You are given two 0-indexed strings s and target. You can take some letters from s and rearrange them to form new strings.
Return the maximum number of copies of target that can be formed by taking letters from s and rearranging them.
\n
Example 1:
\n\n\nInput: s = "ilovecodingonleetcode", target = "code"\nOutput: 2\nExplanation:\nFor the first copy of "code", take the letters at indices 4, 5, 6, and 7.\nFor the second copy of "code", take the letters at indices 17, 18, 19, and 20.\nThe strings that are formed are "ecod" and "code" which can both be rearranged into "code".\nWe can make at most two copies of "code", so we return 2.\n\n\n
Example 2:
\n\n\nInput: s = "abcba", target = "abc"\nOutput: 1\nExplanation:\nWe can make one copy of "abc" by taking the letters at indices 0, 1, and 2.\nWe can make at most one copy of "abc", so we return 1.\nNote that while there is an extra 'a' and 'b' at indices 3 and 4, we cannot reuse the letter 'c' at index 2, so we cannot make a second copy of "abc".\n\n\n
Example 3:
\n\n\nInput: s = "abbaccaddaeea", target = "aaaaa"\nOutput: 1\nExplanation:\nWe can make one copy of "aaaaa" by taking the letters at indices 0, 3, 6, 9, and 12.\nWe can make at most one copy of "aaaaa", so we return 1.\n\n\n
\n
Constraints:
\n\n1 <= s.length <= 1001 <= target.length <= 10s and target consist of lowercase English letters.Compiled with clang 11 using the latest C++ 17 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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\\r\\n\\r\\n1. Adding an item to a hash.\\r\\n
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Some common data structure implementations are provided in the Algorithms module: https://www.rubydoc.info/github/kanwei/algorithms/Algorithms
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