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2707. Extra Characters in a String

Description

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

Solutions

  • class Solution {
    public int minExtraChar(String s, String[] dictionary) {
        Set<String> ss = new HashSet<>();
        for (String w : dictionary) {
            ss.add(w);
        }
        int n = s.length();
        int[] f = new int[n + 1];
        f[0] = 0;
        for (int i = 1; i <= n; ++i) {
            f[i] = f[i - 1] + 1;
            for (int j = 0; j < i; ++j) {
                if (ss.contains(s.substring(j, i))) {
                    f[i] = Math.min(f[i], f[j]);
                }
            }
        }
        return f[n];
    }
}

  • class Solution {
public:
    int minExtraChar(string s, vector<string>& dictionary) {
        unordered_set<string> ss(dictionary.begin(), dictionary.end());
        int n = s.size();
        int f[n + 1];
        f[0] = 0;
        for (int i = 1; i <= n; ++i) {
            f[i] = f[i - 1] + 1;
            for (int j = 0; j < i; ++j) {
                if (ss.count(s.substr(j, i - j))) {
                    f[i] = min(f[i], f[j]);
                }
            }
        }
        return f[n];
    }
};

  • class Solution:
    def minExtraChar(self, s: str, dictionary: List[str]) -> int:
        ss = set(dictionary)
        n = len(s)
        f = [0] * (n + 1)
        for i in range(1, n + 1):
            f[i] = f[i - 1] + 1
            for j in range(i):
                if s[j:i] in ss and f[j] < f[i]:
                    f[i] = f[j]
        return f[n]


  • func minExtraChar(s string, dictionary []string) int {
	ss := map[string]bool{}
	for _, w := range dictionary {
		ss[w] = true
	}
	n := len(s)
	f := make([]int, n+1)
	for i := 1; i <= n; i++ {
		f[i] = f[i-1] + 1
		for j := 0; j < i; j++ {
			if ss[s[j:i]] && f[j] < f[i] {
				f[i] = f[j]
			}
		}
	}
	return f[n]
}

  • function minExtraChar(s: string, dictionary: string[]): number {
    const ss = new Set(dictionary);
    const n = s.length;
    const f = new Array(n + 1).fill(0);
    for (let i = 1; i <= n; ++i) {
        f[i] = f[i - 1] + 1;
        for (let j = 0; j < i; ++j) {
            if (ss.has(s.substring(j, i))) {
                f[i] = Math.min(f[i], f[j]);
            }
        }
    }
    return f[n];
}


  • use std::collections::HashSet;

impl Solution {
    #[allow(dead_code)]
    pub fn min_extra_char(s: String, dictionary: Vec<String>) -> i32 {
        let n = s.len();
        let mut set = dictionary
            .iter()
            .map(|s| s.into())
            .collect::<HashSet<String>>();
        let mut dp = vec![0; n + 1];

        // Initialize the dp vector
        dp[0] = 0;

        // Begin the actual dp process
        for i in 1..=n {
            dp[i] = dp[i - 1] + 1;
            for j in 0..i {
                if set.contains(&s[j..i]) {
                    dp[i] = std::cmp::min(dp[i], dp[j]);
                }
            }
        }

        dp[n]
    }
}

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