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Leetcode 2535. Difference Between Element Sum and Digit Sum of an Array

Problem Statement

Given an integer array nums, you need to compute:

  1. The sum of all elements in the array (elementSum).
  2. The sum of all digits of each element in the array (digitSum).

The task is to return the absolute difference between the elementSum and the digitSum.

Clarifying Questions

  1. Range of Elements: What is the range of values for integers in the array? (Clarifying for edge cases)
    • Answer: The elements can be any integers within the typical constraints for array problems, say within the range of [-1,000,000, 1,000,000].
  2. Array Size: What is the maximum size of the array?
    • Answer: Assume the array can have up to 10,000 elements for this problem.
  3. Negative Numbers: How should we handle the digits of negative numbers?
    • Answer: Usually, only the absolute values of digits are considered.

Strategy

  1. Compute the elementSum by iterating through the array and summing all elements.
  2. Compute the digitSum by iterating through the array, converting each element to its absolute value, iterating through its digits, and summing these digits.
  3. Return the absolute difference between elementSum and digitSum.

Code

Here’s the C++ code to solve the problem:

#include <vector>
#include <cmath> // For std::abs

int differenceOfSum(const std::vector<int>& nums) {
    int elementSum = 0;
    int digitSum = 0;
    
    for (int num : nums) {
        elementSum += num;
        
        int absNum = std::abs(num);
        while (absNum > 0) {
            digitSum += absNum % 10;
            absNum /= 10;
        }
    }
    
    return std::abs(elementSum - digitSum);
}

Time Complexity

The time complexity of this solution is (O(n \cdot k)), where:

Given the constraints (elements range from -1,000,000 to 1,000,000), (k) can be considered as a constant (since it would be around 6-7 digits for most cases), making the overall complexity close to (O(n)).

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