229. Majority Element II

Problem Statement

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?

Intuition

Basically:

In Moore Voting for n/2 
For every second element we pop
XXXXYYY
For Every X we pop Y

But here there is XYZ for n/3
For every Z we pop X and Y

Hence in last else we pop or reduce Count1 and count2 for a different element

Edge case el1 and el2 can become same hence
we check el2 != it


At end we check count for elements again
1 2 3) 4 5 6) 7 8

In the end 7 and 8 remains but not answer

Links

https://leetcode.com/problems/majority-element-ii/description/

Video Links

https://www.youtube.com/watch?v=Eua-UrQ_ANo&ab_channel=NeetCodeIO https://www.youtube.com/watch?v=vwZj1K0e9U8&t=1535s&ab_channel=takeUforward

Approach 1:

C++

class Solution {
public:
    vector<int> majorityElement(vector<int>& nums) {
        int el1, el2, count1=0, count2=0;
        for(auto &it: nums){
            if(count1 == 0 and el2 != it){
                el1 = it;
                count1=1;
            }
            else if(count2 == 0 and el1 != it){
                el2 = it;
                count2=1;
            }
            else if(el1 == it)
                count1++;
            else if(el2 == it)
                count2++;
            else{
                count1--;   count2--;
            }
        }
        
        vector<int> ans;
        int ct1=0, ct2=0;
        for(auto &it: nums){
            if(el1 == it)
                ct1++;
            if(el2 == it)
                ct2++;
        }

        int idx = nums.size()/3 +1;
        if(ct1 >= idx)
            ans.push_back(el1);
        if(ct2 >= idx)
            ans.push_back(el2);

        return ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++