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3292. Minimum Number of Valid Strings to Form Target II

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 * 104
  • The input is generated such that sum(words[i].length) <= 105.
  • words[i] consists only of lowercase English letters.
  • 1 <= target.length <= 5 * 104
  • target consists only of lowercase English letters.

Solutions

Solution 1: String Hashing + Binary Search + Greedy

Due to the large data scale of this problem, using the "Trie + Memoization" method will time out. We need to find a more efficient solution.

Consider starting from the $i$-th character of the string $\textit{target}$ and finding the maximum matching substring length, denoted as $\textit{dist}$. For any $j \in [i, i + \textit{dist} - 1]$, we can find a string in $\textit{words}$ such that $\textit{target}[i..j]$ is a prefix of this string. This has a monotonic property, so we can use binary search to determine $\textit{dist}$.

Specifically, we first preprocess the hash values of all prefixes of strings in $\textit{words}$ and store them in the array $\textit{s}$ grouped by prefix length. Additionally, we preprocess the hash values of $\textit{target}$ and store them in $\textit{hashing}$ to facilitate querying the hash value of any $\textit{target}[l..r]$.

Next, we design a function $\textit{f}(i)$ to represent the maximum matching substring length starting from the $i$-th character of the string $\textit{target}$. We can determine $\textit{f}(i)$ using binary search.

Define the left boundary of the binary search as $l = 0$ and the right boundary as $r = \min(n - i, m)$, where $n$ is the length of the string $\textit{target}$ and $m$ is the maximum length of strings in $\textit{words}$. During the binary search, we need to check if $\textit{target}[i..i+\textit{mid}-1]$ is one of the hash values in $\textit{s}[\textit{mid}]$. If it is, update the left boundary $l$ to $\textit{mid}$; otherwise, update the right boundary $r$ to $\textit{mid} - 1$. After the binary search, return $l$.

After calculating $\textit{f}(i)$, the problem becomes a classic greedy problem. Starting from $i = 0$, for each position $i$, the farthest position we can move to is $i + \textit{f}(i)$. We need to find the minimum number of moves to reach the end.

We define $\textit{last}$ to represent the last moved position and $\textit{mx}$ to represent the farthest position we can move to from the current position. Initially, $\textit{last} = \textit{mx} = 0$. We traverse from $i = 0$. If $i$ equals $\textit{last}$, it means we need to move again. If $\textit{last} = \textit{mx}$, it means we cannot move further, so we return $-1$. Otherwise, we update $\textit{last}$ to $\textit{mx}$ and increment the answer by one.

After the traversal, return the answer.

The time complexity is $O(n \times \log n + L)$, 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.

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class Hashing:
    __slots__ = ["mod", "h", "p"]

    def __init__(self, s: List[str], base: int, mod: int):
        self.mod = mod
        self.h = [0] * (len(s) + 1)
        self.p = [1] * (len(s) + 1)
        for i in range(1, len(s) + 1):
            self.h[i] = (self.h[i - 1] * base + ord(s[i - 1])) % mod
            self.p[i] = (self.p[i - 1] * base) % mod

    def query(self, l: int, r: int) -> int:
        return (self.h[r] - self.h[l - 1] * self.p[r - l + 1]) % self.mod


class Solution:
    def minValidStrings(self, words: List[str], target: str) -> int:
        def f(i: int) -> int:
            l, r = 0, min(n - i, m)
            while l < r:
                mid = (l + r + 1) >> 1
                sub = hashing.query(i + 1, i + mid)
                if sub in s[mid]:
                    l = mid
                else:
                    r = mid - 1
            return l

        base, mod = 13331, 998244353
        hashing = Hashing(target, base, mod)
        m = max(len(w) for w in words)
        s = [set() for _ in range(m + 1)]
        for w in words:
            h = 0
            for j, c in enumerate(w, 1):
                h = (h * base + ord(c)) % mod
                s[j].add(h)
        ans = last = mx = 0
        n = len(target)
        for i in range(n):
            dist = f(i)
            mx = max(mx, i + dist)
            if i == last:
                if i == mx:
                    return -1
                last = mx
                ans += 1
        return ans

