checkIfDoubleExist.py

# Check If N and Its Double Exist
# 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.
 

# Constraints:

# 2 <= arr.length <= 500
# -10^3 <= arr[i] <= 10^3
   # Hide Hint #1  
# Loop from i = 0 to arr.length, maintaining in a hashTable the array elements from [0, i - 1].
   # Hide Hint #2  
# On each step of the loop check if we have seen the element 2 * arr[i] so far or arr[i] / 2 was seen if arr[i] % 2 == 0.

class Solution:
    def checkIfExist(self, arr: List[int]) -> bool:
        for i in arr:
            if i == 0:
                pass
            elif i*2 in arr:
                return True
        
        if arr.count(0) > 1:
            return True
        else:
            return False