390. Elimination Game LeetCode Solution

In this guide, you will get 390. Elimination Game LeetCode Solution with the best time and space complexity. The solution to Elimination Game problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Elimination Game solution in C++
Elimination Game solution in Java
Elimination Game solution in Python
Additional Resources

390. Elimination GameLeetCode Solution image

Problem Statement of Elimination Game

You have a list arr of all integers in the range [1, n] sorted in a strictly increasing order. Apply the following algorithm on arr:

Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.
Repeat the previous step again, but this time from right to left, remove the rightmost number and every other number from the remaining numbers.
Keep repeating the steps again, alternating left to right and right to left, until a single number remains.

Given the integer n, return the last number that remains in arr.

Example 1:

Input: n = 9
Output: 6
Explanation:
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
arr = [2, 4, 6, 8]
arr = [2, 6]
arr = [6]

Example 2:

Input: n = 1
Output: 1

Constraints:

1 <= n <= 109

Complexity Analysis

Time Complexity: O(\log n)
Space Complexity: O(\log n)

390. Elimination Game LeetCode Solution in C++

class Solution {
 public:
  int lastRemaining(int n) {
    return n == 1 ? 1 : 2 * (1 + n / 2 - lastRemaining(n / 2));
  }
};
/* code provided by PROGIEZ */

390. Elimination Game LeetCode Solution in Java

class Solution {
  public int lastRemaining(int n) {
    return n == 1 ? 1 : 2 * (1 + n / 2 - lastRemaining(n / 2));
  }
}
// code provided by PROGIEZ

390. Elimination Game LeetCode Solution in Python

class Solution:
  def lastRemaining(self, n: int) -> int:
    return 1 if n == 1 else 2 * (1 + n // 2 - self.lastRemaining(n // 2))
 # code by PROGIEZ