How do you solve Find All Good Strings on LeetCode?

Find All Good Strings (LeetCode #1397) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m * 2 * 2) time complexity and O(n * m * 2 * 2) space complexity. Key insight: Dynamic programming is crucial for efficiently solving problems with overlapping subproblems.

What is the brute force solution for Find All Good Strings?

The brute force approach for Find All Good Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible strings of length n and check each one against the conditions. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n * m * 2 * 2).

What algorithm or data structure is used to solve Find All Good Strings?

Find All Good Strings uses the following concepts: String, Dynamic Programming, String Matching. The recommended patterns to study are: Dynamic Programming, Backtracking, String Matching.

What are the interview tips for Find All Good Strings?

Clearly explain your thought process while coding; this helps interviewers understand your reasoning. Don't hesitate to ask clarifying questions about the problem constraints. Practice writing clean, modular code to improve readability and maintainability.

What are common mistakes when solving Find All Good Strings?

Not considering edge cases where evil is a prefix of s1 or s2. Failing to properly initialize the DP table can lead to incorrect results.

#1397

Find All Good Strings

Hard

StringDynamic ProgrammingString MatchingDynamic ProgrammingBacktrackingString Matching

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m * 2 * 2)
Space	O(1)	O(n * m * 2 * 2)

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Intuition

Time O(n * m * 2 * 2)Space O(n * m * 2 * 2)

We can use dynamic programming to efficiently count the valid strings by keeping track of the current position, the position in the evil string, and whether we are still bound by s1 and s2.

⚙️

Algorithm

3 steps

1Step 1: Use a DP table where dp[pos][posEvil][equalToS1][equalToS2] stores the count of valid strings of length 'pos'.
2Step 2: Initialize the DP table and iterate through each character position.
3Step 3: For each character, decide whether to match s1 or s2 or to use any character from 'a' to 'z'. Update the position in the evil string based on the current character.

solution.py24 lines

1def countGoodStrings(n, s1, s2, evil):
2    MOD = 10**9 + 7
3    m = len(evil)
4    dp = [[[-1 for _ in range(2)] for _ in range(2)] for _ in range(n + 1)]
5    def dfs(pos, posEvil, equalToS1, equalToS2):
6        if pos == n:
7            return 1
8        if dp[pos][posEvil][equalToS1][equalToS2] != -1:
9            return dp[pos][posEvil][equalToS1][equalToS2]
10        count = 0
11        limit = s2 if equalToS2 else 'z'
12        for c in range(ord('a'), ord(limit) + 1):
13            nextEqualToS1 = equalToS1 and chr(c) == s1[pos]
14            nextEqualToS2 = equalToS2 and chr(c) == s2[pos]
15            nextPosEvil = posEvil
16            while nextPosEvil > 0 and evil[nextPosEvil] != chr(c):
17                nextPosEvil -= 1
18            if evil[nextPosEvil] == chr(c):
19                nextPosEvil += 1
20            if nextPosEvil < m:
21                count = (count + dfs(pos + 1, nextPosEvil, nextEqualToS1, nextEqualToS2)) % MOD
22        dp[pos][posEvil][equalToS1][equalToS2] = count
23        return count
24    return dfs(0, 0, True, True)

ℹ

Complexity note: The complexity is derived from the DP table which has dimensions based on the string length, the length of evil, and two boolean states for bounds.

1Dynamic programming is crucial for efficiently solving problems with overlapping subproblems.
2Understanding the constraints and how they affect string generation can lead to optimized solutions.

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