How do you solve Shortest Uncommon Substring in an Array on LeetCode?

Shortest Uncommon Substring in an Array (LeetCode #3076) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m), where n is the number of strings and m is the average length of the strings. time complexity and O(n * m) space complexity. Key insight: Using a Trie allows efficient substring checks.

What is the time complexity of Shortest Uncommon Substring in an Array?

The optimal solution for Shortest Uncommon Substring in an Array runs in O(n * m), where n is the number of strings and m is the average length of the strings. time and uses O(n * m) space. The time complexity is O(n * m) because we insert each substring into the Trie, which can take linear time relative to the length of the string. The space complexity is also O(n * m) due to the storage of all substrings in the Trie.

What is the brute force solution for Shortest Uncommon Substring in an Array?

The brute force approach for Shortest Uncommon Substring in an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible substring of each string in the array and seeing if it exists in any other string. This is straightforward but inefficient, as it requires a lot of comparisons. The optimal Optimal Solution approach improves this to O(n * m), where n is the number of strings and m is the average length of the strings..

What algorithm or data structure is used to solve Shortest Uncommon Substring in an Array?

Shortest Uncommon Substring in an Array uses the following concepts: Array, Hash Table, String, Trie. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Shortest Uncommon Substring in an Array?

Explain your thought process clearly while coding. Be mindful of edge cases, such as single-character strings. Discuss time and space complexity as you develop your solution.

What are common mistakes when solving Shortest Uncommon Substring in an Array?

Not considering all substrings, especially edge cases. Overlooking the need for lexicographical order when multiple substrings are found.

#3076

Shortest Uncommon Substring in an Array

Medium

ArrayHash TableStringTrieHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m), where n is the number of strings and m is the average length of the strings.
Space	O(1)	O(n * m)

💡

Intuition

Time O(n * m), where n is the number of strings and m is the average length of the strings.Space O(n * m)

By using a Trie (prefix tree), we can efficiently store and check for the existence of substrings across all strings. This allows us to quickly find the shortest uncommon substring without generating all possible substrings explicitly.

⚙️

Algorithm

3 steps

1Step 1: Build a Trie from all strings in the array to store all substrings.
2Step 2: For each string, traverse the Trie to find the shortest substring that is not present in any other string.
3Step 3: Return the results based on the findings.

solution.py38 lines

1class TrieNode:
2    def __init__(self):
3        self.children = {}
4        self.count = 0
5
6class Trie:
7    def __init__(self):
8        self.root = TrieNode()
9
10    def insert(self, word):
11        node = self.root
12        for char in word:
13            if char not in node.children:
14                node.children[char] = TrieNode()
15            node = node.children[char]
16            node.count += 1
17
18    def search_shortest_uncommon(self, word):
19        node = self.root
20        shortest = ''
21        for char in word:
22            node = node.children[char]
23            if node.count == 1:
24                shortest += char
25                return shortest
26            shortest += char
27        return ''
28
29def shortestUncommonSubstring(arr):
30    trie = Trie()
31    for s in arr:
32        for i in range(len(s)):
33            trie.insert(s[i:])
34
35    answer = []
36    for s in arr:
37        answer.append(trie.search_shortest_uncommon(s))
38    return answer

ℹ

Complexity note: The time complexity is O(n * m) because we insert each substring into the Trie, which can take linear time relative to the length of the string. The space complexity is also O(n * m) due to the storage of all substrings in the Trie.

1Using a Trie allows efficient substring checks.
2Lexicographical order can be maintained during substring checks.

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