The “Minimum Value to Get Positive Step by Step Sum” problem on Leetcode is an intriguing issue that showcases the importance of iterative algorithms and the subtleties of managing sequence data. This exhaustive article aims to cover everything you need to know to solve this problem, with a particular focus on Python programming techniques.

## Table of Contents

- Problem Statement & Constraints
- Initial Thoughts
- Brute-Force Approach
- Optimized Approach
- Analyzing Time and Space Complexity
- Testing and Debugging
- Further Optimizations and Insights
- Conclusion

## 1. Problem Statement & Constraints

You are given an array of integers `nums`

. You start with an initial positive value `startValue`

. Calculate the sum at each step, adding `startValue`

with the i-th element in `nums`

at the i-th step. Return the minimum positive value of `startValue`

such that the sum at each step will always be greater than or equal to 1.

### Constraints and Assumptions

- The array length is between 1 and 100.
- Each integer in the array is between -100 and 100.

## 2. Initial Thoughts

Since the problem asks for a minimum value for `startValue`

, we can immediately think of a looping approach to keep track of the sum at each step. Our goal is to minimize `startValue`

such that the step-by-step sum is always greater than or equal to 1.

## 3. Brute-Force Approach

A naive approach is to iterate from `startValue = 1`

onwards, and for each `startValue`

, iterate through the `nums`

array to check if the sum at each step is positive. If it is, we found our answer.

Python Code:

```
def minStartValue(nums):
startValue = 1
while True:
sum = startValue
for num in nums:
sum += num
if sum < 1:
break
else:
return startValue
startValue += 1
```

## 4. Optimized Approach

We can solve this problem in one pass through the `nums`

array. Keep track of the sum and the minimum sum encountered during the loop. If the minimum sum is less than 1, we need to offset it with `startValue`

.

Python Code:

```
def minStartValue(nums):
sum = 0
min_sum = 0
for num in nums:
sum += num
min_sum = min(min_sum, sum)
return 1 - min_sum if min_sum < 0 else 1
```

## 5. Analyzing Time and Space Complexity

**Brute-Force Approach**: Time Complexity is O(n^2) because of nested loops, and Space Complexity is O(1).**Optimized Approach**: Time Complexity is O(n) because of one pass through the array, and Space Complexity is O(1).

## 6. Testing and Debugging

Important test cases to consider:

- Arrays with all positive integers
- Arrays with all negative integers
- Arrays containing zeros
- Arrays with alternating positive and negative integers

## 7. Further Optimizations and Insights

The optimized approach seems like the most efficient way to solve this problem. However, you can further optimize by using Python’s built-in `min()`

function with a list comprehension.

## 8. Conclusion

The “Minimum Value to Get Positive Step by Step Sum” problem offers a good opportunity to practice optimizing nested loops and understanding how to use iterative methods effectively. We started with a brute-force solution but quickly found a more optimized solution with a single loop. This problem teaches the importance of keeping track of state variables like `sum`

and `min_sum`

while iterating through an array, showcasing a classic technique in solving array-related problems.