Minimum Value to Get Positive Step by Step Sum
The idea to solve this problem is to keep track of the step by step sum, and find the smallest sum. This smallest sum represents the lowest point the step by step sum will get, starting from the initial positive value. The minimum initial positive value should be such that it compensates for this lowest point to maintain the sum at least 1.
Here’s way to implement this in Python:


In this solution, we go through each number in nums
list, update the step_by_step_sum
and keep updating the min_sum
if the current step_by_step_sum
is smaller. Finally, if min_sum
is less than 0, we return 1 minus min_sum
(since min_sum
is negative, subtracting it is equivalent to adding its absolute value), otherwise we just return 1. This ensures that the minimum starting value is always positive and the step by step sum is never less than 1.
Coding Constructs
At a high level, this problem involves two key constructs:
Running Sum: Throughout the loop, we maintain a running sum which keeps track of the cumulative sum of the elements from the array as we iterate through it.
Running Minimum: Simultaneously, we also keep track of the running minimum. The goal here is to monitor the smallest sum we encounter as we keep adding the elements of the array to our running sum. The Running Minimum helps us determine the lowest point our sum can get, which is crucial in this problem to ensure our stepbystep sum never dips below 1.
These two constructs capture the essential operations in this problem’s solution. It’s essentially a combination of maintaining a running sum and keeping track of its lowest value as you iterate over the array.
Dry Run
Let’s say our input array nums
is [3, 2, 3, 4, 2]
. We can set up a table to illustrate the prefix sum and minimum value encountered at each step:
nums  Prefix Sum  Min Val 

3  3  3 
2  1  3 
3  4  4 
4  0  4 
2  2  4 
In this table, “Prefix Sum” is the running sum from the start to the current position, and “Min Val” is the smallest prefix sum we’ve encountered so far.
From this table, we can see that the smallest value is 4, so we need a starting value that ensures the sum is never below 1. That value is the absolute value of the minimum value (4) plus 1, giving us 5.
Brute Force
A brute force approach to this problem would involve iterating over a range of potential starting values and checking each one to see if it allows the stepbystep sum to stay at or above 1 for the entire list of nums
. Here’s a simplified pseudocode example of how this might look:


This solution is inefficient because it potentially requires checking a large number of startValues before finding one that works. As the size of nums
or the negative values in nums
increases, the number of startValues we need to check grows as well. This leads to a time complexity of O(N^2), where N is the size of the input list. Therefore, it’s preferable to use the more efficient solution that leverages the running minimum and prefix sum concepts.
Simple Explanation of the Proof
Let’s break down the proof of why this algorithm works.
In the context of this problem, the objective is to find the smallest starting value such that at any point in time, the running sum of the starting value and the numbers in the list is never less than one.
Our algorithm iterates over the list, accumulating a sum as it goes along. As it does this, it keeps track of the smallest value that this running sum reaches. This smallest sum is crucial because it will tell us how much “debt” we accrued while summing the numbers (i.e., how far below 1 we went).
If the smallest sum is positive or zero, it means that our sum never went below 1 (since we start with 1), so the minimum starting value is 1.
However, if the smallest sum is negative (say x), it means that our sum dipped below 1 at some point and we need to cover that “debt”. We add 1 to that “debt” to ensure that we’re not just covering the debt but also ensuring that the running sum stays at least at 1. So the minimum starting value becomes x+1.
This is essentially what the algorithm is doing. It computes the running sum, tracks the smallest value it encounters, and uses that information to determine the minimum starting value.
It works because it directly translates our objective into algorithmic steps. The running sum simulates the process of summing up the numbers with a starting value, and the minimum sum tells us the “debt” we need to cover to keep the running sum above or equal to 1.
Q&A
Why not return the absolute value instead of having an if else condition in the return?
The minimum start value needs to be at least 1, even when the minimum cumulative sum is nonnegative.
If we simply returned the absolute value of the minimum cumulative sum, we might end up with a start value of 0 if the minimum cumulative sum is nonnegative. However, since the problem specifies that the start value must be positive, we need to return at least 1, hence the condition in the return statement.
For example, if the minimum cumulative sum is 3, we need to start with 4 to ensure the running sum never drops below 1. But if the minimum cumulative sum is 2, we only need to start with 1, not 0 (which is the absolute value of 2  1). Thus, the absolute value wouldn’t work in this context.
What is Step by Step Sum?
“Step by step sum” refers to the running or cumulative sum of an array from the first element to the current element.
Here’s how it works:
 You start with an initial value,
startValue
.  In each iteration, you add the current element of the array to the running sum of the previous elements.
 This forms a new sum which includes the
startValue
and all elements in the array up to the current element.
In the context of this problem, you need to find the minimum startValue
such that the step by step sum is never less than 1.
Let’s consider a simple Python example to illustrate the concept of Step by Step Sum. This is just an illustration and is not directly related to the problem we discussed earlier.
Let’s consider an array nums = [4, 2, 8, 5, 2]
. Now, let’s compute the step by step sum of this array.


