Problem Description

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

A string x is considered valid if it is a prefix of at least one string in the words array. For example, if words = ["abc", "def"], then "a", "ab", "abc", "d", "de", "def" are all valid strings, but "b", "ac", "e" are not valid.

Your task is to find the minimum number of valid strings that can be concatenated together to form the target string. The valid strings can be used multiple times and in any order.

If it's impossible to form the target string using valid strings, return -1.

For example:

• If words = ["abc", "ab", "def"] and target = "abcdef", you could use "abc" + "def" (2 valid strings) to form the target.
• If words = ["a", "abc"] and target = "aaa", you could use "a" + "a" + "a" (3 valid strings) to form the target.
• If words = ["abc"] and target = "def", it's impossible to form the target since no valid string can match any part of "def", so return -1.

The problem requires finding the optimal (minimum) way to partition the target string into segments where each segment is a valid string (prefix of some word in words).

Intuition

The key insight is that we need to greedily match the longest possible prefix at each position to minimize the total number of strings used.

Think of it like jumping across a river using stepping stones. At each position in the target string, we want to make the longest possible "jump" using a valid prefix. The longer each jump, the fewer total jumps we need.

For any position i in the target string, we need to find the maximum length substring starting at i that is a valid prefix of some word in words. Let's call this length f(i). For example, if at position i we can match prefixes of length 1, 2, and 5, then f(i) = 5.

The challenge is efficiently finding f(i) for each position. Since we're looking for prefixes, there's a monotonic property: if target[i..i+k] is a valid prefix, then target[i..i+j] is also valid for all j ≤ k (because a prefix of a prefix is also a prefix). This monotonicity suggests we can use binary search to find the maximum matching length.

To quickly check if a substring is a valid prefix, we can use string hashing. We precompute hash values for all prefixes of words in words and group them by length. Then, for any substring of target, we can compute its hash value and check if it exists in our precomputed set in O(1) time.

Once we can efficiently compute f(i) for each position, the problem becomes a classic greedy jumping game: starting from position 0, at each step we can jump to any position within our current reach. We want to reach the end with minimum jumps. The strategy is to always extend our reach as far as possible before making the next jump.

The algorithm tracks:

• last: the position where we made our last jump
• mx: the farthest position we can reach from our current range
• When we reach last, we must make another jump to mx

If at any point last == mx and we haven't reached the end, it means we're stuck and cannot proceed further, so we return -1.

Learn more about Segment Tree, Binary Search and Dynamic Programming patterns.

Solution Approach

The solution implements the intuition using string hashing, binary search, and greedy strategy. Let's walk through each component:

1. String Hashing Setup

First, we create a Hashing class to efficiently compute hash values of any substring of target:

• Uses polynomial rolling hash with base 13331 and modulo 998244353
• Precomputes prefix hashes h[i] and powers p[i] for the target string
• The query(l, r) method returns the hash value of target[l-1..r-1] in O(1) time using the formula: (h[r] - h[l-1] * p[r-l+1]) % mod

2. Preprocessing Word Prefixes

We build a data structure to quickly check if a substring is a valid prefix:

s = [set() for _ in range(m + 1)]  # s[i] contains hash values of all prefixes of length i

For each word in words, we compute hash values of all its prefixes and store them in s grouped by length:

• For word "abc", we store hash("a") in s[1], hash("ab") in s[2], hash("abc") in s[3]
• This allows O(1) lookup to check if a substring of given length is a valid prefix

3. Binary Search for Maximum Match Length

The function f(i) finds the maximum length of valid prefix starting at position i:

def f(i: int) -> int:
    l, r = 0, min(n - i, m)  # search range: [0, min(remaining_length, max_word_length)]
    while l < r:
        mid = (l + r + 1) >> 1
        sub = hashing.query(i + 1, i + mid)  # get hash of target[i..i+mid-1]
        if sub in s[mid]:  # check if this substring is a valid prefix of length mid
            l = mid  # can match at least mid characters, try longer
        else:
            r = mid - 1  # cannot match mid characters, try shorter
    return l

The binary search exploits the monotonic property: if we can match k characters, we can also match any j < k characters.

