Closest Divisors

Closest Divisors - Problem

Math Medium

Given an integer num, find the closest two integers in absolute difference whose product equals num + 1 or num + 2.

Return the two integers in any order.

Example: If num = 8, we need to find two integers whose product is either 9 or 10. For 9: divisors are [1,9] and [3,3]. For 10: divisors are [1,10], [2,5]. The closest pair is [3,3] with difference 0.

Input & Output

Example 1 — Perfect Square Case

$ Input: num = 8

› Output: [3,3]

💡 Note: For num+1=9: divisors are [1,9] and [3,3]. For num+2=10: divisors are [1,10], [2,5]. The closest pair is [3,3] with difference 0.

Example 2 — Small Number

$ Input: num = 2

› Output: [2,2]

💡 Note: For num+1=3: divisors are [1,3]. For num+2=4: divisors are [1,4], [2,2]. The closest pair is [2,2] with difference 0.

Example 3 — Prime Result

$ Input: num = 10

› Output: [3,4]

💡 Note: For num+1=11: divisors are [1,11] (11 is prime). For num+2=12: divisors include [3,4]. The closest pair is [3,4] with difference 1.

Constraints

1 ≤ num ≤ 10⁹

Visualization

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Asked in

F Facebook 25 a Amazon 20

The key insight is that the closest divisors of any number are found near its square root. Start from floor(sqrt(target)) and work downward - the first divisor found gives the optimal pair. Check both num+1 and num+2, return the pair with minimum difference. Best approach is Square Root Optimization. Time: O(√n), Space: O(1).

Common Approaches

✓ Binary Search

⏱️ Time: N/A Space: N/A

Brute Force - Check All Divisors

⏱️ Time: O(n) Space: O(1)

For both num+1 and num+2, iterate through all possible divisors from 1 to the number. For each divisor i, check if the number is divisible by i, then calculate the corresponding divisor j = number/i. Track the pair with minimum absolute difference.

Square Root Optimization

⏱️ Time: O(√n) Space: O(1)

Instead of checking all divisors, start from the square root of each target number and work downward. The first divisor found will give us the closest pair since divisors come in pairs (i, n/i) and they get farther apart as i gets smaller.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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