Is Keep Multiplying Found Values by Two easy or hard?

LeetCode classifies this problem as Easy with an acceptance rate above 75%. The main concept is recognizing that repeated array scans are inefficient and replacing them with constant-time lookups using a hash set.

Keep Multiplying Found Values by Two Python/Java solution

Most implementations build a hash set from the input array, then repeatedly double the value while it exists in the set. This logic is straightforward to implement in Python, Java, JavaScript, C++, and other languages using built-in set or hash set structures.

How to solve Keep Multiplying Found Values by Two in O(n)?

Store all elements from the array in a hash set. While the set contains the current value, multiply the value by two and repeat the lookup. Hash table membership checks run in O(1), so the full algorithm runs in O(n) time.

What is the best approach for Keep Multiplying Found Values by Two?

The hash set approach is the most efficient. Insert all numbers into a set, then repeatedly check if the current value exists and double it. Each lookup is O(1), giving a total time complexity of O(n) and space complexity of O(n).

Is Keep Multiplying Found Values by Two asked at Google/Amazon/Meta?

This problem represents a common interview pattern involving hash tables and repeated lookups. Variants of similar membership-check problems appear in interviews at companies like Amazon and Google where candidates must optimize array searches using hashing.

What data structure is used in Keep Multiplying Found Values by Two?

The primary data structure is a hash set. It allows constant-time membership checks when verifying whether the current value exists in the array. An alternative approach uses a sorted array with binary search.

What is the time complexity of Keep Multiplying Found Values by Two?

The optimal solution runs in O(n) time using a hash set. Building the set takes O(n), and each membership check while doubling the value is constant time. The sorting-based alternative takes O(n log n) due to the initial sort.

Description Hints 4 Solution Video 3:19

2154. Keep Multiplying Found Values by Two

Easy75.1% acceptance4 companies

You are given an array of integers nums. You are also given an integer original which is the first number that needs to be searched for in nums.

You then do the following steps:

If original is found in nums, multiply it by two (i.e., set original = 2 * original).
Otherwise, stop the process.
Repeat this process with the new number as long as you keep finding the number.

Return the final value of original.

Example 1:

Input: nums = [5,3,6,1,12], original = 3
Output: 24
Explanation: 
- 3 is found in nums. 3 is multiplied by 2 to obtain 6.
- 6 is found in nums. 6 is multiplied by 2 to obtain 12.
- 12 is found in nums. 12 is multiplied by 2 to obtain 24.
- 24 is not found in nums. Thus, 24 is returned.

Example 2:

Input: nums = [2,7,9], original = 4
Output: 4
Explanation:
- 4 is not found in nums. Thus, 4 is returned.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i], original <= 1000

Asked at 4 companies

Goldman Sachs

Google Meta Bloomberg

Topics

Hints open one at a time

Hint 1

Repeatedly iterate through the array and check if the current value of original is in the array.

Hint 2

If original is not found, stop and return its current value.

Hint 3

Otherwise, multiply original by 2 and repeat the process.

Hint 4

Use set data structure to check the existence faster.

3:19

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