1346. Check If N and Its Double Exist

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Array Hash Table Two Pointers Binary Search

Time: O(n)

Space: O(n)

Intuition

Check if any element in the array equals 2 * another element. Use a HashSet.

Algorithm

1Add all elements to a set.
2For each element x: if 2*x in set, return true.

Common Pitfalls

•Edge case: if x=0, need two 0s in the array (2*0=0). Check count of 0s or handle separately.

1346.cs

// Approach: HashSet; for each element check if its double or (if even) its half is already seen.
// Time: O(n) Space: O(n)

public class Solution
{
    public bool CheckIfExist(int[] arr)
    {
        HashSet<int> seen = new HashSet<int>();

        foreach (int a in arr)
        {
            if (seen.Contains(a * 2) || (a % 2 == 0 && seen.Contains(a / 2)))
                return true;
            seen.Add(a);
        }

        return false;
    }
}

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