What is the time complexity of Last Remaining Integer After Alternating Deletion Operations?

The optimal solution for Last Remaining Integer After Alternating Deletion Operations runs in O(n) time and uses O(n) space. The recursive depth can go up to n, leading to linear time complexity, while the space complexity is due to the call stack.

What is the brute force solution for Last Remaining Integer After Alternating Deletion Operations?

The brute force approach for Last Remaining Integer After Alternating Deletion Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach simulates the deletion process step by step. We repeatedly remove every second number from the sequence until only one number remains. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Last Remaining Integer After Alternating Deletion Operations?

Last Remaining Integer After Alternating Deletion Operations uses the following concepts: Math, Recursion. The recommended patterns to study are: Recursion, Divide and Conquer.

What are the interview tips for Last Remaining Integer After Alternating Deletion Operations?

Explain your thought process clearly. Consider edge cases, such as small values of n. Practice similar recursive problems to build intuition.

What are common mistakes when solving Last Remaining Integer After Alternating Deletion Operations?

Not handling the alternating direction properly. Overcomplicating the simulation instead of recognizing patterns.

#3782

Last Remaining Integer After Alternating Deletion Operations

Hard

MathRecursionRecursionDivide and Conquer

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

This approach uses recursion to efficiently determine the last remaining integer without simulating the entire deletion process. It leverages the pattern of remaining integers based on the current direction of deletion.

⚙️

Algorithm

3 steps

1Step 1: Define a recursive function that takes the current range and direction.
2Step 2: If only one number remains, return it.
3Step 3: Depending on the direction, calculate the new range after deleting every second number.

solution.py9 lines

1def last_remaining_optimal(n):
2    def helper(start, length, left):
3        if length == 1:
4            return start
5        if left:
6            return helper(start, (length + 1) // 2, not left)
7        else:
8            return helper(start + (length % 2), length // 2, not left)
9    return helper(1, n, True)

ℹ

Complexity note: The recursive depth can go up to n, leading to linear time complexity, while the space complexity is due to the call stack.

1The direction of deletion alternates, affecting which numbers remain.
2The problem can be solved recursively by tracking the start and length.

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