[LeetCode] 2707. Extra Characters in a String

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

字符串中的额外字符。

给你一个下标从 0 开始的字符串 s 和一个单词字典 dictionary 。你需要将 s 分割成若干个 互不重叠 的子字符串，每个子字符串都在 dictionary 中出现过。s 中可能会有一些 额外的字符 不在任何子字符串中。

请你采取最优策略分割 s ，使剩下的字符 最少 。

思路是动态规划。这道题有些类似139题。思路可以直接参考139题，这道题 dp 的定义是长度为 [i, end) 的子串被分割后剩下的字符的个数。

DP自上而下

class Solution {
    int[] dp = new int[51];

    public int minExtraChar(String s, String[] dictionary) {
        Arrays.fill(dp, -1);
        return helper(s, dictionary, 0);
    }

    private int helper(String s, String[] dictionary, int i) {
        if (i == s.length()) {
            return 0;
        }
        if (dp[i] == -1) {
            dp[i] = 1 + helper(s, dictionary, i + 1);
            for (String w : dictionary) {
                if (i + w.length() <= s.length() && s.substring(i, i + w.length()).equals(w)) {
                    dp[i] = Math.min(dp[i], helper(s, dictionary, i + w.length()));
                }
            }
        }
        return dp[i];
    }
}

 

DP自下而上

class Solution {
    public int minExtraChar(String s, String[] dictionary) {
        int[] dp = new int[51];
        int n = s.length();
        for (int i = n - 1; i >= 0; i--) {
            dp[i] = 1 + dp[i + 1];
            for (String w : dictionary) {
                if (i + w.length() <= s.length() && s.substring(i, i + w.length()).equals(w)) {
                    dp[i] = Math.min(dp[i], dp[i + w.length()]);
                }
            }
        }
        return dp[0];
    }
}

 

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