Cycle Detection and Memoization
Intuition\nThe key challenge is efficiently handling large input sizes, where brute-force methods would be too slow. The intuition here is that there might be r
# Intuition\nThe key challenge is efficiently handling large input sizes, where brute-force methods would be too slow. The intuition here is that there might be repeating patterns (or cycles) in how `s1` and `s2` align with each other. Detecting these cycles can allow us to skip unnecessary work and directly calculate the result.\n\n# Approach\nThe approach begins by iterating through the characters of `s1` while matching them against `s2`. We use a dictionary to memoize the state (the current positions within `s1` and `s2`, and how many full repetitions of `s2` have been matched so far). When a previously seen state is encountered, a cycle is detected, which allows us to skip forward in the sequence by calculating how many times this cycle can repeat. This drastically reduces the time complexity.\n\n# Complexity\n- Time complexity:\nThe time complexity is approximately $$O(n1 \\times |s1|)$$ in the worst case. However, due to cycle detection and skipping, it effectively becomes much less in practice.\n\n- Space complexity:\nThe space complexity is $$O(|s1| \\times |s2|)$$ due to the storage of states in the memoization dictionary.\n\n# Code\n```python3 []\nclass Solution:\n    def getMaxRepetitions(self, s1: str, n1: int, s2: str, n2: int) -> int:\n        indexDict = {}\n        s1Count, s2Count, s1Index, s2Index = 0, 0, 0, 0\n\n        while s1Count < n1:\n            if s1[s1Index] == s2[s2Index]:\n                s2Index += 1\n                if s2Index == len(s2):\n                    s2Index = 0\n                    s2Count += 1\n            \n            s1Index += 1\n            if s1Index == len(s1):\n                s1Index = 0\n                s1Count += 1\n\n                if (s1Index, s2Index) in indexDict:\n                    prevS1Count, prevS2Count = indexDict[(s1Index, s2Index)]\n                    cycleLength = s1Count - prevS1Count\n                    cycleCount = (n1 - s1Count) // cycleLength\n                    s1Count += cycleCount * cycleLength\n                    s2Count += cycleCount * (s2Count - prevS2Count)\n                \n                indexDict[(s1Index, s2Index)] = (s1Count, s2Count)\n\n        return s2Count // n2\n\n```
['String', 'Dynamic Programming', 'Python3']