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

这题比较简单，我们只要在不断的相加中，把最小的和找出来，只要最小的和不小于1,就能符合题意。

class MinimumValueToGetPositiveStepByStepSum : public Solution {
public:
    void Exec() {

    }
    int minStartValue(vector<int>& nums) {
        int sum = nums[0];
        int minSum = sum;

        for (int i = 1; i < nums.size(); i++) {
            sum += nums[i];
            if (sum < minSum) {
                minSum = sum;
            }
        }
        if (minSum >= 1) {
            return 1;
        }
        return 1 - minSum;
    }
};