524. Longest Word in Dictionary through Deleting

Description

Given a string s and a string array dictionary, return the longest string in the dictionary that can be formed by deleting some of the given string characters. If there is more than one possible result, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.

 

Example 1:

Input: s = "abpcplea", dictionary = ["ale","apple","monkey","plea"]
Output: "apple"

Example 2:

Input: s = "abpcplea", dictionary = ["a","b","c"]
Output: "a"

 

Constraints:

  • 1 <= s.length <= 1000
  • 1 <= dictionary.length <= 1000
  • 1 <= dictionary[i].length <= 1000
  • s and dictionary[i] consist of lowercase English letters.

Solutions

Solution 1

Python Code
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class Solution:
    def findLongestWord(self, s: str, dictionary: List[str]) -> str:
        def check(a, b):
            m, n = len(a), len(b)
            i = j = 0
            while i < m and j < n:
                if a[i] == b[j]:
                    j += 1
                i += 1
            return j == n

        ans = ''
        for a in dictionary:
            if check(s, a) and (len(ans) < len(a) or (len(ans) == len(a) and ans > a)):
                ans = a
        return ans

Java Code
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class Solution {
    public String findLongestWord(String s, List<String> dictionary) {
        String ans = "";
        for (String a : dictionary) {
            if (check(s, a)
                && (ans.length() < a.length()
                    || (ans.length() == a.length() && a.compareTo(ans) < 0))) {
                ans = a;
            }
        }
        return ans;
    }

    private boolean check(String a, String b) {
        int m = a.length(), n = b.length();
        int i = 0, j = 0;
        while (i < m && j < n) {
            if (a.charAt(i) == b.charAt(j)) {
                ++j;
            }
            ++i;
        }
        return j == n;
    }
}

C++ Code
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class Solution {
public:
    string findLongestWord(string s, vector<string>& dictionary) {
        string ans = "";
        for (string& a : dictionary)
            if (check(s, a) && (ans.size() < a.size() || (ans.size() == a.size() && a < ans)))
                ans = a;
        return ans;
    }

    bool check(string& a, string& b) {
        int m = a.size(), n = b.size();
        int i = 0, j = 0;
        while (i < m && j < n) {
            if (a[i] == b[j]) ++j;
            ++i;
        }
        return j == n;
    }
};

Go Code
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func findLongestWord(s string, dictionary []string) string {
	ans := ""
	check := func(a, b string) bool {
		m, n := len(a), len(b)
		i, j := 0, 0
		for i < m && j < n {
			if a[i] == b[j] {
				j++
			}
			i++
		}
		return j == n
	}
	for _, a := range dictionary {
		if check(s, a) && (len(ans) < len(a) || (len(ans) == len(a) && a < ans)) {
			ans = a
		}
	}
	return ans
}

TypeScript Code
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function findLongestWord(s: string, dictionary: string[]): string {
    dictionary.sort((a, b) => {
        if (a.length === b.length) {
            return b < a ? 1 : -1;
        }
        return b.length - a.length;
    });
    const n = s.length;
    for (const target of dictionary) {
        const m = target.length;
        if (m > n) {
            continue;
        }
        let i = 0;
        let j = 0;
        while (i < n && j < m) {
            if (s[i] === target[j]) {
                j++;
            }
            i++;
        }
        if (j === m) {
            return target;
        }
    }
    return '';
}

Rust Code
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impl Solution {
    pub fn find_longest_word(s: String, mut dictionary: Vec<String>) -> String {
        dictionary.sort_unstable_by(|a, b| (b.len(), a).cmp(&(a.len(), b)));
        for target in dictionary {
            let target: Vec<char> = target.chars().collect();
            let n = target.len();
            let mut i = 0;
            for c in s.chars() {
                if i == n {
                    break;
                }
                if c == target[i] {
                    i += 1;
                }
            }
            if i == n {
                return target.iter().collect();
            }
        }
        String::new()
    }
}