检查两个排序的字符串在 O(log n) 时间内是否相等

Question

I need to write a Python function which takes two sorted strings (the characters in each string are in increasing alphabetical order) containing only lowercase letters, and checks whether or not the strings are equal. The function's time complexity needs to be O(log n) , where n is the length of each string.

I can't figure out how to check it without comparing each character in the first string with the parallel character of the second string.

Answer 1

This is, in fact, possible in O(log n) time in the worst case, since the strings are formed from an alphabet of constant size.

You can do 26 binary searches on each string to find the left-most occurrence of each letter. If the strings are equal, then all 26 binary searches will give the same results; either that the letter exists in neither string, or that its left-most occurrence is the same in both strings.

Conversely, if all of the binary searches give the same result, then the strings must be equal, because (1) the alphabet is fixed, (2) the indices of the left-most occurrences determine the frequency of each letter in the string, and (3) the strings are sorted, so the letter frequencies uniquely determine the string.

I'm assuming here that the strings have the same length. If they might not, then check that first and return False if the lengths are different. Getting the length of a string takes O(1) time.

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As @wim notes in the comments, this solution cannot be generalised to lists of numbers; it specifically only works with strings. When you have an algorithmic problem involving strings, the alphabet size is usually a constant, and this fact can often be exploited to achieve a better time complexity than would otherwise be possible.