Minimum value to get positive step by step sum

Time: O(N); Space: O(1); 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 <= len(nums) <= 100

  • -100 <= nums[i] <= 100

Hints:

  1. Find the minimum prefix sum.

[4]:

class Solution1(object):
    """
    Time: O(N)
    Space: O(1)
    """
    def minStartValue(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        min_prefix, prefix = 0, 0

        for num in nums:
            prefix += num
            min_prefix = min(min_prefix, prefix)

        return 1 - min_prefix

[5]:

s = Solution1()
nums = [-3,2,-3,4,2]
assert s.minStartValue(nums) == 5
nums = [1,2]
assert s.minStartValue(nums) == 1
nums = [1,-2,-3]
assert s.minStartValue(nums) == 5