LeetCode in Kotlin
1413. Minimum Value to Get Positive Step by Step Sum
Easy
Given an array of integers nums, you start with an initial positive value startValue__.
In each iteration, you calculate the step by step sum of startValue plus elements in nums (from left to right).
Return the minimum positive value of startValue such that the step by step sum is never less than 1.
Example 1:
Input: nums = [-3,2,-3,4,2]
Output: 5
Explanation: If you choose startValue = 4, in the third iteration your step by step sum is less than 1.
step by step sum
startValue = 4 | startValue = 5 | nums
(4 -3 ) = 1 | (5 -3 ) = 2 | -3
(1 +2 ) = 3 | (2 +2 ) = 4 | 2
(3 -3 ) = 0 | (4 -3 ) = 1 | -3
(0 +4 ) = 4 | (1 +4 ) = 5 | 4
(4 +2 ) = 6 | (5 +2 ) = 7 | 2
Example 2:
Input: nums = [1,2]
Output: 1
Explanation: Minimum start value should be positive.
Example 3:
Input: nums = [1,-2,-3]
Output: 5
Constraints:
1 <= nums.length <= 100-100 <= nums[i] <= 100
Solution
class Solution {
fun minStartValue(nums: IntArray): Int {
var min = Int.MAX_VALUE
var sum = 0
for (num in nums) {
sum += num
min = Math.min(sum, min)
}
return if (min > 0) 1 else Math.abs(min) + 1
}
}