algoadvance

Given a list of integers nums and an integer k, return the difference between the sum of elements in the list that are divisible by k and the sum of elements that are not divisible by k.

Clarifying Questions

  1. Are there any constraints on the size of the list?
    • No specific constraints are mentioned; assume it fits in memory.
  2. What should be returned if the list is empty?
    • Return 0 since the sums of both divisible and non-divisible elements would be 0.
  3. Can the numbers in the list be negative?
    • Yes, integers can be both positive and negative.
  4. Is k guaranteed to be non-zero?
    • Yes, assume k is always non-zero to avoid division by zero.

Strategy

  1. Initialize two sums: sum_divisible and sum_non_divisible.
  2. Iterate through each number in nums.
  3. If the number is divisible by k, add it to sum_divisible.
  4. Otherwise, add it to sum_non_divisible.
  5. Return the difference between sum_divisible and sum_non_divisible.
def difference_of_sums(nums, k):
    sum_divisible = 0
    sum_non_divisible = 0

    for num in nums:
        if num % k == 0:
            sum_divisible += num
        else:
            sum_non_divisible += num

    return sum_divisible - sum_non_divisible

# Example usage
nums = [1, 2, 3, 4, 5, 6, 7]
k = 3
print(difference_of_sums(nums, k))  # Output: -9 (0 because (3+6)-(sum(1, 2, 4, 5, 7)) equals -17)

Time Complexity

The time complexity of this solution is (O(n)), where (n) is the length of the list nums. This is because we iterate through the list exactly once, performing a constant amount of work for each element.

Space Complexity

The space complexity is (O(1)) since we use only a fixed amount of additional space regardless of the input size.

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