LeetCode in Kotlin
2778. Sum of Squares of Special Elements
Easy
You are given a 1-indexed integer array nums of length n.
An element nums[i] of nums is called special if i divides n, i.e. n % i == 0.
Return the sum of the squares of all special elements of nums.
Example 1:
Input: nums = [1,2,3,4]
Output: 21
Explanation:
There are exactly 3 special elements in nums: nums[1] since 1 divides 4, nums[2] since 2 divides 4, and nums[4] since 4 divides 4.
Hence, the sum of the squares of all special elements of nums is nums[1] * nums[1] + nums[2] * nums[2] + nums[4] * nums[4] = 1 * 1 + 2 * 2 + 4 * 4 = 21.
Example 2:
Input: nums = [2,7,1,19,18,3]
Output: 63
Explanation:
There are exactly 4 special elements in nums: nums[1] since 1 divides 6, nums[2] since 2 divides 6, nums[3] since 3 divides 6, and nums[6] since 6 divides 6.
Hence, the sum of the squares of all special elements of nums is nums[1] * nums[1] + nums[2] * nums[2] + nums[3] * nums[3] + nums[6] * nums[6] = 2 * 2 + 7 * 7 + 1 * 1 + 3 * 3 = 63.
Constraints:
1 <= nums.length == n <= 501 <= nums[i] <= 50
Solution
class Solution {
fun sumOfSquares(nums: IntArray): Int {
var sum = 0
for (i in nums.indices) {
if (nums.size % (i + 1) == 0) sum += nums[i] * nums[i]
}
return sum
}
}