LeetCode #3291 — MEDIUM

Minimum Number of Valid Strings to Form Target I

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Dynamic Programming · Trie · Segment Tree · String Matching

Example 1

["abc","aaaaa","bcdef"]
"aabcdabc"

Example 2

["abababab","ab"]
"ababaababa"

Example 3

["abcdef"]
"xyz"

Core Insight

What unlocks the optimal approach

Let <code>dp[i]</code> be the minimum cost to form the prefix of length <code>i</code> of <code>target</code>.
If <code>target[(i + 1)..j]</code> matches any prefix, update the range <code>dp[(i + 1)..j]</code> to minimum between original value and <code>dp[i] + 1</code>.
Use a Trie to check prefix matching.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3291: Minimum Number of Valid Strings to Form Target I
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];
    }
}

// Accepted solution for LeetCode #3291: Minimum Number of Valid Strings to Form Target I
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
}

# Accepted solution for LeetCode #3291: Minimum Number of Valid Strings to Form Target I
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

// Accepted solution for LeetCode #3291: Minimum Number of Valid Strings to Form Target I
/**
 * [3291] Minimum Number of Valid Strings to Form Target I
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn min_valid_strings(words: Vec<String>, target: String) -> i32 {
        let words: Vec<Vec<char>> = words.into_iter().map(|x| x.chars().collect()).collect();
        let target: Vec<char> = target.chars().collect();

        let calculate_prefix = |word: &Vec<char>, target: &Vec<char>| {
            let s: Vec<char> = word
                .iter()
                .chain(['#'].iter())
                .chain(target.iter())
                .map(|x| x.clone())
                .collect();
            let mut result = vec![0; s.len()];

            for i in 1..s.len() {
                let mut j = result[i - 1];
                while j > 0 && s[i] != s[j] {
                    j = result[j - 1];
                }

                if s[i] == s[j] {
                    j += 1;
                }
                result[i] = j;
            }

            result
        };

        let n = target.len();
        let mut back = vec![0; n];
        for word in words.iter() {
            let pi = calculate_prefix(word, &target);
            let m = word.len();

            for i in 0..n {
                back[i] = back[i].max(pi[m + 1 + i]);
            }
        }

        let mut dp = vec![0; n + 1];

        for i in 1..=n {
            dp[i] = 1e9 as i32;
        }

        for i in 0..n {
            dp[i + 1] = dp[i + 1 - back[i]] + 1;
            if dp[i + 1] > n as i32 {
                return -1;
            }
        }

        dp[n]
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3291() {
        assert_eq!(
            3,
            Solution::min_valid_strings(
                vec_string!("abc", "aaaaa", "bcdef"),
                "aabcdabc".to_owned()
            )
        );
        assert_eq!(
            2,
            Solution::min_valid_strings(vec_string!("abababab", "ab"), "ababaababa".to_owned())
        );
        assert_eq!(
            -1,
            Solution::min_valid_strings(vec_string!("abcdef"), "xyz".to_owned())
        );
    }
}

// Accepted solution for LeetCode #3291: Minimum Number of Valid Strings to Form Target I
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2 + L)

Space

O(n + L)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.