5 Best Ways to Find the Number of Subsequences with Specific Counts of Letters in Python

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πŸ’‘ Problem Formulation: We aim to determine the number of subsequences in a given string where the number of occurrences of letters ‘x’, ‘y’, and ‘z’ is exactly i, j, and k respectively. For example, if our string is “xyxxz” and we want to find subsequences with 2 ‘x’, 1 ‘y’, and 1 ‘z’, the desired output would be 2, corresponding to the subsequences “xyxz” and “xyxz”.

Method 1: Recursive Approach

The recursive approach explores each subsequence possibility by either including or excluding the current character, and recursively calling the function while adjusting the counts for ‘x’, ‘y’, and ‘z’.

Here’s an example:

def find_subsequences(s, i, j, k, idx=0):
    if i == j == k == 0:
        return 1
    if idx == len(s):
        return 0
    count = 0
    if s[idx] == 'x' and i > 0:
        count += find_subsequences(s, i-1, j, k, idx+1)
    elif s[idx] == 'y' and j > 0:
        count += find_subsequences(s, i, j-1, k, idx+1)
    elif s[idx] == 'z' and k > 0:
        count += find_subsequences(s, i, j, k-1, idx+1)
    count += find_subsequences(s, i, j, k, idx+1)
    return count

# Example String and Subsequence Requirement
print(find_subsequences("xyxxz", 2, 1, 1))

Output: 2

This function find_subsequences() takes the input string s, and the required counts i, j, and k of ‘x’, ‘y’, and ‘z’. The function checks if the subsequence is complete (i, j, and k are all 0) and counts it, or if the end of the string is reached. It adds to count by including the current character if it matches one of ‘x’, ‘y’, or ‘z’ and decrements the respective counter before recursively calling. Then, it also increments count by considering excluding the current character, continuing both possibilities in the recursion tree.

Method 2: Dynamic Programming

Dynamic programming involves creating a 3D array to memoize the count of subsequences for every position in the string and every possible count of ‘x’, ‘y’, and ‘z’ up to i, j, and k. This method avoids recalculating the same subproblems as in the recursive method.

Here’s an example:

def count_subsequences(s, i, j, k):
    dp = [[[0]*(k+1) for _ in range(j+1)] for __ in range(i+1)]
    dp[0][0][0] = 1
    for char in s:
        for x in range(i, -1, -1):
            for y in range(j, -1, -1):
                for z in range(k, -1, -1):
                    if char == 'x' and x > 0:
                        dp[x][y][z] += dp[x-1][y][z]
                    elif char == 'y' and y > 0:
                        dp[x][y][z] += dp[x][y-1][z]
                    elif char == 'z' and z > 0:
                        dp[x][y][z] += dp[x][y][z-1]
                    else:
                        dp[x][y][z] += dp[x][y][z]
    return dp[i][j][k]

# Example Usage
print(count_subsequences("xyxxz", 2, 1, 1))

Output: 2

This function count_subsequences() initializes a 3-dimensional list to store the counts of subsequences. The list dp is indexed by the number of ‘x’, ‘y’, and ‘z’ remaining to form a subsequence. We iterate through the string and update dp based on the current character, cumulatively adding the counts of subsequences that could be formed. Because it uses memoization, the dynamic programming method is more efficient than plain recursion.

Method 3: Iterative Approach With Bitmasking

Bitmasking uses bit operations to represent subsets, iterating over all subsets and calculating subsequences that satisfy our condition while only iterating over the string once.

Here’s an example:

def bitmask_subseq_count(s, x, y, z):
    n = len(s)
    count = 0
    for mask in range(1 << n):
        cx, cy, cz = 0, 0, 0
        for i in range(n):
            if mask & (1 << i):
                if s[i] == 'x': cx += 1
                if s[i] == 'y': cy += 1
                if s[i] == 'z': cz += 1
        if cx == x and cy == y and cz == z:
            count += 1
    return count

# Example Usage
print(bitmask_subseq_count("xyxxz", 2, 1, 1))

Output: 2

In this method bitmask_subseq_count(), we generate all possible subsets of the string using bitmasking, and then count how many meet the requirement of having ‘x’, ‘y’, ‘z’ in the amount of i, j, k. Each bit in the mask represents whether to include a character at that position. This method is not as efficient for long strings but can be very illustrative and easy to understand for small cases.

Method 4: Using Python Libraries

Python libraries like itertools can be used to generate all possible subsequences. Then, we can simply iterate over these and count the ones that meet our criteria.

Here’s an example:

from itertools import combinations

def library_subseq_count(s, x, y, z):
    count = 0
    for length in range(1, len(s)+1):
        for subseq in combinations(s, length):
            if subseq.count('x') == x and subseq.count('y') == y and subseq.count('z') == z:
                count += 1
    return count

# Example Usage
print(library_subseq_count("xyxxz", 2, 1, 1))

Output: 2

The function library_subseq_count() uses the combinations function from the itertools to generate all possible subsequences. Then by iterating through these combinations, it counts the subsequences meeting the specifed requirements. This method is the easiest to implement but can also be less efficient on longer strings due to the number of combinations generated.

Bonus One-Liner Method 5: Comprehension with itertools

This concise method utilizes a generator expression within the sum function to count the subsequences directly. It is a compact version of Method 4 using itertools.

Here’s an example:

from itertools import combinations
print(sum(1 for length in range(1, len("xyxxz")+1) for subseq in combinations("xyxxz", length) if (subseq.count('x'), subseq.count('y'), subseq.count('z')) == (2, 1, 1)))

Output: 2

The one-liner utilizes a generator expression to iterate over all possible lengths and subsequences created by the combinations function, counting those that meet our criteria. It is a nifty way to achieve the solution without defining a separate function but should be used with caution on longer strings due to potential inefficiency.

Summary/Discussion

  • Method 1: Recursive Approach. Mimics the natural way of solving the problem with simplicity. Strengths: Easy to implement and understand. Weaknesses: Can be inefficient for larger strings due to extensive recursion and potential for stack overflow.
  • Method 2: Dynamic Programming. Utilizes memoization to boost efficiency. Strengths: Faster than plain recursion, handles bigger strings better. Weaknesses: Can be harder to conceptualize and requires more memory.
  • Method 3: Iterative Approach With Bitmasking. Leverages bit operations for subset generation. Strengths: Intuitive for small strings and educational for learning about bitmasking. Weaknesses: Exponential time complexity makes it impractical for large strings.
  • Method 4: Using Python Libraries. Takes advantage of built-in Python functionalities which reduces code. Strengths: Easy to write and concise. Weaknesses: Like bitmasking, it doesn’t scale well with input size.
  • Bonus Method 5: Comprehension with itertools. Combines the itertools library with comprehension for compact code. Strengths: One-liner solution is elegant. Weaknesses: Same as Method 4 with respect to inefficiency on larger inputs.