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  1. Are there any constraints on the length of the string or the length of the array?
  2. Are the elements of the array always non-empty strings?
  3. What should be the output if either the string or the array is empty? Should we assume such cases will not occur?

Once these questions are clarified, we can proceed with a more detailed strategy and implementation. Assuming typical constraints found on LeetCode:

  1. The length of the string and the combined length of the array’s strings can be reasonably large (up to 10^4 characters).
  2. Elements in the array are non-empty strings.
  3. We need to handle edge cases where either the string or the array might be empty.

Strategy

To determine if the given string s is a prefix of the concatenated elements of the array words, we can follow these steps:

  1. Concatenation of Words: Concatenate the elements of words until their combined length is at least as long as s.
  2. Prefix Check: Check if the resulting concatenated string starts with s.
  3. Efficiency Considerations: Avoid concatenating more characters than necessary to minimize time and space usage.

Time Complexity

The approach involves iterating through each word and possibly part of each word until the combined length is sufficient. Thus, the time complexity is O(n + m), where n is the length of the string s and m is the total length of all words in words up to the necessary length.

Code

def isPrefixString(s: str, words: List[str]) -> bool:
    concatenated = ""
    for word in words:
        concatenated += word
        if len(concatenated) >= len(s):
            return concatenated.startswith(s)
    return False

# Example Use Case
s = "iloveleetcode"
words = ["i", "love", "leetcode", "apples"]
print(isPrefixString(s, words))  # Output: True

Explanation

  1. Concatenate: Start with an empty string concatenated.
  2. Iterate: Iterate through each word in words.
    • Add the current word to concatenated.
    • If the length of concatenated is greater than or equal to the length of s, check if concatenated starts with s. Return True if it does.
  3. Return False: If the loop completes without finding a match, return False.

This solution ensures we only concatenate what is necessary and check for the prefix as early as possible, keeping the operation efficient.

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