g2701_2800.s2707_extra_characters_in_a_string.Solution Maven / Gradle / Ivy

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package g2701_2800.s2707_extra_characters_in_a_string;

// #Medium #Array #String #Hash_Table #Dynamic_Programming #Trie
// #2023_09_15_Time_37_ms_(74.40%)_Space_44.2_MB_(74.60%)

import java.util.Arrays;

/**
 * 2707 - Extra Characters in a String\.
 *
 * 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]` and `s` consists of only lowercase English letters
 * *   `dictionary` contains distinct words
**/
public class Solution {
    public int minExtraChar(String s, String[] dictionary) {
        return tabulationApproach(s, dictionary);
    }

    private int tabulationApproach(String s, String[] dictionary) {
        int m = s.length();
        int[] dp = new int[m + 1];
        for (int i = m - 1; i >= 0; i--) {
            dp[i] = dp[i + 1] + 1;
            int finalI = i;
            Arrays.stream(dictionary)
                    .filter(word -> s.startsWith(word, finalI))
                    .mapToInt(String::length)
                    .map(n -> dp[finalI + n])
                    .forEach(prev -> dp[finalI] = Math.min(dp[finalI], prev));
        }
        return dp[0];
    }
}