How do you solve Minimum Number of Valid Strings to Form Target I on LeetCode?

Minimum Number of Valid Strings to Form Target I (LeetCode #3291) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n) space complexity. Key insight: Using dynamic programming allows us to efficiently compute the minimum valid strings needed without redundant checks.

What is the time complexity of Minimum Number of Valid Strings to Form Target I?

The optimal solution for Minimum Number of Valid Strings to Form Target I runs in O(n * m) time and uses O(n) space. The time complexity is O(n * m) where n is the length of the target and m is the average length of the words, as we check each word for each position in the target.

What is the brute force solution for Minimum Number of Valid Strings to Form Target I?

The brute force approach for Minimum Number of Valid Strings to Form Target I is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will try to build the target string by checking all possible prefixes of the words array. We will keep track of how many valid strings we need to concatenate to form the target, which can be inefficient. The optimal Optimal Solution approach improves this to O(n * m).

What are the interview tips for Minimum Number of Valid Strings to Form Target I?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as when the target cannot be formed at all. Explain your thought process as you code; it shows your understanding and can earn you points even if you make a mistake.

What are common mistakes when solving Minimum Number of Valid Strings to Form Target I?

Failing to initialize the dp array correctly, leading to incorrect results. Not considering all possible prefixes when checking for matches.

#3291

Minimum Number of Valid Strings to Form Target I

Medium

ArrayStringBinary SearchDynamic ProgrammingGreedyTrieSegment TreeRolling HashString MatchingHash FunctionDynamic ProgrammingPrefix Matching

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n)

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Intuition

Time O(n * m)Space O(n)

In the optimal solution, we utilize dynamic programming to keep track of the minimum number of valid strings needed to form each prefix of the target string. This approach is more efficient as it avoids redundant checks.

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Algorithm

5 steps

1Step 1: Initialize a dp array where dp[i] represents the minimum number of valid strings needed to form the prefix of target of length i.
2Step 2: Set dp[0] = 0 (no strings needed to form an empty prefix) and all other values to infinity.
3Step 3: For each position in the target, check all words to see if any prefix can match the substring starting from the current position.
4Step 4: If a match is found, update the dp array to reflect the minimum number of valid strings needed.
5Step 5: Return dp[target.length()] if it's not infinity; otherwise, return -1.

solution.py9 lines

1def min_valid_strings(words, target):
2    n = len(target)
3    dp = [float('inf')] * (n + 1)
4    dp[0] = 0
5    for i in range(n):
6        for word in words:
7            if target.startswith(word, i):
8                dp[i + len(word)] = min(dp[i + len(word)], dp[i] + 1)
9    return dp[n] if dp[n] != float('inf') else -1

ℹ

Complexity note: The time complexity is O(n * m) where n is the length of the target and m is the average length of the words, as we check each word for each position in the target.

1Using dynamic programming allows us to efficiently compute the minimum valid strings needed without redundant checks.
2Prefix matching can be optimized using data structures like Trie for faster lookups.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.