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