LeetCode #390 — MEDIUM

Elimination Game

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

Example 2

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #390: Elimination Game
class Solution {
    public int lastRemaining(int n) {
        int a1 = 1, an = n, step = 1;
        for (int i = 0, cnt = n; cnt > 1; cnt >>= 1, step <<= 1, ++i) {
            if (i % 2 == 1) {
                an -= step;
                if (cnt % 2 == 1) {
                    a1 += step;
                }
            } else {
                a1 += step;
                if (cnt % 2 == 1) {
                    an -= step;
                }
            }
        }
        return a1;
    }
}

// Accepted solution for LeetCode #390: Elimination Game
func lastRemaining(n int) int {
	a1, an, step := 1, n, 1
	for i, cnt := 0, n; cnt > 1; cnt, step, i = cnt>>1, step<<1, i+1 {
		if i%2 == 1 {
			an -= step
			if cnt%2 == 1 {
				a1 += step
			}
		} else {
			a1 += step
			if cnt%2 == 1 {
				an -= step
			}
		}
	}
	return a1
}

# Accepted solution for LeetCode #390: Elimination Game
class Solution:
    def lastRemaining(self, n: int) -> int:
        a1, an = 1, n
        i, step, cnt = 0, 1, n
        while cnt > 1:
            if i % 2:
                an -= step
                if cnt % 2:
                    a1 += step
            else:
                a1 += step
                if cnt % 2:
                    an -= step
            cnt >>= 1
            step <<= 1
            i += 1
        return a1

// Accepted solution for LeetCode #390: Elimination Game
struct Solution;

impl Solution {
    fn last_remaining(n: i32) -> i32 {
        if n == 1 {
            1
        } else {
            2 * (1 + n / 2 - Self::last_remaining(n / 2))
        }
    }
}

#[test]
fn test() {
    let n = 9;
    let res = 6;
    assert_eq!(Solution::last_remaining(n), res);
}

// Accepted solution for LeetCode #390: Elimination Game
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #390: Elimination Game
// class Solution {
//     public int lastRemaining(int n) {
//         int a1 = 1, an = n, step = 1;
//         for (int i = 0, cnt = n; cnt > 1; cnt >>= 1, step <<= 1, ++i) {
//             if (i % 2 == 1) {
//                 an -= step;
//                 if (cnt % 2 == 1) {
//                     a1 += step;
//                 }
//             } else {
//                 a1 += step;
//                 if (cnt % 2 == 1) {
//                     an -= step;
//                 }
//             }
//         }
//         return a1;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.