Count Ways To Build Good Strings

class Solution {
    int[] memo;
    int modulo = 1_000_000_007;
    
    private int countSequences(int length, int zeros, int ones) {
        if (memo[length] != -1) {
            return memo[length];
        }
        
        int result = 0;
        if (length >= ones) {
            result += countSequences(length - ones, zeros, ones);
        }
        if (length >= zeros) {
            result += countSequences(length - zeros, zeros, ones);
        }
        memo[length] = result % modulo;
        return memo[length];
    }
    
    public int findNumberOfGoodStrings(int minLen, int maxLen, int zeros, int ones) {
        memo = new int[maxLen + 1];
        Arrays.fill(memo, -1);
        memo[0] = 1;

        int sum = 0;
        for (int i = minLen; i <= maxLen; ++i) {
            sum += countSequences(i, zeros, ones);
            sum %= modulo;
        }
        return sum;
    }
}

The provided Java code defines a class Solution with methods to count ways to build "good" strings given certain constraints on the number of zeros and ones that can be used. The main functionality is split between two methods.

• countSequences(int length, int zeros, int ones): This method calculates the number of valid sequences of a given length using a recursive approach with memoization. It checks if the provided memoization array already contains the solution for a given length to avoid redundant calculations. The recursion considers strings formed by subtracting the number of ones and zeros from the current length and sums these recursive calls, considering them modulo 1_000_000_007 to handle large numbers.

• findNumberOfGoodStrings(int minLen, int maxLen, int zeros, int ones): This method initializes the memoization array and iterates through all lengths from minLen to maxLen. For each length, it adds up the count of valid sequences obtained from countSequences, again taking results modulo 1_000_000_007. The sum of all these counts for each length provides the total number of good strings within the specified range.

The solution utilizes dynamic programming to efficiently compute the number of sequences for different lengths by storing intermediate results and avoiding repetitive calculations. Each call to countSequences ensures results are modulo 1_000_000_007, a common approach in problems dealing with large numbers to prevent overflow and maintain performance. The memoization helps in significantly reducing the time complexity that would otherwise result from the naive recursive solution.

class Solution:
    def countValidStrings(self, minLen: int, maxLen: int, lenX: int, lenY: int) -> int:
        # Utilize memoization to store counts of valid strings for different lengths
        memo = [1] + [-1] * (maxLen)
        modulus = 1000000007
        
        # Recursive function to determine counts for specific lengths
        def computeCounts(length):
            if memo[length] != -1:
                return memo[length]
            currentCount = 0
            if length >= lenX:
                currentCount += computeCounts(length - lenX)
            if length >= lenY:
                currentCount += computeCounts(length - lenY)
            memo[length] = currentCount % modulus
            return memo[length]
            
        # Sum up counts for all lengths in the required range
        return sum(computeCounts(length) for length in range(minLen, maxLen + 1)) % modulus

This Python solution involves counting the number of "good" strings that can be built within given constraints on their length. The task is approached using dynamic programming with a memoization technique to efficiently solve the problem without recalculating results for the same input parameters.

• Initialize a memoization list memo where each index corresponds to a length of the string and stores the number of valid strings of that length. All indices are initially set to -1, except for the 0th index, which is 1 (representing the count of an empty string).
• Define a modulus modulus = 1000000007 to ensure the results for string counts are within manageable size by taking results modulo this large prime.
• Implement a recursive function computeCounts that:
  • Checks if the result for a given length is already computed by looking up memo.
  • Recursively calculates the number of valid strings of a particular length by summing the counts of strings reduced by lenX and lenY (if applicable), ensuring the problem's constraints (minimum and maximum lengths) are honored.
• Sum the valid string counts between minLen and maxLen, inclusive, using a generator expression that invokes computeCounts for each desired length.
• Use the modulo operation to ensure the final sum remains within numeric limits.

Overall, this solution leverages memoization and recursion to tackle the potential exponential growth in computation that could otherwise result from the many possible string combinations, effectively reducing complexity and increasing performance.