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# Majority Element II LeetCode Solution

## Problem – Majority Element II LeetCode Solution

Given an integer array of size `n`, find all elements that appear more than `⌊ n/3 ⌋` times.

Example 1:

``````Input: nums = [3,2,3]
Output: [3]
``````

Example 2:

``````Input: nums = [1]
Output: [1]
``````

Example 3:

``````Input: nums = [1,2]
Output: [1,2]
``````

Constraints:

• `1 <= nums.length <= 5 * 104`
• `-109 <= nums[i] <= 109`

Follow up: Could you solve the problem in linear time and in `O(1)` space?

## Majority Element II LeetCode Solution in Java

``````	public List<Integer> majorityElement(int[] nums) {
if (nums == null || nums.length == 0)
return new ArrayList<Integer>();
List<Integer> result = new ArrayList<Integer>();
int number1 = nums[0], number2 = nums[0], count1 = 0, count2 = 0, len = nums.length;
for (int i = 0; i < len; i++) {
if (nums[i] == number1)
count1++;
else if (nums[i] == number2)
count2++;
else if (count1 == 0) {
number1 = nums[i];
count1 = 1;
} else if (count2 == 0) {
number2 = nums[i];
count2 = 1;
} else {
count1--;
count2--;
}
}
count1 = 0;
count2 = 0;
for (int i = 0; i < len; i++) {
if (nums[i] == number1)
count1++;
else if (nums[i] == number2)
count2++;
}
if (count1 > len / 3)
if (count2 > len / 3)
return result;
}
``````

## Majority Element II LeetCode Solution in C++

``````class Solution {
public:
vector<int> majorityElement(vector<int>& nums)
{
int sz = nums.size();
int num1 = -1, num2 = -1, count1 = 0, count2 = 0, i;
for (i = 0; i < sz; i++)
{
if (nums[i] == num1)
count1++;
else if (nums[i] == num2)
count2++;
else if (count1 == 0)
{
num1 = nums[i];
count1 = 1;
}
else if (count2 == 0)
{
num2 = nums[i];
count2 = 1;
}
else
{
count1--;
count2--;
}
}
vector<int> ans;
count1 = count2 = 0;
for (i = 0; i < sz; i++)
{
if (nums[i] == num1)
count1++;
else if (nums[i] == num2)
count2++;
}
if (count1 > sz/3)
ans.push_back(num1);
if (count2 > sz/3)
ans.push_back(num2);
return ans;
}
};
``````

## Majority Element II LeetCode Solution in Python

``````class Solution:
def majorityElement(self, nums):
count = Counter()
for num in nums:
count[num] += 1
if len(count) == 3:
new_count = Counter()
for elem, freq in count.items():
if freq != 1: new_count[elem] = freq - 1
count = new_count

cands = Counter(num for num in nums if num in count)
return [num for num in cands if cands[num] > len(nums)/3]
``````
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