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class Hashing {
    private final long[] p;
    private final long[] h;
    private final long mod;

    public Hashing(String word, long base, int mod) {
        int n = word.length();
        p = new long[n + 1];
        h = new long[n + 1];
        p[0] = 1;
        this.mod = mod;
        for (int i = 1; i <= n; i++) {
            p[i] = p[i - 1] * base % mod;
            h[i] = (h[i - 1] * base + word.charAt(i - 1)) % mod;
        }
    }

    public long query(int l, int r) {
        return (h[r] - h[l - 1] * p[r - l + 1] % mod + mod) % mod;
    }
}

class Solution {
    private Hashing hashing;
    private Set<Long>[] s;

    public int minValidStrings(String[] words, String target) {
        int base = 13331, mod = 998244353;
        hashing = new Hashing(target, base, mod);
        int m = Arrays.stream(words).mapToInt(String::length).max().orElse(0);
        s = new Set[m + 1];
        Arrays.setAll(s, k -> new HashSet<>());
        for (String w : words) {
            long h = 0;
            for (int j = 0; j < w.length(); j++) {
                h = (h * base + w.charAt(j)) % mod;
                s[j + 1].add(h);
            }
        }

        int ans = 0;
        int last = 0;
        int mx = 0;
        int n = target.length();
        for (int i = 0; i < n; i++) {
            int dist = f(i, n, m);
            mx = Math.max(mx, i + dist);
            if (i == last) {
                if (i == mx) {
                    return -1;
                }
                last = mx;
                ans++;
            }
        }
        return ans;
    }

    private int f(int i, int n, int m) {
        int l = 0, r = Math.min(n - i, m);
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            long sub = hashing.query(i + 1, i + mid);
            if (s[mid].contains(sub)) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }
}

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class Hashing {
private:
    vector<long long> p;
    vector<long long> h;
    long long mod;

public:
    Hashing(const string& word, long long base, int mod) {
        int n = word.size();
        p.resize(n + 1);
        h.resize(n + 1);
        p[0] = 1;
        this->mod = mod;
        for (int i = 1; i <= n; i++) {
            p[i] = (p[i - 1] * base) % mod;
            h[i] = (h[i - 1] * base + word[i - 1]) % mod;
        }
    }

    long long query(int l, int r) {
        return (h[r] - h[l - 1] * p[r - l + 1] % mod + mod) % mod;
    }
};

class Solution {
public:
    int minValidStrings(vector<string>& words, string target) {
        int base = 13331, mod = 998244353;
        Hashing hashing(target, base, mod);
        int m = 0, n = target.size();
        for (const string& word : words) {
            m = max(m, (int) word.size());
        }

        vector<unordered_set<long long>> s(m + 1);
        for (const string& w : words) {
            long long h = 0;
            for (int j = 0; j < w.size(); j++) {
                h = (h * base + w[j]) % mod;
                s[j + 1].insert(h);
            }
        }

        auto f = [&](int i) -> int {
            int l = 0, r = min(n - i, m);
            while (l < r) {
                int mid = (l + r + 1) >> 1;
                long long sub = hashing.query(i + 1, i + mid);
                if (s[mid].count(sub)) {
                    l = mid;
                } else {
                    r = mid - 1;
                }
            }
            return l;
        };

        int ans = 0, last = 0, mx = 0;
        for (int i = 0; i < n; i++) {
            int dist = f(i);
            mx = max(mx, i + dist);
            if (i == last) {
                if (i == mx) {
                    return -1;
                }
                last = mx;
                ans++;
            }
        }
        return ans;
    }
};

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type Hashing struct {
    p   []int64
    h   []int64
    mod int64
}

func NewHashing(word string, base int64, mod int64) *Hashing {
    n := len(word)
    p := make([]int64, n+1)
    h := make([]int64, n+1)
    p[0] = 1
    for i := 1; i <= n; i++ {
        p[i] = (p[i-1] * base) % mod
        h[i] = (h[i-1]*base + int64(word[i-1])) % mod
    }
    return &Hashing{p, h, mod}
}

func (hashing *Hashing) Query(l, r int) int64 {
    return (hashing.h[r] - hashing.h[l-1]*hashing.p[r-l+1]%hashing.mod + hashing.mod) % hashing.mod
}

func minValidStrings(words []string, target string) (ans int) {
    base, mod := int64(13331), int64(998244353)
    hashing := NewHashing(target, base, mod)

    m, n := 0, len(target)
    for _, w := range words {
        m = max(m, len(w))
    }

    s := make([]map[int64]bool, m+1)

    f := func(i int) int {
        l, r := 0, int(math.Min(float64(n-i), float64(m)))
        for l < r {
            mid := (l + r + 1) >> 1
            sub := hashing.Query(i+1, i+mid)
            if s[mid][sub] {
                l = mid
            } else {
                r = mid - 1
            }
        }
        return l
    }

    for _, w := range words {
        h := int64(0)
        for j := 0; j < len(w); j++ {
            h = (h*base + int64(w[j])) % mod
            if s[j+1] == nil {
                s[j+1] = make(map[int64]bool)
            }
            s[j+1][h] = true
        }
    }

    var last, mx int

    for i := 0; i < n; i++ {
        dist := f(i)
        mx = max(mx, i+dist)
        if i == last {
            if i == mx {
                return -1
            }
            last = mx
            ans++
        }
    }
    return ans
}

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