4. Greedy Jump Strategy

With f(i) computed for each position, we solve the minimum jumps problem:

ans = last = mx = 0
for i in range(n):
    dist = f(i)
    mx = max(mx, i + dist)  # update the farthest reachable position
  
    if i == last:  # reached the end of current jump range
        if i == mx:  # cannot move further
            return -1
        last = mx  # make a new jump to the farthest position
        ans += 1

The algorithm maintains:

• last: the position where we must make the next jump
• mx: the farthest position reachable from current range
• ans: the number of jumps made

At each position i, we:

1. Calculate how far we can jump from i using f(i)
2. Update the maximum reachable position mx
3. If we reach last, we must jump again:
   • If last == mx, we're stuck (return -1)
   • Otherwise, jump to mx and increment the answer

Time Complexity

• Preprocessing: O(L) where L is the total length of all words
• For each position in target: O(log m) for binary search with O(1) hash checks
• Total: O(L + n log m) where n is the length of target and m is the maximum word length

Space Complexity

• O(L) for storing hash values of all prefixes
• O(n) for the hashing data structure of target

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example: words = ["abc", "ab", "def"] and target = "abdef"

Step 1: Preprocessing - Build Hash Sets for Valid Prefixes

First, we compute hash values for all prefixes of words in words:

• From "abc": hash("a") → s[1], hash("ab") → s[2], hash("abc") → s[3]
• From "ab": hash("a") → s[1], hash("ab") → s[2]
• From "def": hash("d") → s[1], hash("de") → s[2], hash("def") → s[3]

Our hash sets become:

• s[1] = {hash("a"), hash("d")}
• s[2] = {hash("ab"), hash("de")}
• s[3] = {hash("abc"), hash("def")}

Step 2: Compute Maximum Match Length at Each Position

For each position i in target = "abdef", we use binary search to find f(i):

• i = 0 (looking at "abdef"):

  • Binary search range: [0, 5]
  • Check mid=3: "abd" → NOT in s[3] → search [0, 2]
  • Check mid=2: "ab" → IS in s[2] → search [2, 2]
  • Result: f(0) = 2 (can match "ab")

• i = 1 (looking at "bdef"):

  • Binary search range: [0, 4]
  • Check mid=2: "bd" → NOT in s[2] → search [0, 1]
  • Check mid=1: "b" → NOT in s[1] → search [0, 0]
  • Result: f(1) = 0 (no valid prefix starts with "b")

• i = 2 (looking at "def"):

  • Binary search range: [0, 3]
  • Check mid=2: "de" → IS in s[2] → search [2, 3]
  • Check mid=3: "def" → IS in s[3] → search [3, 3]
  • Result: f(2) = 3 (can match "def")

• i = 3 (looking at "ef"):

  • Binary search range: [0, 2]
  • Check mid=1: "e" → NOT in s[1] → search [0, 0]
  • Result: f(3) = 0 (no valid prefix starts with "e")

• i = 4 (looking at "f"):

  • Binary search range: [0, 1]
  • Check mid=1: "f" → NOT in s[1] → search [0, 0]
  • Result: f(4) = 0 (no valid prefix starts with "f")

Step 3: Greedy Jump Strategy

Now we simulate the jumping process:

Initial state: ans = 0, last = 0, mx = 0

• i = 0:

  • f(0) = 2, so from position 0 we can reach position 0+2 = 2
  • Update: mx = max(0, 2) = 2
  • Since i == last (both 0), we need to jump:
    • last ≠ mx (0 ≠ 2), so we can proceed
    • Update: last = 2, ans = 1
    • We've made our first jump, using "ab" to cover positions [0,1]

• i = 1:

  • f(1) = 0, so from position 1 we can't jump anywhere
  • mx remains 2
  • i < last (1 < 2), so no jump needed yet

• i = 2:

  • f(2) = 3, so from position 2 we can reach position 2+3 = 5
  • Update: mx = max(2, 5) = 5
  • Since i == last (both 2), we need to jump:
    • last ≠ mx (2 ≠ 5), so we can proceed
    • Update: last = 5, ans = 2
    • We've made our second jump, using "def" to cover positions [2,4]

• i = 3 and i = 4:

  • Both have f(i) = 0
  • mx remains 5
  • i < last, so no jumps needed

Final result: We've reached the end (position 5) with ans = 2 jumps.

