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Check If N and Its Double Exist

Difficulty: Easy

Problem Description

Given an array arr of integers, check if there exist two indices i and j such that :

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

Key Insights

  • The problem requires checking for pairs of elements in the array where one element is double the other.
  • A hash table (or set) can be used to efficiently check for the existence of required values.
  • The solution must ensure that the indices are different when checking for the relationship.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(n)

Solution

To solve the problem, we can use a hash set to store each element of the array as we iterate through it. For each element, we will check if either half (if even) or double of that element exists in the hash set. This allows us to determine if there are two different indices i and j that satisfy the condition arr[i] == 2 * arr[j].

Code Solutions

def checkIfExist(arr):
    seen = set()
    for num in arr:
        if num * 2 in seen or (num % 2 == 0 and num // 2 in seen):
            return True
        seen.add(num)
    return False
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