The description of the problem

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:

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.

The intuition for this

1. suppose the tempory summarization is sum
2. we get the iterative formula: $startValue-sum\ge 1$
3. therefore we just find the greatest value in sums composed by $1-tempSum$

The codes for this

#include <vector>
#include <iostream>
using namespace std;
class Solution {
    
public:
    int minStartValue(vector<int>& nums) {
    
        int res;
        vector<int> sums;
        for (int i = 0; i < nums.size(); i++) {
    
            int sum = 0;
            for (int j = i; j>=0; j--) {
    
                sum += nums[j];
            }
            sums.emplace_back(1 - sum);
        }
        res = *max_element(sums.begin(), sums.end());
        return res > 0 ? res : 1;
    }
};
int main()
{
    
    Solution s;
    vector<int> nums = {
    -3, 2, -3, 4, 2};
    cout << "minStartValue: " << s.minStartValue(nums) << endl;
    return 0;
}

The results

(base) [email protected]-Air testForCpp % ./test
minStartValue: 5