05. Majority Element II

The problem can be found at the following link: Question Link

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++)

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;
    }
};

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