2707. Extra Characters in a String

Problem

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

Solution (Java)

class Solution {
    public int minExtraChar(String s, String[] dictionary) {
        Set<String> dict = new HashSet<>();
        for (String word : dictionary) {
            dict.add(word);
        }

        int n = s.length();
        int[] dp = new int[n + 1];

        for (int i = 1; i <= n; i++) {
            dp[i] = dp[i - 1] + 1;

            for (int j = i - 1; j >= 0; j--) {
                String substring = s.substring(j, i);
                if (dict.contains(substring)) {
                    dp[i] = Math.min(dp[i], dp[j]);
                }
            }
        }

        return dp[n];
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).