1413. Minimum Value to Get Positive Step by Step Sum

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

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这道题说是从一个startValue开始，从左往右加数组里的数字，要保证每次加完这个值都不小于1。翻译过来就是，求这个数组的prefix sum最小的那个数，如果最小的就不小于1，那就return最小的startValue which is 1，如果最小的小于1，那就return (-sum) + 1

class Solution {
    public int minStartValue(int[] nums) {
        int min = Integer.MAX_VALUE;
        int sum = 0;
        for (int num : nums) {
            sum += num;
            min = Math.min(min, sum);
        }
        return min >= 1 ? 1 : 1 - min;
    }
}