LeetCode in Kotlin
3427. Sum of Variable Length Subarrays
Easy
You are given an integer array nums of size n. For each index i where 0 <= i < n, define a non-empty subarrays nums[start ... i] where start = max(0, i - nums[i]).
Return the total sum of all elements from the subarray defined for each index in the array.
Example 1:
Input: nums = [2,3,1]
Output: 11
Explanation:
i
Subarray
Sum
0
nums[0] = [2]
2
1
nums[0 ... 1] = [2, 3]
5
2
nums[1 ... 2] = [3, 1]
4
Total Sum
11
The total sum is 11. Hence, 11 is the output.
Example 2:
Input: nums = [3,1,1,2]
Output: 13
Explanation:
i
Subarray
Sum
0
nums[0] = [3]
3
1
nums[0 ... 1] = [3, 1]
4
2
nums[1 ... 2] = [1, 1]
2
3
nums[1 ... 3] = [1, 1, 2]
4
Total Sum
13
The total sum is 13. Hence, 13 is the output.
Constraints:
1 <= n == nums.length <= 1001 <= nums[i] <= 1000
Solution
class Solution {
fun subarraySum(nums: IntArray): Int {
var res = nums[0]
for (i in 1..<nums.size) {
val j = i - nums[i] - 1
nums[i] += nums[i - 1]
res += nums[i] - (if (j < 0) 0 else nums[j])
}
return res
}
}