The solution correctly identifies that we need 2 valid strings: "ab" + "def" = "abdef"

Key Insights from the Example:

1. The binary search efficiently finds the longest valid prefix at each position
2. The greedy strategy ensures we make the minimum number of jumps by always extending our reach as far as possible
3. Positions where f(i) = 0 (like positions 1, 3, 4) don't affect the result as long as they're covered by previous jumps

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n + L)

The time complexity breaks down as follows:

• Building the hash sets for all prefixes of words takes O(L) time, where L is the total length of all words
• The main loop iterates through each position in the target string (n iterations)
• For each position i, the function f(i) performs a binary search:
  • The binary search range is from 0 to min(n - i, m), where m is the maximum word length
  • Each binary search iteration calls hashing.query() which runs in O(1) time
  • The binary search takes O(log min(n - i, m)) which is bounded by O(log n)
• Therefore, the total time for all binary searches is O(n × log n)
• Overall time complexity: O(n × log n + L)

Space Complexity: O(n + L)

The space complexity consists of:

• The Hashing class stores two arrays h and p, each of size n + 1, using O(n) space
• The hash sets s store hash values for all prefixes of all words:
  • There are at most m + 1 sets
  • The total number of hash values stored across all sets is bounded by the total length of all words L
  • This uses O(L) space
• Other variables (ans, last, mx, etc.) use O(1) space
• Overall space complexity: O(n + L)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Off-by-One Error in Binary Search Range

A common mistake is incorrectly setting the binary search range when finding the maximum prefix match. The issue often occurs when determining the upper bound.

Pitfall Code:

def find_max_prefix_match(start_index: int) -> int:
    # WRONG: May cause index out of bounds
    left, right = 1, min(target_length - start_index, max_word_length)
    # Starting from 1 assumes we always have at least 1 character match

Why it's wrong:

• Starting left from 1 assumes we can always match at least one character, but this isn't guaranteed
• If no prefix matches at position start_index, the function would incorrectly return 1

Solution:

def find_max_prefix_match(start_index: int) -> int:
    # CORRECT: Start from 0 to handle no-match case
    left, right = 0, min(target_length - start_index, max_word_length)

2. Hash Collision Not Considered

While the code uses a large prime modulo (998244353), hash collisions can still occur in edge cases, leading to false positive matches.

Pitfall: Relying on a single hash function without verification.

Solution: Use double hashing or verify matches with actual string comparison:

# Enhanced version with double hashing
class DoubleHashing:
    def __init__(self, s, base1=13331, mod1=998244353, base2=31, mod2=10**9+7):
        self.hash1 = Hashing(s, base1, mod1)
        self.hash2 = Hashing(s, base2, mod2)
  
    def query(self, l, r):
        return (self.hash1.query(l, r), self.hash2.query(l, r))

3. Incorrect Greedy Jump Logic

A subtle but critical error occurs when updating the jump position without properly checking if progress is being made.

Pitfall Code:

for current_pos in range(target_length):
    match_length = find_max_prefix_match(current_pos)
    max_reachable = max(max_reachable, current_pos + match_length)
  
    if current_pos == last_covered:
        # WRONG: Not checking if we're stuck before incrementing answer
        last_covered = max_reachable
        answer += 1

Why it's wrong:

• This would increment answer even when max_reachable equals current_pos (no progress)
• Results in returning a positive number instead of -1 for impossible cases

Solution:

if current_pos == last_covered:
    # CORRECT: Check if we're stuck first
    if current_pos == max_reachable:
        return -1  # Cannot progress further
    last_covered = max_reachable
    answer += 1

4. Empty Words or Target Edge Cases

The code doesn't explicitly handle edge cases like empty strings in the words array or empty target.

Pitfall: Not validating input can cause unexpected behavior.

Solution:

def minValidStrings(self, words: List[str], target: str) -> int:
    # Add input validation
    if not target:
        return 0
    if not words or all(not word for word in words):
        return -1
  
    # Filter out empty words
    words = [word for word in words if word]
    if not words:
        return -1
  
    # Continue with main logic...

5. Integer Overflow in Hash Computation

When computing hash values for very long strings, intermediate calculations might overflow even with modulo operations.

Pitfall Code:

# In preprocessing
hash_value = (hash_value * BASE + ord(char)) % MOD

Why it's a concern:

• Python handles big integers automatically, but in other languages this could overflow
• Even in Python, very large intermediate values can slow down computation

Solution:

# More careful modulo application
hash_value = ((hash_value * BASE) % MOD + ord(char)) % MOD

These pitfalls highlight the importance of careful boundary condition handling, proper validation, and understanding the nuances of the greedy approach in this problem.