1413. Minimum Value to Get Positive Step by Step Sum.md

1413. Minimum Value to Get Positive Step by Step Sum

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

代码

• 记录逐步求和的最小值，若为负数，则最小大于1的初始值就是1 - totalMin，否则选1即可。

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