Problem Description

You are given an array of strings arr and an integer k. A distinct string is defined as a string that appears exactly once in the array (not duplicated).

Your task is to find and return the kth distinct string from the array, considering the strings in the order they appear. If there are fewer than k distinct strings in the array, return an empty string "".

For example, if arr = ["a", "b", "a", "c", "b", "d"] and k = 2:

• The distinct strings in order are: "c" (appears once) and "d" (appears once)
• The 1st distinct string is "c"
• The 2nd distinct string is "d"
• So the answer would be "d"

The solution uses a hash table to count occurrences of each string using Counter(arr). It then iterates through the original array maintaining the order, checks if each string has a count of 1 (meaning it's distinct), and decrements k each time a distinct string is found. When k reaches 0, that string is the kth distinct string and is returned. If the loop completes without finding k distinct strings, an empty string is returned.

Intuition

The key insight is that we need to identify which strings are "distinct" (appear only once) while preserving their original order in the array. This immediately suggests a two-step approach:

1. First, we need to know which strings qualify as distinct - this requires counting how many times each string appears
2. Then, we need to find the kth such string in the original order

Why use a hash table? When we need to count occurrences of elements, a hash table (or Counter in Python) is the natural choice because it gives us O(1) lookup time for checking how many times a string appears.

The clever part is realizing we don't need to create a separate list of distinct strings. Instead, we can iterate through the original array again, which naturally preserves the order. As we encounter each string, we check if it's distinct (count equals 1), and if so, we decrement our counter k. This way, when k reaches 0, we've found exactly the kth distinct string.

This approach avoids the need for extra space to store distinct strings separately and handles the ordering requirement elegantly. If we exhaust the array without k reaching 0, it means there aren't enough distinct strings, so we return an empty string.

The time complexity is O(n) for counting and O(n) for finding the kth distinct string, giving us O(n) overall. The space complexity is O(n) for the hash table in the worst case where all strings are unique.

Solution Approach

The implementation follows a straightforward counting and traversal strategy using a hash table.

Step 1: Count Occurrences

cnt = Counter(arr)

We use Python's Counter from the collections module to create a hash table that maps each string to its frequency in the array. This takes O(n) time where n is the length of the array.

Step 2: Find the kth Distinct String

for s in arr:
    if cnt[s] == 1:
        k -= 1
        if k == 0:
            return s

We iterate through the original array arr to maintain the order of appearance. For each string s:

• Check if cnt[s] == 1, which means the string is distinct (appears only once)
• If it's distinct, decrement k by 1
• When k reaches 0, we've found the kth distinct string and return it immediately

Step 3: Handle Edge Case

return ""

If we complete the loop without finding k distinct strings (meaning there are fewer than k distinct strings in the array), we return an empty string as required.

Example Walkthrough: For arr = ["d", "b", "c", "b", "c", "a"] and k = 2:

1. cnt = {"d": 1, "b": 2, "c": 2, "a": 1}
2. Traverse array:
   • "d": count is 1, so k becomes 1
   • "b": count is 2, skip
   • "c": count is 2, skip
   • "b": count is 2, skip
   • "c": count is 2, skip
   • "a": count is 1, so k becomes 0, return "a"

The algorithm efficiently solves the problem in O(n) time with O(n) space complexity.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example: arr = ["cat", "dog", "cat", "bird", "dog", "fish"] and k = 2

Step 1: Count occurrences using Counter

cnt = {"cat": 2, "dog": 2, "bird": 1, "fish": 1}

• "cat" appears 2 times (not distinct)
• "dog" appears 2 times (not distinct)
• "bird" appears 1 time (distinct)
• "fish" appears 1 time (distinct)

Step 2: Traverse the original array to find the kth distinct string

Result: The function returns "fish" as it's the 2nd distinct string in order.

The key insight is that we maintain the original ordering by iterating through the array as it was given, only counting strings that appear exactly once. When we've seen exactly k such strings, we return the current one.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the number of strings in the array arr. This is because:

• Creating the Counter takes O(n) time to iterate through all elements
• The second loop iterates through the array once in the worst case, which is O(n)
• Overall: O(n) + O(n) = O(n)

The space complexity is O(n), where n is the number of strings in the array arr. This is because:

• The Counter dictionary stores at most n unique strings as keys
• Each string reference in the Counter points to the same string objects in the original array, so we're only storing references, not duplicating the strings themselves

Note: The reference answer suggests O(L) where L is the total length of all strings. This would be accurate if we consider the space needed to store the actual string data. However, in Python's implementation with Counter, we're storing references to existing strings rather than creating new copies. The time complexity could also be considered O(L) if we account for string comparison operations during hashing, as comparing strings takes time proportional to their length. Under this interpretation:

• Time complexity: O(L) accounting for string comparison in hash operations
• Space complexity: O(L) accounting for the total space occupied by all unique strings

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Losing Original Order When Using Only the Counter

A common mistake is trying to iterate through the Counter object directly instead of the original array:

# WRONG APPROACH
def kthDistinct(self, arr: List[str], k: int) -> str:
    frequency_map = Counter(arr)
  
    # This loses the original order!
    for string, count in frequency_map.items():
        if count == 1:
            k -= 1
            if k == 0:
                return string
    return ""

Why it's wrong: The Counter object doesn't preserve the insertion order of the original array. When you iterate through frequency_map.items(), you might get strings in a different order than they appeared in the input array.

Solution: Always iterate through the original array arr to maintain the order of appearance, and use the Counter only for frequency lookups.

Pitfall 2: Modifying k Without Creating a Copy

While the current solution correctly modifies the parameter k directly, in some contexts this could be problematic if you need to preserve the original value:

# Potentially problematic if k is needed elsewhere
def kthDistinct(self, arr: List[str], k: int) -> str:
    frequency_map = Counter(arr)
    original_k = k  # If you need the original value later
  
    for string in arr:
        if frequency_map[string] == 1:
            k -= 1
            if k == 0:
                return string
    return ""

Solution: If you need to preserve the original k value for any reason (debugging, logging, etc.), create a copy before modifying it.

Pitfall 3: Not Handling Edge Cases Properly

Developers might forget to handle cases where k <= 0 or when the array is empty:

# MORE ROBUST VERSION
def kthDistinct(self, arr: List[str], k: int) -> str:
    # Handle edge cases explicitly
    if not arr or k <= 0:
        return ""
  
    frequency_map = Counter(arr)
  
    for string in arr:
        if frequency_map[string] == 1:
            k -= 1
            if k == 0:
                return string
  
    return ""

Why it matters: While the current solution works correctly even without explicit edge case handling (it will naturally return "" for these cases), explicitly handling them makes the code more readable and can prevent unnecessary computation for invalid inputs.