You are given a 0-indexed string s and a dictionary of words dictionary. You have to break s into one or more non-overlapping substrings such that each substring is present in dictionary. There may be some extra characters in s which are not present in any of the substrings.
Return the minimum number of extra characters left over if you break up s optimally.
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Example 1:
\n\n\nInput: s = "leetscode", dictionary = ["leet","code","leetcode"]\nOutput: 1\nExplanation: We can break s in two substrings: "leet" from index 0 to 3 and "code" from index 5 to 8. There is only 1 unused character (at index 4), so we return 1.\n\n\n\n
Example 2:
\n\n\nInput: s = "sayhelloworld", dictionary = ["hello","world"]\nOutput: 3\nExplanation: We can break s in two substrings: "hello" from index 3 to 7 and "world" from index 8 to 12. The characters at indices 0, 1, 2 are not used in any substring and thus are considered as extra characters. Hence, we return 3.\n\n\n
\n
Constraints:
\n\n1 <= s.length <= 501 <= dictionary.length <= 501 <= dictionary[i].length <= 50dictionary[i] and s consists of only lowercase English lettersdictionary contains distinct wordsCompiled 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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