3291. Minimum Number of Valid Strings to Form Target I

Description

You are given an array of strings words and a string target.

A string x is called valid if x is a prefix of any string in words.

Return the minimum number of valid strings that can be concatenated to form target. If it is not possible to form target, return -1.

Example 1:

Input: words = ["abc","aaaaa","bcdef"], target = "aabcdabc"

Output: 3

Explanation:

The target string can be formed by concatenating:

Prefix of length 2 of words[1], i.e. "aa".
Prefix of length 3 of words[2], i.e. "bcd".
Prefix of length 3 of words[0], i.e. "abc".

Example 2:

Input: words = ["abababab","ab"], target = "ababaababa"

Output: 2

Explanation:

The target string can be formed by concatenating:

Prefix of length 5 of words[0], i.e. "ababa".
Prefix of length 5 of words[0], i.e. "ababa".

Example 3:

Input: words = ["abcdef"], target = "xyz"

Output: -1

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 5 * 10³
The input is generated such that sum(words[i].length) <= 10⁵.
words[i] consists only of lowercase English letters.
1 <= target.length <= 5 * 10³
target consists only of lowercase English letters.

Solutions

Solution 1: Trie + Memoization

We can use a trie to store all valid strings and then use memoization to calculate the answer.

We design a function $\textit{dfs}(i)$, which represents the minimum number of strings needed to concatenate starting from the $i$-th character of the string $\textit{target}$. The answer is $\textit{dfs}(0)$.

The function $\textit{dfs}(i)$ is calculated as follows:

If $i \geq n$, it means the string $\textit{target}$ has been completely traversed, so we return $0$;
Otherwise, we can find valid strings in the trie that start with $\textit{target}[i]$, and then recursively calculate $\textit{dfs}(i + \text{len}(w))$, where $w$ is the valid string found. We take the minimum of these values and add $1$ as the return value of $\textit{dfs}(i)$.

To avoid redundant calculations, we use memoization.

The time complexity is $O(n^2 + L)$, and the space complexity is $O(n + L)$. Here, $n$ is the length of the string $\textit{target}$, and $L$ is the total length of all valid strings.

Python3JavaC++GoTypeScript

def min(a: int, b: int) -> int:
    return a if a < b else b


class Trie:
    def __init__(self):
        self.children: List[Optional[Trie]] = [None] * 26

    def insert(self, w: str):
        node = self
        for i in map(lambda c: ord(c) - 97, w):
            if node.children[i] is None:
                node.children[i] = Trie()
            node = node.children[i]


class Solution:
    def minValidStrings(self, words: List[str], target: str) -> int:
        @cache
        def dfs(i: int) -> int:
            if i >= n:
                return 0
            node = trie
            ans = inf
            for j in range(i, n):
                k = ord(target[j]) - 97
                if node.children[k] is None:
                    break
                node = node.children[k]
                ans = min(ans, 1 + dfs(j + 1))
            return ans

        trie = Trie()
        for w in words:
            trie.insert(w)
        n = len(target)
        ans = dfs(0)
        return ans if ans < inf else -1

class Trie {
    Trie[] children = new Trie[26];

    void insert(String w) {
        Trie node = this;
        for (int i = 0; i < w.length(); ++i) {
            int j = w.charAt(i) - 'a';
            if (node.children[j] == null) {
                node.children[j] = new Trie();
            }
            node = node.children[j];
        }
    }
}

class Solution {
    private Integer[] f;
    private char[] s;
    private Trie trie;
    private final int inf = 1 << 30;

    public int minValidStrings(String[] words, String target) {
        trie = new Trie();
        for (String w : words) {
            trie.insert(w);
        }
        s = target.toCharArray();
        f = new Integer[s.length];
        int ans = dfs(0);
        return ans < inf ? ans : -1;
    }

    private int dfs(int i) {
        if (i >= s.length) {
            return 0;
        }
        if (f[i] != null) {
            return f[i];
        }
        Trie node = trie;
        f[i] = inf;
        for (int j = i; j < s.length; ++j) {
            int k = s[j] - 'a';
            if (node.children[k] == null) {
                break;
            }
            f[i] = Math.min(f[i], 1 + dfs(j + 1));
            node = node.children[k];
        }
        return f[i];
    }
}

class Trie {
public:
    Trie* children[26]{};

    void insert(string& word) {
        Trie* node = this;
        for (char& c : word) {
            int i = c - 'a';
            if (!node->children[i]) {
                node->children[i] = new Trie();
            }
            node = node->children[i];
        }
    }
};

class Solution {
public:
    int minValidStrings(vector<string>& words, string target) {
        int n = target.size();
        Trie* trie = new Trie();
        for (auto& w : words) {
            trie->insert(w);
        }
        const int inf = 1 << 30;
        int f[n];
        memset(f, -1, sizeof(f));
        auto dfs = [&](this auto&& dfs, int i) -> int {
            if (i >= n) {
                return 0;
            }
            if (f[i] != -1) {
                return f[i];
            }
            f[i] = inf;
            Trie* node = trie;
            for (int j = i; j < n; ++j) {
                int k = target[j] - 'a';
                if (!node->children[k]) {
                    break;
                }
                node = node->children[k];
                f[i] = min(f[i], 1 + dfs(j + 1));
            }
            return f[i];
        };
        int ans = dfs(0);
        return ans < inf ? ans : -1;
    }
};

type Trie struct {
    children [26]*Trie
}

func (t *Trie) insert(word string) {
    node := t
    for _, c := range word {
        idx := c - 'a'
        if node.children[idx] == nil {
            node.children[idx] = &Trie{}
        }
        node = node.children[idx]
    }
}

func minValidStrings(words []string, target string) int {
    n := len(target)
    trie := &Trie{}
    for _, w := range words {
        trie.insert(w)
    }
    const inf int = 1 << 30
    f := make([]int, n)
    var dfs func(int) int
    dfs = func(i int) int {
        if i >= n {
            return 0
        }
        if f[i] != 0 {
            return f[i]
        }
        node := trie
        f[i] = inf
        for j := i; j < n; j++ {
            k := int(target[j] - 'a')
            if node.children[k] == nil {
                break
            }
            f[i] = min(f[i], 1+dfs(j+1))
            node = node.children[k]
        }
        return f[i]
    }
    if ans := dfs(0); ans < inf {
        return ans
    }
    return -1
}

class Trie {
    children: (Trie | null)[] = Array(26).fill(null);

    insert(word: string): void {
        let node: Trie = this;
        for (const c of word) {
            const i = c.charCodeAt(0) - 'a'.charCodeAt(0);
            if (!node.children[i]) {
                node.children[i] = new Trie();
            }
            node = node.children[i];
        }
    }
}

function minValidStrings(words: string[], target: string): number {
    const n = target.length;
    const trie = new Trie();
    for (const w of words) {
        trie.insert(w);
    }
    const inf = 1 << 30;
    const f = Array(n).fill(0);

    const dfs = (i: number): number => {
        if (i >= n) {
            return 0;
        }
        if (f[i]) {
            return f[i];
        }
        f[i] = inf;
        let node: Trie | null = trie;
        for (let j = i; j < n; ++j) {
            const k = target[j].charCodeAt(0) - 'a'.charCodeAt(0);
            if (!node?.children[k]) {
                break;
            }
            node = node.children[k];
            f[i] = Math.min(f[i], 1 + dfs(j + 1));
        }
        return f[i];
    };

    const ans = dfs(0);
    return ans < inf ? ans : -1;
}

3291. Minimum Number of Valid Strings to Form Target I

Description

Solutions

Solution 1: Trie + Memoization

Comments