How do you solve Kth Distinct String in an Array on LeetCode?

Kth Distinct String in an Array (LeetCode #2053) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Distinct strings are those that appear only once.

What is the brute force solution for Kth Distinct String in an Array?

The brute force approach for Kth Distinct String in an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each string in the array and counting how many times it appears. We then collect the distinct strings and return the k-th one. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Kth Distinct String in an Array?

Kth Distinct String in an Array uses the following concepts: Array, Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Kth Distinct String in an Array?

Clarify the definition of distinct strings before starting. Think about the data structures that can help you count efficiently. Always consider edge cases, like when k is greater than the number of distinct strings.

What are common mistakes when solving Kth Distinct String in an Array?

Not using a data structure to count occurrences, leading to inefficient solutions. Confusing the index for k with the count of distinct strings.

#2053

Kth Distinct String in an Array

Easy

ArrayHash TableStringCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Using a HashMap to count occurrences of each string allows us to efficiently determine which strings are distinct. We can then iterate through the array a second time to collect the distinct strings in order.

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Algorithm

5 steps

1Step 1: Create a HashMap to count occurrences of each string.
2Step 2: Iterate through the array and populate the HashMap with counts.
3Step 3: Initialize an empty list for distinct strings.
4Step 4: Iterate through the array again, adding strings with a count of 1 to the distinct list.
5Step 5: Return the k-th distinct string if it exists, otherwise return an empty string.

solution.py6 lines

1from collections import Counter
2
3def kthDistinct(arr, k):
4    count = Counter(arr)
5    distinct = [s for s in arr if count[s] == 1]
6    return distinct[k-1] if len(distinct) >= k else ''

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Complexity note: The time complexity is O(n) because we traverse the array twice: once for counting and once for collecting distinct strings. The space complexity is O(n) due to the HashMap storing counts.

1Distinct strings are those that appear only once.
2The order of appearance matters when determining the k-th distinct string.

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