1492. The kth Factor of n

Description

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.

 

Example 1:

Input: n = 12, k = 3
Output: 3
Explanation: Factors list is [1, 2, 3, 4, 6, 12], the 3rd factor is 3.

Example 2:

Input: n = 7, k = 2
Output: 7
Explanation: Factors list is [1, 7], the 2nd factor is 7.

Example 3:

Input: n = 4, k = 4
Output: -1
Explanation: Factors list is [1, 2, 4], there is only 3 factors. We should return -1.

 

Constraints:

• 1 <= k <= n <= 1000

 

Follow up:

Could you solve this problem in less than O(n) complexity?

Solutions

Solution 1: Brute Force Enumeration

A "factor" is a number that can divide another number. Therefore, we only need to enumerate from $1$ to $n$, find all numbers that can divide $n$, and then return the $k$-th one.

The time complexity is $O(n)$, and the space complexity is $O(1)$.

Python3JavaC++GoTypeScript

class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        for i in range(1, n + 1):
            if n % i == 0:
                k -= 1
                if k == 0:
                    return i
        return -1

class Solution {
    public int kthFactor(int n, int k) {
        for (int i = 1; i <= n; ++i) {
            if (n % i == 0 && (--k == 0)) {
                return i;
            }
        }
        return -1;
    }
}

class Solution {
public:
    int kthFactor(int n, int k) {
        for (int i = 1; i <= n; ++i) {
            if (n % i == 0 && (--k == 0)) {
                return i;
            }
        }
        return -1;
    }
};

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

function kthFactor(n: number, k: number): number {
    for (let i = 1; i <= n; ++i) {
        if (n % i === 0 && --k === 0) {
            return i;
        }
    }
    return -1;
}

Solution 2: Optimized Enumeration

We can observe that if $n$ has a factor $x$, then $n$ must also have a factor $n/x$.

Therefore, we first need to enumerate $[1,2,...\left \lfloor \sqrt{n} \right \rfloor]$, find all numbers that can divide $n$. If we find the $k$-th factor, then we can return it directly. If we do not find the $k$-th factor, then we need to enumerate $[\left \lfloor \sqrt{n} \right \rfloor ,..1]$ in reverse order, and find the $k$-th factor.

The time complexity is $O(\sqrt{n})$, and the space complexity is $O(1)$.

Python3JavaC++GoTypeScript

class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        i = 1
        while i * i < n:
            if n % i == 0:
                k -= 1
                if k == 0:
                    return i
            i += 1
        if i * i != n:
            i -= 1
        while i:
            if (n % (n // i)) == 0:
                k -= 1
                if k == 0:
                    return n // i
            i -= 1
        return -1

class Solution {
    public int kthFactor(int n, int k) {
        int i = 1;
        for (; i < n / i; ++i) {
            if (n % i == 0 && (--k == 0)) {
                return i;
            }
        }
        if (i * i != n) {
            --i;
        }
        for (; i > 0; --i) {
            if (n % (n / i) == 0 && (--k == 0)) {
                return n / i;
            }
        }
        return -1;
    }
}

class Solution {
public:
    int kthFactor(int n, int k) {
        int i = 1;
        for (; i < n / i; ++i) {
            if (n % i == 0 && (--k == 0)) {
                return i;
            }
        }
        if (i * i != n) {
            --i;
        }
        for (; i > 0; --i) {
            if (n % (n / i) == 0 && (--k == 0)) {
                return n / i;
            }
        }
        return -1;
    }
};

func kthFactor(n int, k int) int {
    i := 1
    for ; i < n/i; i++ {
        if n%i == 0 {
            k--
            if k == 0 {
                return i
            }
        }
    }
    if i*i != n {
        i--
    }
    for ; i > 0; i-- {
        if n%(n/i) == 0 {
            k--
            if k == 0 {
                return n / i
            }
        }
    }
    return -1
}

function kthFactor(n: number, k: number): number {
    let i: number = 1;
    for (; i < n / i; ++i) {
        if (n % i === 0 && --k === 0) {
            return i;
        }
    }
    if (i * i !== n) {
        --i;
    }
    for (; i > 0; --i) {
        if (n % Math.floor(n / i) === 0 && --k === 0) {
            return Math.floor(n / i);
        }
    }
    return -1;
}

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