#2154

Easy

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Keep Multiplying Found Values by Two

The "Keep Multiplying Found Values by Two" problem involves repeatedly multiplying a number by two if it is found in an array.

array hash table sorting simulation

Problem Statement

You are given an array of integers nums and an integer original. Start by checking if the original value exists in nums. If found, multiply it by 2 and repeat the search with the new value. Continue this process until the current value is no longer found in the array.

Return the final value of the original number after performing these steps. If at any point the current value is not found in the array, return the current number as the result.

Examples

Example 1

Input: nums = [5,3,6,1,12], original = 3

Output: 24

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

4 is not found in nums. Thus, 4 is returned.

Constraints

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

Solution Approach

Array Scanning

The solution involves iterating through the array and checking if the current number exists. If found, multiply it by two and repeat the check. This process continues until the number is not found in the array.

Hash Lookup Optimization

A hash set can be used to optimize the search operation. By adding all the numbers in the array to the set, you can quickly check for membership in constant time, reducing the time complexity of the search.

Iterative Approach

The key part of this problem is the iterative nature of checking and updating the value. By repeating the process until no further multiplication is possible, the solution handles all edge cases and efficiently computes the result.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this solution depends on how the array is scanned and the number of iterations needed to find and multiply the number. Using a hash set for lookups ensures O(1) time complexity for each search. Space complexity depends on storing the array and hash set, which is O(n).

What Interviewers Usually Probe

Can the candidate optimize array lookups using hash sets?
Is the candidate comfortable with iterative problem-solving techniques?
How well does the candidate handle the trade-off between time and space complexity?

Common Pitfalls or Variants

Common pitfalls

Not using a hash set to optimize the search operation.
Missing edge cases where the original value is never found in the array.
Failing to update the value properly after each multiplication step.

Follow-up variants

What if the array is sorted? Can you optimize the solution further?
What if the problem involves multiple values to check and multiply?
How would the solution change if we needed to track the number of iterations?

FAQ

What is the pattern behind the 'Keep Multiplying Found Values by Two' problem?

This problem follows an array scanning plus hash lookup pattern, where you iteratively check and update the value if found in the array.

How does the hash set help in this problem?

The hash set allows for O(1) time complexity for checking if a number exists in the array, improving the efficiency compared to direct array scanning.

Can this problem be solved without using a hash set?

Yes, but using a hash set significantly reduces the time complexity of searching for elements in the array.

What is the time complexity of the solution?

The time complexity depends on the number of iterations and the use of a hash set for O(1) lookups, making it O(n) in the worst case.

How can GhostInterview help me with this problem?

GhostInterview provides step-by-step guidance to solve this problem efficiently by focusing on the array scanning technique and optimizing with a hash set.

terminal

Solution

Solution 1: Hash Table

We use a hash table $\textit{s}$ to record all the numbers in the array $\textit{nums}$.

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class Solution:
    def findFinalValue(self, nums: List[int], original: int) -> int:
        s = set(nums)
        while original in s:
            original <<= 1
        return original

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