LeetCode 390: Elimination Game Solution

Problem Explanation

Explanation

The problem can be solved by simulating the elimination process. We can keep track of the leftmost element, the current step size, and the direction of traversal. At each step, we update the leftmost element and the step size according to the rules of the elimination process until only one element remains.

Algorithmic Idea

Initialize left to 1, step to 1, remaining to n, and isLeftToRight to true.
While remaining > 1:
- If isLeftToRight is true or remaining is odd, update left to left + step.
- Update remaining to remaining / 2.
- Double the step.
- Toggle the value of isLeftToRight.
Return left.

Time Complexity

The time complexity of this algorithm is O(log n) as we reduce the size of the problem by half at each step.

Space Complexity

The space complexity is O(1) as we only use a constant amount of extra space.

Solution Code

class Solution {
    public int lastRemaining(int n) {
        int left = 1, step = 1, remaining = n;
        boolean isLeftToRight = true;
        
        while (remaining > 1) {
            if (isLeftToRight || remaining % 2 == 1) {
                left += step;
            }
            remaining /= 2;
            step *= 2;
            isLeftToRight = !isLeftToRight;
        }
        
        return left;
    }
}

Frequently Asked Questions

How to solve LeetCode 390 (Elimination Game)?

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What is the time complexity of LeetCode 390 (Elimination Game)?

The time complexity for LeetCode 390 (Elimination Game) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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390. Elimination Game