![]() In this article, we explored an algorithmic approach to finding all permutations of a given string A within another string B. The space complexity of this algorithm is O(1), as the frequency arrays freqA and freqB store the frequency of characters and their lengths are constant. This is because we iterate through B only once, incrementing and decrementing the window boundaries. The time complexity of this algorithm is O(N), where N is the length of string B. ![]() Return the stored starting indices as the positions of all permutations of A in B. Repeat steps 4 and 5 until the right pointer reaches the end of B. Store the starting index of the current window. If the frequencies of characters in freqA and freqB match, we have found a permutation.
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