Compare the strings one character position at a time, from left to right. The first mismatch—or the first string that runs out—ends the shared prefix. If that happens at position zero, return "".
What counts as a common prefix?
A prefix is a sequence of characters at the beginning of a string. Every input string must share it from position zero; a matching sequence found later in a string does not count. For example, ["flower", "flow", "flight"] has the longest common prefix "fl". For ["dog", "racecar", "car"], no prefix is shared, so the result is "". LeetCode’s problem statement allows 1 to 200 strings, each 0 to 200 characters long; non-empty strings contain lowercase English letters.
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Compare characters by position
Use the first string as the reference and scan it from left to right. At each position, check that every other string has a character there and that it matches the reference. Return the reference string up to—but not including—the first position that fails.
- Start with the first string as the reference.
- For each character position in that string, compare the character with the character at the same position in every other string.
- If a string has ended or a character differs, return the reference string before that position.
- If every position in the reference matches, return the entire reference string.
In Python:
def longest_common_prefix(strs: list[str]) -> str:
first = strs[0]
for i, char in enumerate(first):
for word in strs[1:]:
if i == len(word) or word[i] != char:
return first[:i]
return firstThe length check must come before word[i]; otherwise, indexing a shorter string can raise an error. The input constraints guarantee at least one string, so strs[0] exists. If the first string is empty, the loop does not run and the function returns "". If a later string is empty, the first comparison returns "".
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Why the first failure is the stopping point
A prefix has to match continuously from the beginning. Once a position differs—or one string has no character at that position—no later characters can make the preceding sequence a longer shared prefix. The function therefore returns immediately rather than checking later positions.
Time and space complexity
Let n be the number of strings and m the length of the shortest string. The character-comparison method takes O(n × m) time in the cited analysis, since it may compare up to m positions across the strings. It uses O(1) auxiliary space for its algorithm state. Creating the returned prefix with a Python slice may allocate the output string; that output is not counted as auxiliary state in this analysis. Doocs’ solution explanation gives this complexity for the comparison approach.
When to use a different approach
A trie is another possible way to represent shared prefixes, but the direct scan is easier to follow for the problem’s stated limits. The cited solution discusses a trie as an alternative but gives no measured runtime comparison, so there is no basis here to claim it is faster. For this task, the position-by-position scan provides a short implementation and a clear stopping condition.
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