:
b, count_b = x, 1
elif x == b:
count_b += 1
else:
count_a -= 1
count_b -= 1
return sorted(x for x in {a, b} if A.count(x) > len(A) // 3)
```

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