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

Minimum Number of Valid Strings to Form Target II (LeetCode #3292) 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: Dynamic programming can significantly reduce redundant calculations.

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

The optimal solution for Minimum Number of Valid Strings to Form Target II runs in O(n * m) time and uses O(n) space. This complexity arises from iterating through the target and checking each word, where n is the length of the target and m is the number of words.

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

The brute force approach for Minimum Number of Valid Strings to Form Target II is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try to form the target string by checking every possible prefix of the words array. This method is straightforward but inefficient, as it may involve many repeated checks. 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 II?

Clarify the problem statement and constraints before jumping into coding. Think aloud to explain your thought process and approach. Test your solution with edge cases to ensure robustness.

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

Not considering all possible prefixes for each character. Failing to initialize the DP array correctly.

#3292

Minimum Number of Valid Strings to Form Target II

Hard

ArrayStringBinary SearchDynamic ProgrammingGreedySegment 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)

💡

Intuition

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

We can use dynamic programming to efficiently track the minimum number of valid strings needed to form each prefix of the target. This approach reduces redundant checks and speeds up the process.

⚙️

Algorithm

5 steps

1Step 1: Create a DP array where dp[i] represents the minimum number of valid strings needed to form the prefix of target up to index i.
2Step 2: Initialize dp[0] = 0 (no strings needed for an empty prefix) and all other dp[i] to infinity.
3Step 3: For each position i in target, check all words to see if they can form a valid prefix ending at i.
4Step 4: If a word can match, update dp[i] to be the minimum of its current value and dp[i - length of word] + 1.
5Step 5: After processing, return dp[target.length()] if it's not infinity, otherwise return -1.

solution.py11 lines

1# Full working Python code
2
3def min_valid_strings(words, target):
4    dp = [float('inf')] * (len(target) + 1)
5    dp[0] = 0
6    for i in range(1, len(target) + 1):
7        for word in words:
8            if target.startswith(word, i - len(word)):
9                dp[i] = min(dp[i], dp[i - len(word)] + 1)
10    return dp[len(target)] if dp[len(target)] != float('inf') else -1
11

ℹ

Complexity note: This complexity arises from iterating through the target and checking each word, where n is the length of the target and m is the number of words.

1Dynamic programming can significantly reduce redundant calculations.
2Understanding prefix matching is crucial for efficient solutions.

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