DevExCode

2652. Sum Multiples

Math

Explanation

To solve this problem, we iterate through all numbers from 1 to n and check if the number is divisible by 3, 5, or 7. If the number is divisible by any of these, we add it to the sum. Finally, we return the sum as the result.

  • We iterate through numbers from 1 to n
  • For each number, we check if it is divisible by 3, 5, or 7
  • If it is divisible by any of these, we add it to the sum
  • Return the final sum as the result

Time Complexity: O(n) Space Complexity: O(1)

View Leetcode Problem
class Solution {
    public int sumMultiples(int n) {
        int sum = 0;
        for (int i = 1; i <= n; i++) {
            if (i % 3 == 0 || i % 5 == 0 || i % 7 == 0) {
                sum += i;
            }
        }
        return sum;
    }
}

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