Round 1 | HR phone screen (30 Mins):
- Introduction about G-P
- What is G-P and what problem it solves?

Round 2 | HR phone screen (1 hour):
 - Company culture

Round 3 | Coding (1 hour):

Question:
Given an array of elements, where all the elements appear two time except one element.
Find the element that appears only one time.

Case 1: Elements in array can appear in any order.
Solution: Unordered HashMap to maintain frequency count.

Case 2: Same elements are always together.
Solution: Binary search.

#include <bits/stdc++.h>
using namespace std;

int fun(vector<int> arr) {
  int n = arr.size();

  int low = 0, high = n-1;
  while(high-low>2) {
    int mid = low + (high-low)/2;

    if(arr[mid]!=arr[mid-1] && arr[mid]!=arr[mid+1])
      return arr[mid];

    if((mid-low+1)%2==0) {
      if(arr[mid]==arr[mid-1]) {
        low = mid+1;
      }
      else {
        high = mid-1;
      }
    }
    else {
      if(arr[mid]==arr[mid+1]) {
        low = mid+1;
      }
      else {
        high = mid-1;
      }
    }
  }
  if(low+2>=n)  low--;
  return (arr[low]==arr[low+1]) ? arr[low+2] : arr[low];
}

int main() {

  vector<int> arr = {2,2,3,3,4,4,5};
  cout<<fun(arr)<<endl;

  arr = {2,2,4,4,5};
  cout<<fun(arr)<<endl;
  
  arr = {2,2,3,3,4,4,5,7,7,8,8};
  cout<<fun(arr)<<endl;
  
  arr = {1,2,2,3,3,4,4,5,5};
  cout<<fun(arr)<<endl;

  arr = {0,0,1,2,2,3,3,4,4,5,5};
  cout<<fun(arr)<<endl;

  arr = {2,2,6,3,3,4,4,5,5};
  cout<<fun(arr)<<endl;
}

Note: Solution can be simplified but this worked for me.

Round 4 | Managerial round (1 hour):
- About your life journey and why G-P you choose to work at G-P?
- Would you choose G-P over other offers and why?