How do you solve Keep Multiplying Found Values by Two on LeetCode?

Keep Multiplying Found Values by Two (LeetCode #2154) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a set allows for faster lookups compared to an array.

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

The optimal solution for Keep Multiplying Found Values by Two runs in O(n) time and uses O(n) space. The time complexity is O(n) because we create a set from the array, which takes O(n) time, and each lookup for 'original' is O(1).

What is the brute force solution for Keep Multiplying Found Values by Two?

The brute force approach for Keep Multiplying Found Values by Two is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we repeatedly check if the current value of 'original' exists in the array. If found, we multiply it by two and continue searching until it's no longer found. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Keep Multiplying Found Values by Two?

Keep Multiplying Found Values by Two uses the following concepts: Array, Hash Table, Sorting, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Keep Multiplying Found Values by Two?

Always consider the data structure you are using for lookups. Explain your thought process clearly during the interview. Be prepared to discuss the time and space complexity of your solution.

What are common mistakes when solving Keep Multiplying Found Values by Two?

Not using a set for efficient lookups, leading to unnecessary time complexity. Failing to handle the case where 'original' is not found at all.

#2154

Keep Multiplying Found Values by Two

Easy

ArrayHash TableSortingSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Using a set for quick lookups allows us to efficiently check if 'original' exists in 'nums', reducing the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Convert 'nums' into a set for O(1) average time complexity for lookups.
2Step 2: Initialize a loop that continues while 'original' is in the set.
3Step 3: Multiply 'original' by 2 each time it is found.
4Step 4: Return the final value of 'original'.

solution.py5 lines

1def findFinalValue(nums, original):
2    num_set = set(nums)
3    while original in num_set:
4        original *= 2
5    return original

ℹ

Complexity note: The time complexity is O(n) because we create a set from the array, which takes O(n) time, and each lookup for 'original' is O(1).

1Using a set allows for faster lookups compared to an array.
2The problem can be viewed as a simulation of multiplying values until a condition fails.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.