Twice as Much

Easy

The list amounts holds the totals of a day's receipts. Check whether two receipts at **different** positions exist where one is exactly twice the other: amounts[i] == 2 * amounts[j] with i != j. Return True if such a pair exists, and False otherwise.

Example 1

amounts=[7, 3, 12, 6] → True

12 is exactly twice 6.

Example 2

amounts=[5, 9, 4, 11] → False

No number is double another one.

Example 3

amounts=[0, 8, 0] → True

Zero is twice zero, and there are two zeros here at different positions.

Constraints

2 ≤ len(amounts) ≤ 500
−1000 ≤ amounts[i] ≤ 1000

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Topics

Array Hash Table Two Pointers Binary Search Sorting

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Input

amounts=[7, 3, 12, 6]

Expected

True

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