How do you solve Check If String Is a Prefix of Array on LeetCode?

Check If String Is a Prefix of Array (LeetCode #1961) 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: Understanding how prefixes work is crucial for string manipulation problems.

What is the brute force solution for Check If String Is a Prefix of Array?

The brute force approach for Check If String Is a Prefix of Array is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check if the string `s` can be formed by concatenating the first few strings from the `words` array. This involves checking all possible prefixes of the `words` array until we either match `s` or exceed its length. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check If String Is a Prefix of Array?

Check If String Is a Prefix of Array uses the following concepts: Array, Two Pointers, String. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Check If String Is a Prefix of Array?

Always clarify the problem statement and constraints before jumping into coding. Think aloud during the interview to show your thought process. Test edge cases, such as empty strings or single-character strings.

What are common mistakes when solving Check If String Is a Prefix of Array?

Not checking the lengths of strings before comparing them, leading to unnecessary operations. Confusing the order of concatenation and comparison.

#1961

Check If String Is a Prefix of Array

Easy

ArrayTwo PointersStringTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Instead of building the prefix string step-by-step, we can directly compare the concatenated result of the first k words with the string `s`. This reduces unnecessary checks and string manipulations.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable `prefixLength` to 0.
2Step 2: Iterate through the `words` array, adding the length of each word to `prefixLength`.
3Step 3: If at any point `prefixLength` equals the length of `s`, check if the concatenated substring of `s` matches the prefix formed by the first k words.
4Step 4: If `prefixLength` exceeds the length of `s`, return false.
5Step 5: If the loop completes and no match is found, return false.

solution.py15 lines

1# Full working Python code
2s = 'iloveleetcode'
3words = ['i', 'love', 'leetcode', 'apples']
4
5def is_prefix_string(s, words):
6    prefixLength = 0
7    for word in words:
8        prefixLength += len(word)
9        if prefixLength == len(s):
10            return s == ''.join(words[:words.index(word) + 1])
11        if prefixLength > len(s):
12            return False
13    return False
14
15print(is_prefix_string(s, words))

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the `words` array once, accumulating lengths and checking conditions, which is efficient compared to the brute force method.

1Understanding how prefixes work is crucial for string manipulation problems.
2Recognizing when to optimize a brute force solution can save time and resources.

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