number which appears more than n/3 times in an array

Jan Dvorak's answer is probably best:

• Start with two empty candidate slots and two counters set to 0.
• for each item:
  • if it is equal to either candidate, increment the corresponding count
  • else if there is an empty slot (i.e. a slot with count 0), put it in that slot and set the count to 1
  • else reduce both counters by 1

At the end, make a second pass over the array to check whether the candidates really do have the required count. This isn't allowed by the question you link to but I don't see how to avoid it for this modified version. If there is a value that occurs more than n/3 times then it will be in a slot, but you don't know which one it is.

If this modified version of the question guaranteed that there were two values with more than n/3 elements (in general, k-1 values with more than n/k) then we wouldn't need the second pass. But when the original question has k=2 and 1 guaranteed majority there's no way to know whether we "should" generalize it as guaranteeing 1 such element or guaranteeing k-1. The stronger the guarantee, the easier the problem.

Using Boyer-Moore Majority Vote Algorithm, we get:

vector<int> majorityElement(vector<int>& nums) {
    int cnt1=0, cnt2=0;
    int a,b;
    for(int n: A){
      if (n == a) cnt1++;
      else if (n == b) cnt2++;
      else if (cnt1 == 0){
        cnt1++;
        a = n;
      }
      else if (cnt2 == 0){
        cnt2++;
        b = n;
      }
      else{
        cnt1--;
        cnt2--;
      }
    }
    cnt1=cnt2=0;
    for(int n: nums){
        if (n==a) cnt1++;
        else if (n==b) cnt2++;
    }
    vector<int> result;
    if (cnt1 > nums.size()/3) result.push_back(a);
    if (cnt2 > nums.size()/3) result.push_back(b);
    return result;
}

Updated, correction from @Vipul Jain