1413. Minimum Value to Get Positive Step by Step Sum

Link: https://leetcode.com/problems/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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題目翻譯:

選擇一個 startvalue 而這個值在跟陣列的從頭到尾的元素相加時不能小於 1。 startvalue 至少要是正數。

程式思路:

很 easy 的題目。

class Solution {
public:
    int minStartValue(vector<int>& nums) {

        int x = 0, step_min = 0;
        for(auto it : nums)
        {
            x += it;
            if(step_min > x)
                step_min = x;
        }
        return  (step_min >= 0) ? 1 : 1 - step_min;
    }
};