Divisible and Non-divisible Sums Difference

Divisible and Non-divisible Sums Difference - Problem

Math Easy

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 by m.
num2: The sum of all integers in the range [1, n] (both inclusive) that are divisible by m.

Return the integer num1 - num2.

Input & Output

Example 1 — Basic Case

$ Input: n = 10, m = 3

› Output: 19

💡 Note: Numbers 1-10: [1,2,3,4,5,6,7,8,9,10]. Divisible by 3: [3,6,9] sum=18. Not divisible: [1,2,4,5,7,8,10] sum=37. Return 37-18=19.

Example 2 — Small Range

$ Input: n = 5, m = 6

› Output: 15

💡 Note: Numbers 1-5: [1,2,3,4,5]. None divisible by 6, so num2=0. num1=1+2+3+4+5=15. Return 15-0=15.

Example 3 — Perfect Divisor

$ Input: n = 5, m = 1

› Output: -15

💡 Note: All numbers [1,2,3,4,5] divisible by 1, so num1=0, num2=15. Return 0-15=-15.

Constraints

1 ≤ n ≤ 1000
1 ≤ m ≤ 1000

Visualization

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Asked in

M Microsoft 15 A Apple 12

The key insight is to use arithmetic sequence formulas instead of iteration. Calculate total sum using n*(n+1)/2, then find sum of multiples using count of multiples times the arithmetic sequence formula. Best approach is Mathematical Formula with Time: O(1), Space: O(1).

Common Approaches

✓ Brute Force - Iterate Through All Numbers

⏱️ Time: O(n) Space: O(1)

Iterate through all numbers from 1 to n, check divisibility by m, and maintain two separate sums. Finally return the difference.

Mathematical Formula - Arithmetic Sequence

⏱️ Time: O(1) Space: O(1)

Instead of iterating, use mathematical formulas. Total sum from 1 to n is n*(n+1)/2. Sum of multiples of m up to n can be calculated using arithmetic sequence formula.

Brute Force - Iterate Through All Numbers — Algorithm Steps

Initialize num1 = 0 and num2 = 0
For each number i from 1 to n, check if i % m == 0
If divisible, add to num2; otherwise add to num1
Return num1 - num2

Visualization

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Step-by-Step Walkthrough

Initialize

Set num1=0, num2=0

Check Each

For each i, check i % m

Calculate

Return num1 - num2

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int n, int m) {
    int num1 = 0; // sum of non-divisible
    int num2 = 0; // sum of divisible
    
    for (int i = 1; i <= n; i++) {
        if (i % m == 0) {
            num2 += i;
        } else {
            num1 += i;
        }
    }
    
    return num1 - num2;
}

int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    printf("%d\n", solution(n, m));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop iterating through n numbers

✓ Linear Growth

Space Complexity

O(1)

Only using two integer variables for sums

✓ Linear Space

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Medium Frequency

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Related Problems

Common Approaches

Brute Force - Iterate Through All Numbers — Algorithm Steps

Visualization

Code -

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