Nth Magical Number - Practice Coding Problems

Nth Magical Number - Problem

Magical Number Finder

Imagine you're a mathematician exploring the mystical world of magical numbers! A positive integer is considered magical if it's divisible by either a or b (or both).

Your mission: Given three integers n, a, and b, find the nth magical number in the infinite sequence of magical numbers.

Example: If a = 2 and b = 3, the magical numbers are: 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18...

Challenge: Since the answer can be astronomically large, return it modulo 10⁹ + 7.

Input & Output

example_1.py — Basic Case

$ Input: n = 1, a = 2, b = 3

› Output: 2

💡 Note: The first magical number divisible by 2 or 3 is 2 itself.

example_2.py — Multiple Magicals

$ Input: n = 4, a = 2, b = 3

› Output: 6

💡 Note: Magical numbers: 2(1st), 3(2nd), 4(3rd), 6(4th). The 4th magical number is 6.

example_3.py — Large Result

$ Input: n = 1000000000, a = 40000, b = 40001

› Output: 999720007

💡 Note: For very large n with coprime a,b, result needs modulo operation. The answer before modulo would be huge.

Constraints

1 ≤ n ≤ 10⁹
2 ≤ a, b ≤ 4 × 10⁴
Return result modulo 10⁹ + 7
Both a and b are positive integers

Visualization

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Understanding the Visualization

Set Up the Telescope

Configure binary search range from 1 to n×min(a,b)

Calculate Star Density

Use inclusion-exclusion to count magical numbers up to any point

Smart Navigation

Adjust telescope focus based on whether we've found enough stars

Pinpoint Target

Converge on the exact location of the nth magical number

Key Takeaway

🎯 Key Insight: By combining binary search with mathematical formulas, we can calculate exactly how many magical numbers exist up to any point without counting them individually!

Asked in

G Google 35 a Amazon 28 f Meta 22 ⊞ Microsoft 18

The optimal solution uses binary search combined with mathematical counting. Instead of checking each number individually, we binary search on the answer range and use the inclusion-exclusion principle to count magical numbers up to any given point in O(1) time. This reduces complexity from O(n) to O(log n), making it efficient even for n up to 10⁹.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Linear Search)	O(n × max(a,b))	O(1)	Check every number sequentially until we find the nth magical number
Binary Search + Math (Optimal)	O(log(n × min(a,b)))	O(1)	Use binary search on the answer with mathematical counting

Brute Force (Linear Search) — Algorithm Steps

Initialize counter = 0 and current number = 1
For each number, check if it's divisible by a or b
If magical, increment counter
When counter reaches n, return current number
Apply modulo 10^9 + 7 to the result

Visualization

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Step-by-Step Walkthrough

Start Counting

Begin with number 1, check if divisible by a or b

Found Magical

When we find a magical number, increment our counter

Continue Search

Keep checking subsequent numbers until counter reaches n

Code -

solution.c — C

#include <stdio.h>

int nthMagicalNumber(int n, int a, int b) {
    const int MOD = 1000000007;
    int count = 0;
    long long num = 1;
    
    while (count < n) {
        if (num % a == 0 || num % b == 0) {
            count++;
            if (count == n) {
                return num % MOD;
            }
        }
        num++;
    }
    
    return -1;
}

int main() {
    int n, a, b;
    scanf("%d %d %d", &n, &a, &b);
    printf("%d\n", nthMagicalNumber(n, a, b));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × max(a,b))

We potentially check every number up to the nth magical number, which could be very large

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to track count and current number

✓ Linear Space

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Medium-High Frequency

~25 min Avg. Time

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Ln 1, Col 1

Smart Actions

💡 Explanation

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Linear Search) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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