In this example, each print statement will output the cumulative sum up to the current element in the array nums
. The startValue
variable holds the step by step sum.
Here’s what each iteration does:
 For the first element, 4,
startValue
becomes 4.  For the second element, 2,
startValue
becomes 4  2 = 2.  For the third element, 8,
startValue
becomes 2  8 = 6.  For the fourth element, 5,
startValue
becomes 6 + 5 = 1.  For the fifth element, 2,
startValue
becomes 1  2 = 3.
So the step by step sum of the array nums
is [4, 2, 6, 1, 3].
This is Sum Accumulator
Correct. A more formal name for “step by step sum” would indeed be an “accumulator”. In computer science, an accumulator is a variable that collects the result of an operation. In this case, the operation is summing or accumulating the values as you iterate through the list.
In terms of Python, this concept is often utilized with a simple for loop, as was demonstrated in the example. This sort of pattern can be useful in a variety of contexts, including when you want to calculate a running total of values in a list.
What if we pick a very large value as the startValue? Will it work?
Technically, yes, picking a very large value for startValue
would prevent the stepbystep sum from ever going below 1. However, the task at hand is to find the minimum positive value of startValue
that meets this condition. Thus, simply picking a large value wouldn’t provide the correct solution as per the problem’s constraints. We need to find the smallest startValue
that prevents the running total from ever going below 1 during the computation. This is why we use the prefix sum and running minimum approach in our solution.
Can you please help me in understanding why 1  minVal is needed? Im finding it hard to understand
Sure, the reasoning behind “1  minVal” is tied to the nature of the problem.
We want to ensure that the “stepbystep sum” is always at least 1. If our minimum sum is already above 0 (i.e., minVal > 0), the smallest startValue we need is just 1, as that keeps the sum positive.
If the minimum sum dips below 0 (i.e., minVal <= 0), we need to adjust the startValue upward to prevent the sum from going below 1. If minVal is 2, for instance, adding just 1 as the startValue would result in a stepbystep sum that goes to 1 (because 2 + 1 = 1). By doing “1  minVal”, we effectively push the sum upward to at least 1 in all cases. In the case of minVal = 2, that gives us “1  (2)” = 3 as the startValue, which ensures the sum stays at least 1 even when we add the 2 from nums.
So the “1  minVal” calculation serves to adjust the startValue upward just enough to keep the running sum at least 1, which is the problem’s requirement.
Hey, can you give some insights by looking at the question of how to figure out to use prefix sum? although i understood solution but not getting why did we use prefix sum please tell how did you figured out
The clues that prefix sum might be helpful in this problem are rooted in the problem’s constraints and requirements.
First, notice that the problem asks about a “stepbystep sum” of elements in the array, which is essentially a running or cumulative sum, that is never less than 1. This should point you towards prefix sum because it’s a technique that involves creating an array where the value at each index is the sum of all previous elements.
Second, prefix sums are particularly useful when dealing with subarrays or a series of numbers in which the order matters. In this case, you’re dealing with a start value and then adding each element of the array to that running total in order. This is another indicator that prefix sum could be a valuable tool.
Finally, using prefix sum provides a straightforward way to identify the smallest cumulative sum, which is crucial to determine the minimum starting value to meet the problem’s requirement.
These are the kinds of insights and thought processes you might use when deciding to use prefix sum for a problem like this. It involves understanding the nature of the problem, the purpose of prefix sum, and how it can be applied in a given context.
Here are 10 distinct problems that use similar underlying concepts:
“Maximal Square”: This problem is related because it requires tracking a running total (sum of square areas) and updating a maximum value, similar to how we kept track of a running sum and updated our minimum value.
“Container With Most Water”: This problem requires an understanding of the principles of keeping track of a maximum value and making comparisons, similar to our running minimum.
“Minimum Path Sum”: This problem requires a similar technique of stepbystep computation, this time finding the path with the smallest sum.
“Product of Array Except Self”: This problem requires running products and keeping track of an array, similar to running sums.
“Minimum Size Subarray Sum”: Like our problem, this one involves finding a subarray that meets a certain criteria, which involves keeping track of a running total.
“Maximum Subarray”: Involves keeping track of running totals and updating a maximum value, similar to our problem but with maximum instead of minimum.
“Maximum Product Subarray”: This problem requires running products instead of sums but uses a very similar approach to track the maximum product.
“Best Time to Buy and Sell Stock”: Here, a running minimum (the lowest price) must be maintained, similar to our problem’s running minimum.
“Trapping Rain Water”: In this problem, running maximums are used to compute how much water can be trapped, analogous to running minimums in our problem.
“Shortest Unsorted Continuous Subarray”: This problem involves manipulating running totals and keeping track of minimum and maximum values to identify the length of the subarray, similar to how we computed running sum and kept track of the minimum value.
Product of Array Except Self
Continuous Subarray Sum
Subarray Sum Equals K
Contiguous Array
Range Sum Query  Immutable
Range Sum Query 2D  Immutable
Max Consecutive Ones III
Subarray Sums Divisible by K
Shortest Subarray with Sum at Least K