Extra Characters in a String
2707. Extra Characters in a String
Difficulty: Medium
Topics: Array, Hash Table, String, Dynamic Programming, Trie
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] and s consists of only lowercase English letters
- dictionary contains distinct words
Hint:
- Can we use Dynamic Programming here?
- Define DP[i] as the min extra character if breaking up s[0:i] optimally.
Solution:
We can define a dp array where dp[i] represents the minimum number of extra characters in the substring s[0:i] after optimal segmentation.
Approach:
-
Dynamic Programming Definition:
- Let dp[i] be the minimum number of extra characters in the substring s[0:i].
- To calculate dp[i], we can:
- Either consider the character s[i-1] as an extra character and move to the next index.
- Or check if some substring ending at index i exists in the dictionary, and if it does, then use it to reduce extra characters.
-
Transition:
- For each index i, we either:
- Add one to dp[i-1] if we treat s[i] as an extra character.
- Check every possible substring s[j:i] (for j < i) and if s[j:i] is in the dictionary, we update dp[i] based on dp[j].
- For each index i, we either:
-
Result:
- The value of dp[len(s)] will give us the minimum number of extra characters in the entire string s.
Let's implement this solution in PHP: 2707. Extra Characters in a String
Explanation:
Base Case:
- dp[0] = 0 since no extra characters exist in an empty string.
Dictionary Lookup:
- We store the dictionary words in a hash map using array_flip() for constant-time lookup.
Transition:
- For each position i, we check all possible substrings s[j:i]. If a substring exists in the dictionary, we update the dp[i] value.
Time Complexity:
- The time complexity is O(n^2) where n is the length of the string s because for each index, we check all previous indices to form substrings.
Test Results:
For the input "leetscode" with dictionary ["leet","code","leetcode"], the function correctly returns 1, as only 1 extra character ("s") remains.
For the input "sayhelloworld" with dictionary ["hello","world"], the function returns 3, as the first three characters ("say") are extra.
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