Arrays: Check If N and Its Double Exist

Problem Statement:

You are given an array, you need to check if a number and its double exist in that array.

Example

Input arr = {10, 2, 5, 3}
Output: True

N = 5 M = 10. 5*2 = 10. So there exist 2 integers, such that the integer and its double exist.

Solution

In this solution, we will use hash map.
At every index, we check for 2 cases:

Case 1: If we find the double of the element in the map, then we return true. This is for odd numbers

Case 2: We divide the current element by 2 and find if the arr[i]/2 element is present then we return true. This is for even numbers.

else we just insert that element into the array.

Solution in C++

#include <algorithm>  
#include <iostream>    
#include <string>
#include <stack>
#include <vector>
#include <unordered_map> 
#include <queue> 

using namespace std;

bool check_if_exist(vector<int>& arr) 
{
    unordered_map<int, int> mp;
    for(int i = 0; i < arr.size(); i++)
    {

        if(mp.find(arr[i] * 2) != mp.end())
            return true;

        if(arr[i] % 2 == 0 && mp.find(arr[i]/2) != mp.end())
            return true;

        mp[arr[i]]++;
    }
    
    return false;
}

int main()
{
  vector<int> arr = {10,2,5,3};

  if(check_if_exist(arr))
  {
    cout<<"There are 2 element exist"<<endl;
  }
  else 
  {
    cout<<"There are no 2 element exist"<<endl;
  }

  return 0;

}

Output:

There are 2 element exist