Leetcode 1492: The kth Factor of n
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 sorted list of all factors of n in ascending order. Return the kth factor from this list or return -1 if n has less than k factors.
Problem
Approach
Steps
Complexity
Input: You are given two integers n and k. The integer n represents the number for which we need to find the factors, and k represents the position of the factor to return.
Example: n = 18, k = 4
Constraints:
• 1 <= k <= n <= 1000
Output: Return the kth factor in the list of factors of n, or -1 if n has fewer than k factors.
Example: Output: 6
Constraints:
• If n has less than k factors, return -1.
Goal: To find the kth factor of n by iterating through its divisors and returning the kth one.
Steps:
• Iterate through integers from 1 to n.
• Check if each integer divides n without a remainder.
• Count the divisors and return the kth divisor if it exists.
• If fewer than k divisors are found, return -1.
Goal: The integer n must be between 1 and 1000, and k is a positive integer less than or equal to n.
Steps:
• 1 <= k <= n <= 1000
Assumptions:
• The input values of n and k will always be valid integers within the specified range.
• Input: n = 18, k = 4
• Explanation: The factors of 18 are [1, 2, 3, 6, 9, 18]. The 4th factor is 6.
• Input: n = 9, k = 3
• Explanation: The factors of 9 are [1, 3, 9]. The 3rd factor is 9.
• Input: n = 10, k = 5
• Explanation: The factors of 10 are [1, 2, 5, 10]. There are only 4 factors, so the result is -1.
Approach: To solve this problem efficiently, iterate through potential factors of n and check for divisibility. Return the kth factor if found.
Observations:
• The factors of n are the integers from 1 to n that divide n without leaving a remainder.
• We can check each integer in this range to see if it's a factor of n.
• We need to find the kth factor, so tracking the count of factors is essential to solving this problem.
Steps:
• Initialize a counter to track the number of factors found.
• Iterate over integers from 1 to n and check divisibility.
• When a factor is found, increment the counter.
• If the counter matches k, return the factor.
• If no kth factor is found by the end, return -1.
Empty Inputs:
• There will always be valid input with n >= 1.
Large Inputs:
• For larger values of n (up to 1000), the solution must be efficient enough to handle the full range of factors.
Special Values:
• If k exceeds the number of factors of n, return -1.
Constraints:
• Ensure the solution handles all edge cases efficiently, especially when k is larger than the number of factors.
int kthFactor(int n, int k) {
for(int i = 1; i <= n; i++) {
if(n % i == 0) k--;
if(k == 0) return i;
}
return -1;
}
1 : Method
int kthFactor(int n, int k) {
The function `kthFactor` is designed to find the k-th divisor of a given integer `n`.
2 : Loop
for(int i = 1; i <= n; i++) {
A loop is used to iterate through all integers from 1 to `n` to check for divisibility.
3 : Divisibility Check
if(n % i == 0) k--;
This condition checks if `i` is a divisor of `n`. If it is, it decrements the value of `k`.
4 : Return Divisor
if(k == 0) return i;
If `k` reaches 0, it means the current number `i` is the k-th divisor of `n`, and it is returned.
5 : Return -1
return -1;
If no k-th divisor is found, the function returns -1 to indicate that the k-th divisor does not exist.
Best Case: O(1)
Average Case: O(n)
Worst Case: O(n)
Description: The worst-case time complexity occurs when we check every integer from 1 to n for divisibility.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant, as we are only using a few variables for tracking the factors.
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