Divisible and Non-divisible Sums Difference - Solution & Explanation
Problem Statement
You are given positive integers n and m.
Define two integers as follows:
num1: The sum of all integers in the range[1, n](both inclusive) that are not divisible bym.num2: The sum of all integers in the range[1, n](both inclusive) that are divisible bym.
Return the integer num1 - num2.
Example 1:
Input: n = 10, m = 3 Output: 19 Explanation: In the given example: - Integers in the range [1, 10] that are not divisible by 3 are [1,2,4,5,7,8,10], num1 is the sum of those integers = 37. - Integers in the range [1, 10] that are divisible by 3 are [3,6,9], num2 is the sum of those integers = 18. We return 37 - 18 = 19 as the answer.
Example 2:
Input: n = 5, m = 6 Output: 15 Explanation: In the given example: - Integers in the range [1, 5] that are not divisible by 6 are [1,2,3,4,5], num1 is the sum of those integers = 15. - Integers in the range [1, 5] that are divisible by 6 are [], num2 is the sum of those integers = 0. We return 15 - 0 = 15 as the answer.
Example 3:
Input: n = 5, m = 1 Output: -15 Explanation: In the given example: - Integers in the range [1, 5] that are not divisible by 1 are [], num1 is the sum of those integers = 0. - Integers in the range [1, 5] that are divisible by 1 are [1,2,3,4,5], num2 is the sum of those integers = 15. We return 0 - 15 = -15 as the answer.
Constraints:
1 <= n, m <= 1000
Approach Overview
Problem Overview: Given two integers n and m, compute the difference between the sum of numbers from 1..n that are not divisible by m and the sum of numbers that are divisible by m. The result is sumNonDivisible - sumDivisible.
Approach 1: Iterative Calculation (O(n) time, O(1) space)
Iterate through every number from 1 to n. For each value i, check the divisibility condition using i % m. If the remainder is zero, add i to the divisible sum; otherwise add it to the non‑divisible sum. After the loop finishes, subtract the two totals. This approach mirrors the problem statement directly and is easy to implement in any language. It works well when n is small or when clarity matters more than micro‑optimizations. The algorithm uses constant memory and a single pass through the range.
Approach 2: Formula-based Optimization (O(1) time, O(1) space)
The optimized approach relies on arithmetic series formulas from math. First compute the sum of all numbers from 1..n using total = n * (n + 1) / 2. Next count how many numbers are divisible by m: k = n / m. Those numbers are m, 2m, 3m ... km. Their sum forms another arithmetic series: sumDivisible = m * (k * (k + 1) / 2). The non‑divisible sum is simply total - sumDivisible. The required difference becomes (total - sumDivisible) - sumDivisible, which simplifies to total - 2 * sumDivisible. This eliminates iteration entirely and runs in constant time using simple arithmetic operations, a common trick in number theory style problems.
Recommended for interviews: Start with the iterative approach to show you understand the requirement and the divisibility check. Then optimize using the arithmetic formula. Interviewers usually expect the constant‑time mathematical solution because it demonstrates pattern recognition and familiarity with sum formulas.
Approach 1: Iterative Approach
This approach involves iterating through every number from 1 to n and checking whether each number is divisible by m. Based on this check, we add the number to either num1 or num2. Finally, calculate the result as num1 - num2.
Iterate through numbers 1 to n. Use modulo to check divisibility by m. Sum numbers accordingly in num1 and num2, then return the difference.
Complexity
Time Complexity: O(n), as we iterate from 1 to n. Space Complexity: O(1), since we use constant extra space.
Approach 2: Formula-based Optimization
This approach uses arithmetic series formulas to calculate the sums directly rather than iterating through each element. The sum of numbers divisible by m can be expressed in terms of another arithmetic series with step m.
Calculates the total sum from 1 to n using n(n+1)/2. Divisible sums are determined using the formula m * (p * (p + 1))/2, where p = n / m. Updates non-divisible sum by subtracting divisible sums from total sums.
Complexity
Time Complexity: O(1) because calculations are done through formulas. Space Complexity: O(1).
Approach 3: Simulation
We traverse every number in the range [1, n]. If it is divisible by m, we subtract it from the answer. Otherwise, we add it to the answer.
After the traversal, we return the answer.
The time complexity is O(n), where n is the given integer. The space complexity is O(1).
Complexity Comparison
| Approach | Complexity |
|---|---|
| Iterative Approach | Time Complexity: O(n), as we iterate from 1 to n. Space Complexity: O(1), since we use constant extra space. |
| Formula-based Optimization | Time Complexity: O(1) because calculations are done through formulas. Space Complexity: O(1). |
| Simulation | — |
Detailed Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Iterative Calculation | O(n) | O(1) | Simple implementation or when directly simulating the condition |
| Formula-based Optimization | O(1) | O(1) | Best for large n and interviews requiring mathematical optimization |
Video Solution
Leetcode | 2894. Divisible and Non-divisible Sums Difference | Easy | Java Solution • Developer Docs • 921 views views
Watch 9 more video solutions →Frequently Asked Questions
Is Divisible and Non-divisible Sums Difference easy or hard?
Divisible and Non-divisible Sums Difference Python/Java solution
How to solve Divisible and Non-divisible Sums Difference in O(1)?
What is the best approach for Divisible and Non-divisible Sums Difference?
Is Divisible and Non-divisible Sums Difference asked at Google/Amazon/Meta?
What data structure is used in Divisible and Non-divisible Sums Difference?
What is the time complexity of Divisible and Non-divisible Sums Difference?
Ready to solve this problem?
Practice Divisible and Non-divisible Sums Difference with our built-in code editor and test cases.
Practice on FleetCode