LeetCode #3782 — HARD

Last Remaining Integer After Alternating Deletion Operations

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer n.

We write the integers from 1 to n in a sequence from left to right. Then, alternately apply the following two operations until only one integer remains, starting with operation 1:

Operation 1: Starting from the left, delete every second number.
Operation 2: Starting from the right, delete every second number.

Return the last remaining integer.

Example 1:

Input: n = 8

Output: 3

Explanation:

Write [1, 2, 3, 4, 5, 6, 7, 8] in a sequence.
Starting from the left, we delete every second number: [1, 2, 3, 4, 5, 6, 7, 8]. The remaining integers are [1, 3, 5, 7].
Starting from the right, we delete every second number: [1, 3, 5, 7]. The remaining integers are [3, 7].
Starting from the left, we delete every second number: [3, 7]. The remaining integer is [3].

Example 2:

Input: n = 5

Output: 1

Explanation:

Write [1, 2, 3, 4, 5] in a sequence.
Starting from the left, we delete every second number: [1, 2, 3, 4, 5]. The remaining integers are [1, 3, 5].
Starting from the right, we delete every second number: [1, 3, 5]. The remaining integers are [1, 5].
Starting from the left, we delete every second number: [1, 5]. The remaining integer is [1].

Example 3:

Input: n = 1

Output: 1

Explanation:

Write [1] in a sequence.
The last remaining integer is 1.

Constraints:

1 <= n <= 10¹⁵

Step 01

Brute Force Baseline

Problem summary: You are given an integer n. We write the integers from 1 to n in a sequence from left to right. Then, alternately apply the following two operations until only one integer remains, starting with operation 1: Operation 1: Starting from the left, delete every second number. Operation 2: Starting from the right, delete every second number. Return the last remaining integer.

Baseline thinking

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

Pattern signal: Math

Example 1

Example 2

Example 3

Step 02

Core Insight

What unlocks the optimal approach

Use divide and conquer.
Maintain the start value, remaining length, and current direction using recursion.
Track the current direction (left or right) and update the start value based on the parity of the remaining length.

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

Largest constraint values

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 #3782: Last Remaining Integer After Alternating Deletion Operations
// Auto-generated Java example from go.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (go):
// // Accepted solution for LeetCode #3782: Last Remaining Integer After Alternating Deletion Operations
// package main
// 
// // https://space.bilibili.com/206214
// func lastInteger1(n int64) int64 {
// 	start, d := int64(1), int64(1) // 等差数列首项，公差
// 	for ; n > 1; n = (n + 1) / 2 {
// 		start += (n - 2 + n%2) * d
// 		d *= -2
// 	}
// 	return start
// }
// 
// // https://oeis.org/A090569
// func lastInteger(n int64) int64 {
// 	const mask = 0xAAAAAAAAAAAAAAA // ...1010
// 	return (n-1)&mask + 1 // 取出 n-1 的从低到高第 2,4,6,... 位，最后再加一（从 1 开始）
// }

// Accepted solution for LeetCode #3782: Last Remaining Integer After Alternating Deletion Operations
package main

// https://space.bilibili.com/206214
func lastInteger1(n int64) int64 {
	start, d := int64(1), int64(1) // 等差数列首项，公差
	for ; n > 1; n = (n + 1) / 2 {
		start += (n - 2 + n%2) * d
		d *= -2
	}
	return start
}

// https://oeis.org/A090569
func lastInteger(n int64) int64 {
	const mask = 0xAAAAAAAAAAAAAAA // ...1010
	return (n-1)&mask + 1 // 取出 n-1 的从低到高第 2,4,6,... 位，最后再加一（从 1 开始）
}

# Accepted solution for LeetCode #3782: Last Remaining Integer After Alternating Deletion Operations
#
# @lc app=leetcode id=3782 lang=python3
#
# [3782] Last Remaining Integer After Alternating Deletion Operations
#

# @lc code=start
class Solution:
    def lastInteger(self, n: int) -> int:
        head = 1
        step = 1
        remaining = n
        is_left = True
        
        while remaining > 1:
            if is_left:
                # Starting from left, we always keep the first element (head)
                # deleting every second number means we keep indices 0, 2, 4...
                # So head remains unchanged.
                pass
            else:
                # Starting from right
                # If remaining count is odd, the first element survives
                # (e.g., [1, 2, 3] -> keep 3, del 2, keep 1)
                # If remaining count is even, the first element is deleted
                # (e.g., [1, 2, 3, 4] -> keep 4, del 3, keep 2, del 1)
                if remaining % 2 == 0:
                    head += step
            
            step *= 2
            remaining = (remaining + 1) // 2
            is_left = not is_left
            
        return head
# @lc code=end

// Accepted solution for LeetCode #3782: Last Remaining Integer After Alternating Deletion Operations
// Rust example auto-generated from go reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (go):
// // Accepted solution for LeetCode #3782: Last Remaining Integer After Alternating Deletion Operations
// package main
// 
// // https://space.bilibili.com/206214
// func lastInteger1(n int64) int64 {
// 	start, d := int64(1), int64(1) // 等差数列首项，公差
// 	for ; n > 1; n = (n + 1) / 2 {
// 		start += (n - 2 + n%2) * d
// 		d *= -2
// 	}
// 	return start
// }
// 
// // https://oeis.org/A090569
// func lastInteger(n int64) int64 {
// 	const mask = 0xAAAAAAAAAAAAAAA // ...1010
// 	return (n-1)&mask + 1 // 取出 n-1 的从低到高第 2,4,6,... 位，最后再加一（从 1 开始）
// }

// Accepted solution for LeetCode #3782: Last Remaining Integer After Alternating Deletion Operations
// Auto-generated TypeScript example from go.
function exampleSolution(): void {
}
// Reference (go):
// // Accepted solution for LeetCode #3782: Last Remaining Integer After Alternating Deletion Operations
// package main
// 
// // https://space.bilibili.com/206214
// func lastInteger1(n int64) int64 {
// 	start, d := int64(1), int64(1) // 等差数列首项，公差
// 	for ; n > 1; n = (n + 1) / 2 {
// 		start += (n - 2 + n%2) * d
// 		d *= -2
// 	}
// 	return start
// }
// 
// // https://oeis.org/A090569
// func lastInteger(n int64) int64 {
// 	const mask = 0xAAAAAAAAAAAAAAA // ...1010
// 	return (n-1)&mask + 1 // 取出 n-1 的从低到高第 2,4,6,... 位，最后再加一（从 1 开始）
// }

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.