Bulb Switcher

Bulb Switcher - Problem

Math Brainteaser Medium

There are n bulbs that are initially off. You first turn on all the bulbs, then you turn off every second bulb.

On the third round, you toggle every third bulb (turning on if it's off or turning off if it's on). For the i-th round, you toggle every i-th bulb.

For the n-th round, you only toggle the last bulb.

Return the number of bulbs that are on after n rounds.

Input & Output

Example 1 — Small Case

$ Input: n = 3

› Output: 1

💡 Note: Initial: [OFF,OFF,OFF]. Round 1: [ON,ON,ON]. Round 2: [ON,OFF,ON]. Round 3: [ON,OFF,OFF]. Only bulb 1 stays on.

Example 2 — Perfect Square

$ Input: n = 9

› Output: 3

💡 Note: Perfect squares 1, 4, 9 have odd number of divisors, so they remain on after all toggles.

Example 3 — Edge Case

$ Input: n = 0

› Output: 0

💡 Note: No bulbs means 0 bulbs are on.

Constraints

0 ≤ n ≤ 10⁹

Visualization

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Asked in

G Google 25 M Microsoft 20 A Apple 15

The key insight is that bulb i gets toggled once for each of its divisors. Only perfect squares have an odd number of divisors, so only they remain on. Best approach is the mathematical solution: return floor(√n). Time: O(1), Space: O(1)

Common Approaches

✓ Bfs

⏱️ Time: N/A Space: N/A

Brute Force Simulation

⏱️ Time: O(n²) Space: O(n)

Create an array to track each bulb's state, then simulate all n rounds. In round i, toggle every i-th bulb by flipping its state.

Mathematical Insight

⏱️ Time: O(1) Space: O(1)

Each bulb i gets toggled once for each of its divisors. Only perfect squares have an odd number of divisors, so only they remain on after all rounds.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>

int integerSqrt(long long n) {
    if (n == 0) return 0;
    
    long long left = 1, right = n > 100000 ? 100000 : n;
    int result = 0;
    
    while (left <= right) {
        long long mid = left + (right - left) / 2;
        if (mid * mid <= n) {
            result = (int)mid;
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    
    return result;
}

int main() {
    long long n;
    scanf("%lld", &n);
    
    int result = integerSqrt(n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Ln 1, Col 1

Smart Actions

💡 Explanation

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