How do you solve Count Ways To Build Good Strings on LeetCode?

Count Ways To Build Good Strings (LeetCode #2466) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Dynamic programming can significantly reduce the number of calculations by storing intermediate results.

What is the brute force solution for Count Ways To Build Good Strings?

The brute force approach for Count Ways To Build Good Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach generates all possible good strings by recursively appending '0's and '1's until the string length exceeds 'high'. We then filter these strings to count only those that fall within the specified range of 'low' to 'high'. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Ways To Build Good Strings?

Count Ways To Build Good Strings uses the following concepts: Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Recursion, Memoization.

What are the interview tips for Count Ways To Build Good Strings?

Always clarify the constraints and edge cases before jumping into coding. Explain your thought process as you code; it shows your understanding and helps the interviewer follow along. Practice writing clean and efficient code; it reflects your coding style and professionalism.

What are common mistakes when solving Count Ways To Build Good Strings?

Not considering the modulo operation which can lead to overflow errors. Failing to initialize the dp array correctly or misunderstanding its indexing.

#2466

Count Ways To Build Good Strings

Medium

Dynamic ProgrammingDynamic ProgrammingRecursionMemoization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to efficiently count the number of good strings by storing previously computed results. This avoids redundant calculations and allows us to build up the count iteratively.

⚙️

Algorithm

4 steps

1Step 1: Create a dp array where dp[i] represents the number of good strings of length i.
2Step 2: Initialize dp[0] = 1 (the empty string).
3Step 3: For each length from 1 to high, calculate dp[i] as the sum of dp[i - zero] and dp[i - one] if those indices are valid.
4Step 4: Sum the values in dp from index low to high to get the final count, and return it modulo 10^9 + 7.

solution.py10 lines

1def countGoodStrings(low, high, zero, one):
2    MOD = 10**9 + 7
3    dp = [0] * (high + 1)
4    dp[0] = 1
5    for i in range(1, high + 1):
6        if i - zero >= 0:
7            dp[i] = (dp[i] + dp[i - zero]) % MOD
8        if i - one >= 0:
9            dp[i] = (dp[i] + dp[i - one]) % MOD
10    return sum(dp[low:high + 1]) % MOD

ℹ

Complexity note: The time complexity is O(n) because we iterate through lengths from 1 to high once, and the space complexity is O(n) due to the dp array storing counts for each length.

1Dynamic programming can significantly reduce the number of calculations by storing intermediate results.
2Understanding the problem constraints helps in formulating the right approach.

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