Class Solution
Medium
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.
Example 1:
Input: s = “leetscode”, dictionary = [“leet”,“code”,“leetcode”]
Output: 1
Explanation: 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.
Example 2:
Input: s = “sayhelloworld”, dictionary = [“hello”,“world”]
Output: 3
Explanation: 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.
Constraints:
1 <= s.length <= 50
1 <= dictionary.length <= 50
1 <= dictionary[i].length <= 50
dictionary[i]
ands
consists of only lowercase English lettersdictionary
contains distinct words
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Constructor Summary
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Method Summary
Modifier and TypeMethodDescriptionint
minExtraChar
(java.lang.String s, java.lang.String[] dictionary) Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
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Constructor Details
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Solution
public Solution()
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Method Details
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minExtraChar
public int minExtraChar(java.lang.String s, java.lang.String[] dictionary)
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