How do you solve K-th Smallest Prime Fraction on LeetCode?

K-th Smallest Prime Fraction (LeetCode #786) can be solved using Brute Force. The optimal approach is Brute Force with O(n² log n) time complexity and O(n²) space complexity. Key insight: We can generate all possible fractions from the given array and sort them to find the k-th smallest. This is straightforward but inefficient for larger arrays.

What is the time complexity of K-th Smallest Prime Fraction?

The optimal solution for K-th Smallest Prime Fraction runs in O(n² log n) time and uses O(n²) space. The time complexity is O(n²) for generating fractions and O(n² log n) for sorting them. Space complexity is O(n²) due to storing all fractions.

What algorithm or data structure is used to solve K-th Smallest Prime Fraction?

K-th Smallest Prime Fraction uses the following concepts: Array, Two Pointers, Binary Search, Sorting, Heap (Priority Queue). The recommended patterns to study are: .

K-th Smallest Prime Fraction — LeetCode #786 (Medium)

Tags: Array, Two Pointers, Binary Search, Sorting, Heap (Priority Queue)

Related patterns:

Brute Force approach

Time complexity: O(n² log n). Space complexity: O(n²).

We can generate all possible fractions from the given array and sort them to find the k-th smallest. This is straightforward but inefficient for larger arrays.

The time complexity is O(n²) for generating fractions and O(n² log n) for sorting them. Space complexity is O(n²) due to storing all fractions.

Step 1: Initialize an empty list to store all fractions.
Step 2: Use two nested loops to iterate through the array, generating fractions arr[i] / arr[j] for all valid pairs (i, j).
Step 3: Sort the list of fractions and return the k-th smallest fraction.

For arr = [1, 2, 3, 5] and k = 3: 1. Generate fractions: [(1, 2), (1, 3), (1, 5), (2, 3), (2, 5), (3, 5)] 2. Sort fractions by value: [(1, 5), (1, 3), (1, 2), (2, 3), (2, 5), (3, 5)] 3. k-th smallest (k=3): (2, 5) 4. Result: [2, 5]

#786

K-th Smallest Prime Fraction

Medium

Array↗Two Pointers↗Binary Search↗Sorting↗Heap (Priority Queue)↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n² log n)Space O(n²)

We can generate all possible fractions from the given array and sort them to find the k-th smallest. This is straightforward but inefficient for larger arrays.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty list to store all fractions.
2Step 2: Use two nested loops to iterate through the array, generating fractions arr[i] / arr[j] for all valid pairs (i, j).
3Step 3: Sort the list of fractions and return the k-th smallest fraction.

solution.py10 lines

1# Full working Python code
2arr = [1, 2, 3, 5]
3k = 3
4fractions = []
5for i in range(len(arr)):
6    for j in range(i + 1, len(arr)):
7        fractions.append((arr[i], arr[j]))
8fractions.sort(key=lambda x: x[0] / x[1])
9result = fractions[k - 1]
10print(result)

ℹ

Complexity note: The time complexity is O(n²) for generating fractions and O(n² log n) for sorting them. Space complexity is O(n²) due to storing all fractions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.