# 05. Majority Element II

The problem can be found at the following link: [Question Link](https://leetcode.com/problems/majority-element-ii/description/)

## My Approach

1. Initialize variables:
   * `n` to store the size of the input vector `nums`.
   * Create an unordered map `cnt` to store the count of occurrences of each element.
   * Initialize an empty vector `ans` to store the majority elements.
2. Iterate through the `nums` vector using a for loop:
   * Use the `cnt` map to count the occurrences of each element in the `nums` vector.
3. Iterate through the elements in the `cnt` map:
   * Check if the count (`j->second`) of an element is greater than one-third of the total elements (`n/3`).
   * If the count meets the condition, add the element (`j->first`) to the `ans` vector.
4. After both loops have finished, the `ans` vector will contain the majority elements (elements that appear more than one-third of the time) in the input vector `nums`.
5. Return the `ans` vector as the result.

## Time and Auxiliary Space Complexity

* **Time Complexity**: `O(n)`
* **Auxiliary Space Complexity**: `O(n)`

## Code (C++)

```cpp

class Solution {
public:
    vector<int> majorityElement(vector<int>& nums) {  
        int n=nums.size();
        unordered_map<int, int> cnt;
        vector<int> ans;
        for(int i=0;i<n;i++){
            cnt[nums[i]]++;
        }
        for(auto j=cnt.begin();j!=cnt.end();j++){
            if(j->second>n/3)ans.push_back(j->first);
        }
        return ans;
    }
};

```

## Contribution and Support

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