LeetCode in Kotlin
3487. Maximum Unique Subarray Sum After Deletion
Easy
You are given an integer array nums.
You are allowed to delete any number of elements from nums without making it empty. After performing the deletions, select a non-empty subarrays of nums such that:
- All elements in the subarray are unique.
- The sum of the elements in the subarray is maximized.
Return the maximum sum of such a subarray.
Example 1:
Input: nums = [1,2,3,4,5]
Output: 15
Explanation:
Select the entire array without deleting any element to obtain the maximum sum.
Example 2:
Input: nums = [1,1,0,1,1]
Output: 1
Explanation:
Delete the element nums[0] == 1, nums[1] == 1, nums[2] == 0, and nums[3] == 1. Select the entire array [1] to obtain the maximum sum.
Example 3:
Input: nums = [1,2,-1,-2,1,0,-1]
Output: 3
Explanation:
Delete the elements nums[2] == -1 and nums[3] == -2, and select the subarray [2, 1] from [1, 2, 1, 0, -1] to obtain the maximum sum.
Constraints:
1 <= nums.length <= 100-100 <= nums[i] <= 100
Solution
class Solution {
fun maxSum(nums: IntArray): Int {
var sum = 0
val st = mutableSetOf<Int>()
var mxNeg = Int.MIN_VALUE
for (num in nums) {
if (num > 0) {
st.add(num)
} else {
mxNeg = maxOf(mxNeg, num)
}
}
for (value in st) {
sum += value
}
return if (st.isNotEmpty()) sum else mxNeg
}
}