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:

1. Can we use Dynamic Programming here?
2. 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:

1. 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.

2. 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].

3. 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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