5 Best Ways to Check if N Can Be Represented as Sum of Integers Chosen from Set A & B in Python

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πŸ’‘ Problem Formulation: We are tackling a classic computational issue where we want to verify if a certain number (n) can be expressed as a sum of elements taken from two sets A and B. For instance, if A = {1, 2} and B = {3, 4}, and we want to check if n = 7, our desired output would be True since 2 (from A) + 5 (from B) equals 7.

Method 1: Recursive Approach

The recursive approach involves defining a function that tries to build up the sum n using elements from sets A and B. It subtracts elements from n until it is reduced to 0 (success case) or it cannot be reduced further (failure case).

Here’s an example:

def can_sum(n, A, B):
    if n == 0:
        return True
    elif n < 0:
        return False
    for a in A:
        for b in B:
            if can_sum(n - a - b, A, B):
                return True
    return False

print(can_sum(7, {1, 2}, {3, 4}))

Output: True

This function attempts to find a pair of numbers from A and B whose sum is equal to n. It does this recursively by subtracting elements of A and B from n, backtracking whenever it oversteps bounds. It’s depth-first search in essence; however, this method can be inefficient due to overlapping subproblems.

Method 2: Iterative Approach with Memoization

Memoization can optimize the recursive solution by storing the results of processed subproblems. This iterative approach with memoization ensures each problem is calculated only once, improving efficiency.

Here’s an example:

def can_sum(n, A, B):
    memo = {}
    def rec(target):
        if target in memo:
            return memo[target]
        if target == 0:
            return True
        if target < 0:
            return False
        for a in A:
            for b in B:
                if rec(target - a - b):
                    memo[target] = True
                    return True
        memo[target] = False
        return False
    
    return rec(n)

print(can_sum(7, {1, 2}, {3, 4}))

Output: True

Here, a helper function rec is defined within can_sum() that operates much like the function from Method 1 but uses a dictionary, memo, to store results. This drastically reduces the number of calculations needed, especially for larger n.

Method 3: Dynamic Programming

With Dynamic Programming, we build a table iteratively and fill it with True or False values to represent if the sum n can be achieved with elements from A and B up to that index. This bottom-up approach avoids recursion and can be more efficient.

Here’s an example:

def can_sum(n, A, B):
    dp = [False] * (n + 1)
    dp[0] = True
    for i in range(1, n + 1):
        for a in A:
            for b in B:
                if i - a - b >= 0 and dp[i - a - b]:
                    dp[i] = True
    return dp[n]

print(can_sum(7, {1, 2}, {3, 4}))

Output: True

This method initializes an array ‘dp’ with False, except for the first element, which represents the base case of n=0. The function updates the dp table based on whether any combination of elements from A and B can sum up to each index. A simple lookup of dp[n] then gives us our answer.

Method 4: Using Set Operations

Python sets provide a way to generate different sums efficiently using set operations. This method computes all possible sums and checks if n is within the resulting set.

Here’s an example:

def can_sum(n, A, B):
    possible_sums = {a + b for a in A for b in B}
    return n in possible_sums

print(can_sum(7, {1, 2}, {3, 4}))

Output: True

In this snippet, a set comprehension is used to create a set of all possible sums between elements of A and B. The function then simply checks if n is a member of this set, providing a very concise and efficient detection method.

Bonus One-Liner Method 5: Inclusive Check Using Generators

This bonus one-liner uses a generator expression within the any() built-in function to check for the sum in a memory-efficient manner.

Here’s an example:

can_sum = lambda n, A, B: any(n == a + b for a in A for b in B)
print(can_sum(7, {1, 2}, {3, 4}))

Output: True

By using the any() function with a generator expression, this approach finds if n can be represented as a sum of any two integers from sets A and B without computing all possibilities ahead of time. It’s succinct and memory-efficient but may not be the fastest for large sets.

Summary/Discussion

  • Method 1: Recursive Approach. Simple but potentially very slow due to redundant calculations.
  • Method 2: Iterative Approach with Memoization. More efficient than pure recursion by avoiding duplicated work. Still has recursive depth limitations.
  • Method 3: Dynamic Programming. Non-recursive and generally efficient for larger inputs. Requires additional space proportional to n.
  • Method 4: Using Set Operations. Concise and fast for smaller sets, but can become less practical as set sizes grow due to rapid growth of possible sums.
  • Method 5: Inclusive Check Using Generators. Compact and memory-efficient, this is a practical choice for smaller problems, but it can be less efficient for larger sets due to its linear search nature.