In this guide, you will get 2652. Sum Multiples LeetCode Solution with the best time and space complexity. The solution to Sum Multiples problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Given a positive integer n, find the sum of all integers in the range [1, n] inclusive that are divisible by 3, 5, or 7.
Return an integer denoting the sum of all numbers in the given range satisfying the constraint.
Example 1:
Input: n = 7
Output: 21
Explanation: Numbers in the range [1, 7] that are divisible by 3, 5, or 7 are 3, 5, 6, 7. The sum of these numbers is 21.
Example 2:
Input: n = 10
Output: 40
Explanation: Numbers in the range [1, 10] that are divisible by 3, 5, or 7 are 3, 5, 6, 7, 9, 10. The sum of these numbers is 40.
Example 3:
Input: n = 9
Output: 30
Explanation: Numbers in the range [1, 9] that are divisible by 3, 5, or 7 are 3, 5, 6, 7, 9. The sum of these numbers is 30.
Constraints:
1 <= n <= 103
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
2652. Sum Multiples LeetCode Solution in C++
class Solution {
public:
int sumOfMultiples(int n) {
int ans = 0;
for (int i = 1; i <= n; ++i)
if (i % 3 == 0 || i % 5 == 0 || i % 7 == 0)
ans += i;
return ans;
}
}; /* code provided by PROGIEZ */
2652. Sum Multiples LeetCode Solution in Java
class Solution {
public int sumOfMultiples(int n) {
int ans = 0;
for (int i = 1; i <= n; ++i)
if (i % 3 == 0 || i % 5 == 0 || i % 7 == 0)
ans += i;
return ans;
}
} // code provided by PROGIEZ
2652. Sum Multiples LeetCode Solution in Python
class Solution:
def sumOfMultiples(self, n: int) -> int:
ans = 0
for i in range(1, n + 1):
if i % 3 == 0 or i % 5 == 0 or i % 7 == 0:
ans += i
return ans # code by PROGIEZ
