LeetCode #672 — MEDIUM

Bulb Switcher II

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

The Problem

Problem Statement

There is a room with n bulbs labeled from 1 to n that all are turned on initially, and four buttons on the wall. Each of the four buttons has a different functionality where:

Button 1: Flips the status of all the bulbs.
Button 2: Flips the status of all the bulbs with even labels (i.e., 2, 4, ...).
Button 3: Flips the status of all the bulbs with odd labels (i.e., 1, 3, ...).
Button 4: Flips the status of all the bulbs with a label j = 3k + 1 where k = 0, 1, 2, ... (i.e., 1, 4, 7, 10, ...).

You must make exactly presses button presses in total. For each press, you may pick any of the four buttons to press.

Given the two integers n and presses, return the number of different possible statuses after performing all presses button presses.

Example 1:

Input: n = 1, presses = 1
Output: 2
Explanation: Status can be:
- [off] by pressing button 1
- [on] by pressing button 2

Example 2:

Input: n = 2, presses = 1
Output: 3
Explanation: Status can be:
- [off, off] by pressing button 1
- [on, off] by pressing button 2
- [off, on] by pressing button 3

Example 3:

Input: n = 3, presses = 1
Output: 4
Explanation: Status can be:
- [off, off, off] by pressing button 1
- [on, off, on] by pressing button 2
- [off, on, off] by pressing button 3
- [off, on, on] by pressing button 4

Constraints:

1 <= n <= 1000
0 <= presses <= 1000

Step 01

Brute Force Baseline

Problem summary: There is a room with n bulbs labeled from 1 to n that all are turned on initially, and four buttons on the wall. Each of the four buttons has a different functionality where: Button 1: Flips the status of all the bulbs. Button 2: Flips the status of all the bulbs with even labels (i.e., 2, 4, ...). Button 3: Flips the status of all the bulbs with odd labels (i.e., 1, 3, ...). Button 4: Flips the status of all the bulbs with a label j = 3k + 1 where k = 0, 1, 2, ... (i.e., 1, 4, 7, 10, ...). You must make exactly presses button presses in total. For each press, you may pick any of the four buttons to press. Given the two integers n and presses, return the number of different possible statuses after performing all presses button presses.

Baseline thinking

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

Pattern signal: Math · Bit Manipulation

Example 1

1
1

Example 2

2
1

Example 3

3
1

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 #672: Bulb Switcher II
class Solution {
    public int flipLights(int n, int presses) {
        int[] ops = new int[] {0b111111, 0b010101, 0b101010, 0b100100};
        Set<Integer> vis = new HashSet<>();
        n = Math.min(n, 6);
        for (int mask = 0; mask < 1 << 4; ++mask) {
            int cnt = Integer.bitCount(mask);
            if (cnt <= presses && cnt % 2 == presses % 2) {
                int t = 0;
                for (int i = 0; i < 4; ++i) {
                    if (((mask >> i) & 1) == 1) {
                        t ^= ops[i];
                    }
                }
                t &= ((1 << 6) - 1);
                t >>= (6 - n);
                vis.add(t);
            }
        }
        return vis.size();
    }
}

// Accepted solution for LeetCode #672: Bulb Switcher II
func flipLights(n int, presses int) int {
	if n > 6 {
		n = 6
	}
	ops := []int{0b111111, 0b010101, 0b101010, 0b100100}
	vis := map[int]bool{}
	for mask := 0; mask < 1<<4; mask++ {
		cnt := bits.OnesCount(uint(mask))
		if cnt <= presses && cnt%2 == presses%2 {
			t := 0
			for i, op := range ops {
				if mask>>i&1 == 1 {
					t ^= op
				}
			}
			t &= 1<<6 - 1
			t >>= (6 - n)
			vis[t] = true
		}
	}
	return len(vis)
}

# Accepted solution for LeetCode #672: Bulb Switcher II
class Solution:
    def flipLights(self, n: int, presses: int) -> int:
        ops = (0b111111, 0b010101, 0b101010, 0b100100)
        n = min(n, 6)
        vis = set()
        for mask in range(1 << 4):
            cnt = mask.bit_count()
            if cnt <= presses and cnt % 2 == presses % 2:
                t = 0
                for i, op in enumerate(ops):
                    if (mask >> i) & 1:
                        t ^= op
                t &= (1 << 6) - 1
                t >>= 6 - n
                vis.add(t)
        return len(vis)

// Accepted solution for LeetCode #672: Bulb Switcher II
struct Solution;

impl Solution {
    fn flip_lights(n: i32, m: i32) -> i32 {
        let n = n.min(3);
        if m == 0 || n == 0 {
            return 1;
        }
        if n == 1 {
            return 2;
        }
        if n == 2 {
            return if m == 1 { 3 } else { 4 };
        }
        if m == 1 {
            return 4;
        }
        if m == 2 {
            7
        } else {
            8
        }
    }
}

#[test]
fn test() {
    let n = 1;
    let m = 1;
    let res = 2;
    assert_eq!(Solution::flip_lights(n, m), res);
    let n = 2;
    let m = 1;
    let res = 3;
    assert_eq!(Solution::flip_lights(n, m), res);
    let n = 3;
    let m = 1;
    let res = 4;
    assert_eq!(Solution::flip_lights(n, m), res);
}

// Accepted solution for LeetCode #672: Bulb Switcher II
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #672: Bulb Switcher II
// class Solution {
//     public int flipLights(int n, int presses) {
//         int[] ops = new int[] {0b111111, 0b010101, 0b101010, 0b100100};
//         Set<Integer> vis = new HashSet<>();
//         n = Math.min(n, 6);
//         for (int mask = 0; mask < 1 << 4; ++mask) {
//             int cnt = Integer.bitCount(mask);
//             if (cnt <= presses && cnt % 2 == presses % 2) {
//                 int t = 0;
//                 for (int i = 0; i < 4; ++i) {
//                     if (((mask >> i) & 1) == 1) {
//                         t ^= ops[i];
//                     }
//                 }
//                 t &= ((1 << 6) - 1);
//                 t >>= (6 - n);
//                 vis.add(t);
//             }
//         }
//         return vis.size();
//     }
// }

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

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

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.