# Solving Leetcode Interviews in Seconds with AI: Minimum Value to Get Positive Step by Step Sum


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1413" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the breakdown of the solution:

*   **Calculate Prefix Sums:** Iterate through the `nums` array, calculating the cumulative sum at each step. This will represent the step-by-step sum if we started with a `startValue` of 0.

*   **Find Minimum Prefix Sum:** Determine the minimum value among all prefix sums. This indicates the largest negative dip the cumulative sum experiences.

*   **Determine Minimum Start Value:** The minimum positive `startValue` needed is the absolute value of the minimum prefix sum, plus 1. This ensures that even at the lowest point, the sum remains at least 1.

*   **Time & Space Complexity:** O(n) time complexity, O(1) space complexity.

	
	# Code
	```python
	def minStartValue(nums):
    """
    Finds the minimum positive start value such that the step by step sum is never less than 1.

    Args:
        nums: A list of integers.

    Returns:
        The minimum positive start value.
    """

    min_prefix_sum = 0
    current_sum = 0

    for num in nums:
        current_sum += num
        min_prefix_sum = min(min_prefix_sum, current_sum)

    return abs(min_prefix_sum) + 1
	```