Difference Between Nondivisible and Divisible Sums

# Difference Between Nondivisible and Divisible Sums You are given positive integers `n` and `m`. Consider every integer in the inclusive range `[1,n]`. Let `num1` be the sum of the integers in that range that are not divisible by `m`. Let `num2` be the sum of the integers in that range that are divisible by `m`. Return `num1 - num2`. ### Examples **Example 1** ```text Input: n = 8, m = 3 Output: 18 ``` The nondivisible integers are `[1,2,4,5,7,8]`, whose sum is `27`. The divisible integers are `[3,6]`, whose sum is `9`. Their difference is `27 - 9 = 18`. **Example 2** ```text Input: n = 4, m = 1 Output: -10 ``` Every integer in `[1,4]` is divisible by `1`. Thus `num1 = 0`, `num2 = 1 + 2 + 3 + 4 = 10`, and the result is `-10`. ### Constraints - `1 <= n, m <= 1000`. ### Clarifying Questions - Are both endpoints included? Yes; include every integer from `1` through `n`. - What is the sum if one of the two groups is empty? Zero. - Can the answer be negative? Yes; return the signed difference in the specified order. - If `m > n`, is any integer in the range divisible by `m`? No; in that case `num2` is zero. ### C++ Interface ```cpp class Solution { public: int differenceOfSums(int n, int m) { } }; ``` ### Reasoning Checkpoints - Partition the inclusive range according to divisibility without losing an endpoint. - Keep the subtraction order and the empty-group cases explicit. - Justify the arithmetic and its range under the given bounds. ```hint Relate the two sums The sum of all integers from one through n is known. The divisible group consists of m times the integers from one through floor(n / m). How can these sums give the requested difference? ```