Majority Element II in C++

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, ]