LeetCode #3842 — EASY

Toggle Light Bulbs

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

Solve on LeetCode

The Problem

Problem Statement

You are given an array bulbs of integers between 1 and 100.

There are 100 light bulbs numbered from 1 to 100. All of them are switched off initially.

For each element bulbs[i] in the array bulbs:

If the bulbs[i]^th light bulb is currently off, switch it on.
Otherwise, switch it off.

Return the list of integers denoting the light bulbs that are on in the end, sorted in ascending order. If no bulb is on, return an empty list.

Example 1:

Input: bulbs = [10,30,20,10]

Output: [20,30]

Explanation:

The bulbs[0] = 10^th light bulb is currently off. We switch it on.
The bulbs[1] = 30^th light bulb is currently off. We switch it on.
The bulbs[2] = 20^th light bulb is currently off. We switch it on.
The bulbs[3] = 10^th light bulb is currently on. We switch it off.
In the end, the 20^th and the 30^th light bulbs are on.

Example 2:

Input: bulbs = [100,100]

Output: []

Explanation:

The bulbs[0] = 100^th light bulb is currently off. We switch it on.
The bulbs[1] = 100^th light bulb is currently on. We switch it off.
In the end, no light bulb is on.

Constraints:

1 <= bulbs.length <= 100
1 <= bulbs[i] <= 100

Step 01

Brute Force Baseline

Problem summary: You are given an array bulbs of integers between 1 and 100. There are 100 light bulbs numbered from 1 to 100. All of them are switched off initially. For each element bulbs[i] in the array bulbs: If the bulbs[i]th light bulb is currently off, switch it on. Otherwise, switch it off. Return the list of integers denoting the light bulbs that are on in the end, sorted in ascending order. If no bulb is on, return an empty list.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

[10,30,20,10]

Example 2

[100,100]

Step 02

Core Insight

What unlocks the optimal approach

Simulate as described

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 #3842: Toggle Light Bulbs
class Solution {
    public List<Integer> toggleLightBulbs(List<Integer> bulbs) {
        int[] st = new int[101];
        for (int x : bulbs) {
            st[x] ^= 1;
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < st.length; ++i) {
            if (st[i] == 1) {
                ans.add(i);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3842: Toggle Light Bulbs
func toggleLightBulbs(bulbs []int) []int {
	st := make([]int, 101)
	for _, x := range bulbs {
		st[x] ^= 1
	}
	ans := make([]int, 0)
	for i := 0; i < 101; i++ {
		if st[i] == 1 {
			ans = append(ans, i)
		}
	}
	return ans
}

# Accepted solution for LeetCode #3842: Toggle Light Bulbs
class Solution:
    def toggleLightBulbs(self, bulbs: list[int]) -> list[int]:
        st = [0] * 101
        for x in bulbs:
            st[x] ^= 1
        return [i for i, x in enumerate(st) if x]

// Accepted solution for LeetCode #3842: Toggle Light Bulbs
fn toggle_light_bulbs(bulbs: Vec<i32>) -> Vec<i32> {
    use std::collections::HashSet;

    let mut s = HashSet::new();
    for b in bulbs {
        if s.contains(&b) {
            s.remove(&b);
        } else {
            s.insert(b);
        }
    }

    let mut ret: Vec<_> = s.into_iter().collect();
    ret.sort_unstable();
    ret
}

fn main() {
    let ret = toggle_light_bulbs(vec![10, 30, 20, 10]);
    println!("ret={ret:?}");
}

#[test]
fn test() {
    assert_eq!(toggle_light_bulbs(vec![10, 30, 20, 10]), [20, 30]);
    assert_eq!(toggle_light_bulbs(vec![100, 100]), []);
}

// Accepted solution for LeetCode #3842: Toggle Light Bulbs
function toggleLightBulbs(bulbs: number[]): number[] {
    const st: number[] = new Array(101).fill(0);
    for (const x of bulbs) {
        st[x] ^= 1;
    }
    const ans: number[] = [];
    for (let i = 0; i < 101; i++) {
        if (st[i] === 1) {
            ans.push(i);
        }
    }
    return ans;
}

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(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.