#1413

Easy

LeetCode Problem Workspace

Minimum Value to Get Positive Step by Step Sum

Find the minimum starting value to ensure the step-by-step sum of an array is always at least 1.

Problem Statement

You are given an array of integers, nums. Starting with an initial positive value, startValue, you need to calculate the cumulative sum in each iteration by adding each element of the array to the current total. Your goal is to return the minimum value of startValue that keeps the cumulative sum never less than 1 during the whole iteration process.

For example, if nums = [-3, 2, -3, 4, 2], after choosing a startValue, you compute the cumulative sum step-by-step. You must ensure that, at no point in the process, the cumulative sum drops below 1. If it does, you need to adjust the start value accordingly.

Examples

Example 1

Input: nums = [-3,2,-3,4,2]

Output: 5

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

Minimum start value should be positive.

Example 3

Input: nums = [1,-2,-3]

Output: 5

Example details omitted.

Constraints

1 <= nums.length <= 100
-100 <= nums[i] <= 100

Solution Approach

Prefix Sum Calculation

Calculate the cumulative prefix sum of the elements in the array as you iterate through it. The key insight here is that by tracking the minimum prefix sum during the iteration, you can determine the minimum start value required to avoid negative sums at any point.

Adjust Start Value

If the minimum cumulative sum calculated is less than 1, the minimum start value should be the difference between 1 and this sum. This ensures the sum never dips below 1 during the iteration process.

Optimization with Linear Time Complexity

The optimal solution involves iterating through the array once, calculating prefix sums, and determining the minimum start value in linear time. This avoids unnecessary recalculations and ensures the solution is efficient for large input sizes.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution is O(n) where n is the length of the array, as we only need to traverse the array once to calculate the prefix sums and determine the minimum start value. The space complexity is O(1) since we only need a few variables to track the current sum and minimum value.

What Interviewers Usually Probe

Understanding of prefix sums and cumulative calculations is required.
The candidate should be able to relate the concept of array adjustments and prefix sum manipulation to real-world scenarios.
The candidate may need to demonstrate efficiency in solving this problem with respect to both time and space.

Common Pitfalls or Variants

Common pitfalls

Forgetting to consider the negative values in the array, which can reduce the cumulative sum below 1.
Confusing the minimum prefix sum with the overall cumulative sum, leading to incorrect start values.
Not optimizing the approach to handle larger arrays efficiently, which may result in time limit exceed errors.

Follow-up variants

Modify the problem to allow negative start values and analyze how the solution adapts.
Introduce more complex arrays, such as those with very large or very small numbers, and observe performance changes.
Add constraints to the problem, like limiting the number of positive or negative values, to test the robustness of the solution.

FAQ

What is the minimum value to get a positive step-by-step sum?

The minimum start value is determined by the difference between 1 and the lowest prefix sum encountered during the iteration.

How does prefix sum help solve this problem?

Prefix sum allows us to track the cumulative total at each step, which helps in determining the minimum start value required to maintain a positive sum.

Why is linear time complexity important in this problem?

Linear time complexity is important because the problem involves iterating through an array, and an optimal solution should handle the array size efficiently, especially when n is large.

What happens if the start value is too small?

If the start value is too small, the cumulative sum may fall below 1 during the iterations, which is not allowed according to the problem's constraints.

What is the primary failure mode in solving this problem?

The primary failure mode is not correctly adjusting the start value when the prefix sum dips below 1, which causes the step-by-step sum to become negative.

terminal

Solution

Solution 1

#### Python3

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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)

Solution 2

#### Python3

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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)

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