The kth Factor of n

The kth Factor of n - Problem

You are given two positive integers n and k. A factor of an integer n is defined as an integer i where n % i == 0.

Consider a list of all factors of n sorted in ascending order, return the kth factor in this list or return -1 if n has less than k factors.

Input & Output

Example 1 — Basic Case

$ Input: n = 12, k = 3

› Output: 3

💡 Note: Factors of 12 in ascending order: [1, 2, 3, 4, 6, 12]. The 3rd factor is 3.

Example 2 — Not Enough Factors

$ Input: n = 7, k = 2

› Output: 7

💡 Note: Factors of 7 are [1, 7]. The 2nd factor is 7.

Example 3 — Perfect Square

$ Input: n = 4, k = 4

› Output: -1

💡 Note: Factors of 4 are [1, 2, 4]. There are only 3 factors, so the 4th factor doesn't exist.

Constraints

1 ≤ k ≤ n ≤ 1000

Visualization

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Asked in

F Facebook 15 a Amazon 12

The key insight is that factors come in pairs (i, n/i), so we only need to check up to √n. Best approach is mathematical early termination with Time: O(√n), Space: O(1).

Common Approaches

✓ Brute Force - Check Every Number

⏱️ Time: O(n) Space: O(1)

Iterate through all numbers from 1 to n and count how many divide n evenly. When we reach the kth factor, return it.

Mathematical - Early Termination

⏱️ Time: O(√n) Space: O(1)

Combine the efficiency of checking up to √n with early termination. Count factors as we find them and return as soon as we reach the kth factor.

Optimized - Square Root Approach

⏱️ Time: O(√n + f log f) Space: O(f)

Since factors come in pairs (i, n/i), we only need to check up to √n. Collect all factors in a list, sort them, and return the kth factor.

Brute Force - Check Every Number — Algorithm Steps

Iterate through numbers 1 to n
Count factors that divide n evenly
Return the kth factor when found

Visualization

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Step-by-Step Walkthrough

Check Each Number

Test if each number divides n evenly

Count Factors

Increment counter when factor found

Return kth Factor

Return when counter reaches k

Code -

solution.c — C

#include <stdio.h>

int solution(int n, int k) {
    int count = 0;
    for (int i = 1; i <= n; i++) {
        if (n % i == 0) {
            count++;
            if (count == k) {
                return i;
            }
        }
    }
    return -1;
}

int main() {
    int n, k;
    scanf("%d %d", &n, &k);
    printf("%d\n", solution(n, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop from 1 to n checking each number

⚡ Linearithmic

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check Every Number — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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