Check If N and Its Double Exist.py

''' https://leetcode.com/explore/featured/card/fun-with-arrays/527/searching-for-items-in-an-array/3250/

Given an array arr of integers, check if there exists two integers N and M such that N is the double of M ( i.e. N = 2 * M).

More formally check if there exists two indices i and j such that :

i != j
0 <= i, j < arr.length
arr[i] == 2 * arr[j]
 

Example 1:

Input: arr = [10,2,5,3]
Output: true
Explanation: N = 10 is the double of M = 5,that is, 10 = 2 * 5.
Example 2:

Input: arr = [7,1,14,11]
Output: true
Explanation: N = 14 is the double of M = 7,that is, 14 = 2 * 7.
Example 3:

Input: arr = [3,1,7,11]
Output: false
Explanation: In this case does not exist N and M, such that N = 2 * M.

'''
from collections import defaultdict
class Solution:
    def checkIfExist(self, arr: List[int]) -> bool:
        dict_ = defaultdict(int)
        for v in arr:
            dict_[v] += 1
        
        if dict_[0] >= 2: # more than two zeros
            return True

        for v in arr:
            if v == 0 or v%2 == 1: # odd number
                continue
            
            if dict_.get(v//2, False):
                return True
        
        return False