# Plain, simple, efficient C++ code with Explanatory Comments - No fancy code show-off

• This question is quite similar to Majority element I. Here, the only trick is to realize that there can be at most 2 majority elements which satisfy the given condition. Then applying the same counting algorithm on any two different candidates as follows:

``````vector<int> majorityElement(vector<int>& nums) {
vector<int> res; // Result vector of majority number if they exist. There can be at most 2 majority elements.
if(nums.size() > 0) { // nums must be non-empty
if(nums.size() == 1) res.push_back(nums[0]);
else {
int maj1 = INT_MIN, maj2 = INT_MAX; // initialize majority elements
int ct1 = 0, ct2 = 0; // initialize their counts
for(int i = 0; i < nums.size(); i++) {
if(nums[i] == maj1) ct1++; // If maj1 is found increase its count
else if(nums[i] == maj2) ct2++; // If maj2 is found increase its count
else if(ct1 == 0) { // maj1 is not valid
maj1 = nums[i]; // Assign the current number to maj1
ct1++; // Update the maj1`s count
}
else if(ct2 == 0) { // maj2 is not valid
maj2= nums[i]; // Assign the current number to maj2
ct2++; // Update the maj2`s count
}
else { // The current number is not equal to any of the majority elements candidates
ct1--;
ct2--;
}
}
ct1 = 0; // Initialize the counts of the candidates
ct2 = 0; // because we will count them to check validity of "greater than floor(n/3)" condition
for(int i = 0; i < nums.size(); i++) {
if(nums[i] == maj1) ct1++;
else if(nums[i] == maj2) ct2++;
}
int floorN3 = nums.size()/3; // Calculate this number only once and use it for both counters
if(ct1 > floorN3) res.push_back(maj1);// If maj1 satisfies the condition
if(ct2 > floorN3) res.push_back(maj2);// If maj2 satisfies the condition
}
}
return res;
}
``````

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