Divisible Drama — Who’s In, Who’s Out?

 Divisible Drama — Who’s In, Who’s Out?

Problem Statement

You’re given two positive integers n and m.

  • num1 = sum of all numbers from 1 to n not divisible by m

  • num2 = sum of all numbers from 1 to n divisible by m

Your mission, should you choose to accept it:
Return num1 - num2

Example

Input: n = 10, m = 3  
Output: 19

Breakdown:
Not divisible by 3 → [1, 2, 4, 5, 7, 8, 10] → num1 = 37  
Divisible by 3 → [3, 6, 9] → num2 = 18  
Answer = 37 - 18 = 19

Best Data Structure for Solving the Problem

This one’s a math. Why overthink it when Gauss already gave us the formula for sum of 1 to n?

Different Approaches (from Brute Force to Optimized)

Approach 1: Brute Force (aka List & Sum Vibes)

Loop from 1 to n, split the nums based on whether they’re divisible by m, and sum them up.

class Solution {
    public int differenceOfSums(int n, int m) {
        int num1 = 0, num2 = 0;
        for (int i = 1; i <= n; i++) {
            if (i % m == 0) num2 += i;
            else num1 += i;
        }
        return num1 - num2;
    }
}

Approach 2: Math

Instead of looping through each number like it's 2010, we flex some math muscles:

  • totalSum = n * (n + 1) / 2 ← classic Gauss move

  • num2 = m * k * (k + 1) / 2 where k = n / m ← sum of all multiples of m

num1 = totalSum - num2  
Return num1 - num2 → totalSum - 2 * num2
class Solution {
    public int differenceOfSums(int n, int m) {
        int totalSum = n * (n + 1) / 2;
        int count = n / m;
        int divisibleSum = m * count * (count + 1) / 2;
        return totalSum - 2 * divisibleSum;
    }
}

⚠️ Common Mistakes and Pitfalls

  • Only subtracting once: totalSum - divisibleSum ← WRONG, needs to be totalSum - 2 * divisibleSum

  • Integer division messing with your results (but we're safe here)

  • Not spotting the math shortcut and brute-forcing everything 

📊 Time and Space Complexity Analysis

Approach          Time Complexity         Space Complexity
Brute Force            O(n)            O(1)
Math Optimized            O(1)            O(1)

🌍 Real-World Applications

  • Financial Planning: Filtering monthly transactions divisible by fixed costs (like rent/utilities)

  • Digital Coupons: Separating discounts applicable on every nth order

  • Productivity Tools: Time tracking, like focusing only on “non-distraction” intervals

  • Educational Apps: Rewarding users for days not divisible by skipped days 


I hope you found this article insightful! Keep exploring, keep learning, and don’t forget to check back daily for more exciting problem breakdowns. If you know someone who would benefit from this, share it with them and help them grow too! That’s it for today—see you in the next post!

                                                                                                                                         Signing off!! 

                                                                                                Master DSA One Problem at a Time :)

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