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

## Description

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

## Solutions

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

• class Solution {
public:
int minStartValue(vector<int>& nums) {
int s = 0, t = INT_MAX;
for (int num : nums) {
s += num;
t = min(t, s);
}
return max(1, 1 - t);
}
};

• class Solution:
def minStartValue(self, nums: List[int]) -> int:
s, t = 0, inf
for num in nums:
s += num
t = min(t, s)
return max(1, 1 - t)


• func minStartValue(nums []int) int {
s, t := 0, 10000
for _, num := range nums {
s += num
if s < t {
t = s
}
}
if t < 0 {
return 1 - t
}
return 1
}

• function minStartValue(nums: number[]): number {
let sum = 0;
let min = Infinity;
for (const num of nums) {
sum += num;
min = Math.min(min, sum);
}
return Math.max(1, 1 - min);
}


• impl Solution {
pub fn min_start_value(nums: Vec<i32>) -> i32 {
let mut sum = 0;
let mut min = i32::MAX;
for num in nums.iter() {
sum += num;
min = min.min(sum);
}
(1).max(1 - min)
}
}