# Majority Element II in C++

C++Server Side ProgrammingProgramming

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Suppose we have one integer array; we have to find those elements that appear more than floor of n/3. Here n is the size of array.

So if the input is like [1,1,1,3,3,2,2,2], then the results will be [1,2]

To solve this, we will follow these steps −

• first := 0, second := 1, cnt1 := 0, cnt2 := 0, n := size of array nums

• for i in range 0 to size of n – 1

• x := nums[i]

• if x is first, then increase cnt by 1,

• otherwise when x is second, then increase cnt2 by 1

• otherwise when cnt1 is 0, then set first as x and cnt1 := 1

• otherwise when cnt2 is 0, then set second as x and cnt2 := 1

• otherwise decrease cnt1 and cnt2 by 1

• set cnt1 := 0 and cnt2 := 0

• for i in range 0 to n – 1

• if nums[i] = first, then increase cnt1 by 1, otherwise when nums[i] is second, then increase cnt2 by 1

• make an array called ret

• if cnt1 > n / 3, then insert first into ret

• if cnt2 > n / 3, then insert second into ret

• return ret.

## Example(C++)

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
class Solution {
public:
vector<int> majorityElement(vector<int>& nums) {
int first = 0;
int second = 1;
int cnt1 = 0;
int cnt2 = 0;
int n = nums.size();
for(int i = 0; i < n; i++){
int x = nums[i];
if(x == first){
cnt1++;
}
else if(x == second){
cnt2++;
}
else if(cnt1 == 0){
first = x;
cnt1 = 1;
}
else if(cnt2 == 0){
second = x;
cnt2 = 1;
} else {
cnt1--;
cnt2--;
}
}
cnt1 = 0;
cnt2 = 0;
for(int i = 0; i < n; i++){
if(nums[i] == first)cnt1++;
else if(nums[i] == second)cnt2++;
}
vector <int> ret;
if(cnt1 > n / 3)ret.push_back(first);
if(cnt2 > n / 3)ret.push_back(second);
return ret;
}
};
main(){
Solution ob;
vector<int> v = {1, 1, 1, 3, 3, 2, 2, 2};
print_vector(ob.majorityElement(v));
}

## Input

[1,1,1,3,3,2,2,2]

## Output

[2, 1, ]