Introduction

In this blog post, we will explore how to solve the LeetCode problem "2053" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

A distinct string is a string that is present only once in an array. Given an array of strings arr, and an integer k, return the kth distinct string present in arr. If there are fewer than k distinct strings, return an empty string "". Note that the strings are considered in the order in which they appear in the array. Example 1: Input: arr = ["d","b","c","b","c","a"], k = 2 Output: "a" Explanation: The only distinct strings in arr are "d" and "a". "d" appears 1st, so it is the 1st distinct string. "a" appears 2nd, so it is the 2nd distinct string. Since k == 2, "a" is returned. Example 2: Input: arr = ["aaa","aa","a"], k = 1 Output: "aaa" Explanation: All strings in arr are distinct, so the 1st string "aaa" is returned. Example 3: Input: arr = ["a","b","a"], k = 3 Output: "" Explanation: The only distinct string is "b". Since there are fewer than 3 distinct strings, we return an empty string "". Constraints: 1 <= k <= arr.length <= 1000 1 <= arr[i].length <= 5 arr[i] consists of lowercase English letters.

Explanation

• Use a hash map (dictionary in Python) to store the frequency of each string in the input array.
  • Iterate through the input array, and for each string, check if its frequency in the hash map is 1. If it is, increment a counter.
  • If the counter equals k, return the current string. If after iterating through the entire array the counter is less than k, return an empty string.

• Runtime Complexity: O(n), Storage Complexity: O(n) where n is the length of the array.

Code

    def kthDistinct(arr, k):
    """
    Finds the kth distinct string in an array of strings.

    Args:
        arr: A list of strings.
        k: An integer representing the kth distinct string to find.

    Returns:
        The kth distinct string in the array, or an empty string if there are fewer than k distinct strings.
    """

    counts = {}
    for s in arr:
        counts[s] = counts.get(s, 0) + 1

    distinct_count = 0
    for s in arr:
        if counts[s] == 1:
            distinct_count += 1
            if distinct_count == k:
                return s

    return ""