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Minimum Value to Get Positive Step by Step Sum

Easy

Swiggy

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Topics:

ArraysGreedy Algorithms

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

Try coding it in LeetCode

Minimum Value to Get Positive Step by Step Sum

Easy

Swiggy

3 views

Topics:

ArraysGreedy Algorithms

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the integers within the `nums` array?
Can the input array `nums` be empty or null?
What is the maximum possible length of the `nums` array?
Is there a guarantee that a valid `startValue` will always exist for any given input array?
Are we looking for the absolute *minimum* `startValue`, or is any `startValue` that satisfies the condition acceptable?

Brute Force Solution

Approach

The problem asks us to find the smallest starting number that guarantees a positive sum at each step when we sequentially add numbers from a given list. A brute-force strategy involves guessing a starting number and verifying if it works, then adjusting the guess until the smallest possible starting number is found.

Here's how the algorithm would work step-by-step:

Begin by guessing a starting number, like 1.
Walk through the list of numbers, adding each one to your current sum.
At each addition, check if the current sum is positive.
If, at any point, the sum becomes zero or negative, your initial guess was too low.
Increase the starting number guess (e.g., try 2, 3, and so on).
Repeat steps 2-4 with the new starting number.
Keep doing this until you find a starting number that keeps the sum positive throughout the entire list.
Once a valid starting number is found, check the prior guess to see if you can decrement it to reach the minimal value, while ensuring positive step by step sum.

Code Implementation

def minimum_value_to_get_positive_step_by_step_sum(numbers):
    starting_number = 1

    while True:
        current_sum = starting_number
        is_valid = True

        for number in numbers:
            current_sum += number

            # If the current sum is not positive, mark this starting number as invalid.
            if current_sum <= 0:
                is_valid = False
                break

        # If the starting number is valid, try decrementing it.
        if is_valid:
            if starting_number == 1:
                return starting_number

            test_starting_number = starting_number - 1
            test_current_sum = test_starting_number
            test_is_valid = True

            for number in numbers:
                test_current_sum += number

                # Check if decrementing the number is valid or not
                if test_current_sum <= 0:
                    test_is_valid = False
                    break

            if test_is_valid:
                starting_number = test_starting_number
            else:
                # If we can't decrement, return the current valid value.
                return starting_number
        else:
            # If the starting number is invalid, increment it and try again.
            starting_number += 1

Big(O) Analysis

Time Complexity

O(n*m) – The algorithm iterates through potential starting numbers in the worst case until a suitable one is found. For each starting number, it iterates through the array of size 'n' to check if the running sum remains positive. Let 'm' be the number of starting number attempts required to find the minimum valid starting number. Thus, the outer loop runs 'm' times and the inner loop runs 'n' times. The overall time complexity is therefore O(n*m).

Space Complexity

O(1) – The described algorithm in the plain English explanation iteratively guesses a starting number and checks the sum. It uses only a few variables to store the current sum and the starting number guess. There are no auxiliary data structures like arrays, hash maps, or recursion. Therefore, the space required remains constant, irrespective of the input list size N.

Optimal Solution

Approach

We need to find the smallest starting number that ensures the running total stays positive. The trick is to figure out how low the running total ever gets, and then calculate what starting number would prevent it from going negative or zero.

Here's how the algorithm would work step-by-step:

Keep a running total as you go through the numbers one by one.
At each step, update the running total by adding the current number.
Keep track of the lowest running total you've seen so far.
The minimum starting value we need is the opposite of that lowest running total, plus one. This makes sure we are always positive.

Code Implementation

def minimum_value_to_get_positive_step_by_step_sum(numbers):
    running_total = 0
    minimum_running_total = 0

    for number in numbers:
        running_total += number

        # Track the lowest running total seen so far.
        minimum_running_total = min(minimum_running_total, running_total)

    # Need to start with at least this value.
    minimum_start_value = abs(minimum_running_total) + 1

    return minimum_start_value

Big(O) Analysis

Time Complexity

O(n) – The solution iterates through the input array of size n exactly once to calculate the running total and track the minimum running total encountered. Each operation inside the loop (addition, comparison, assignment) takes constant time. Therefore, the time complexity is directly proportional to the number of elements in the array, resulting in a linear time complexity of O(n).

Space Complexity

O(1) – The algorithm described maintains only two variables: a running total and the minimum running total seen so far. These variables hold single numerical values and do not scale with the input size. Therefore, the auxiliary space required remains constant, regardless of the number of integers (N) in the input array. Consequently, the space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array	Return 1 since an empty array requires only a starting value of 1 to maintain a positive sum.
Array with all positive numbers	The minimum startValue will be 1, as the sum will always be positive.
Array with all negative numbers	The minimum startValue will need to offset the cumulative sum of the negative numbers.
Array with mixed positive and negative numbers	Track the minimum cumulative sum and calculate the required startValue to keep the sum above 0.
Array containing zero values	Zero values do not impact the cumulative sum, so they should be processed normally without special handling.
Integer overflow when calculating cumulative sum	Use long data type to store the cumulative sum to prevent overflow.
Array with extremely large negative numbers such that the required startValue also overflows	Consider adding a check for cases where even a long data type can't hold the negative sums, and either return an error or throw an exception.
Very large array size impacting performance	The solution iterates through the array once, achieving O(n) time complexity which should be efficient enough even for large inputs.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the integers within the `nums` array?
Can the input array `nums` be empty or null?
What is the maximum possible length of the `nums` array?
Is there a guarantee that a valid `startValue` will always exist for any given input array?
Are we looking for the absolute *minimum* `startValue`, or is any `startValue` that satisfies the condition acceptable?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Begin by guessing a starting number, like 1.
Walk through the list of numbers, adding each one to your current sum.
At each addition, check if the current sum is positive.
If, at any point, the sum becomes zero or negative, your initial guess was too low.
Increase the starting number guess (e.g., try 2, 3, and so on).
Repeat steps 2-4 with the new starting number.
Keep doing this until you find a starting number that keeps the sum positive throughout the entire list.
Once a valid starting number is found, check the prior guess to see if you can decrement it to reach the minimal value, while ensuring positive step by step sum.

Code Implementation

def minimum_value_to_get_positive_step_by_step_sum(numbers):
    starting_number = 1

    while True:
        current_sum = starting_number
        is_valid = True

        for number in numbers:
            current_sum += number

            # If the current sum is not positive, mark this starting number as invalid.
            if current_sum <= 0:
                is_valid = False
                break

        # If the starting number is valid, try decrementing it.
        if is_valid:
            if starting_number == 1:
                return starting_number

            test_starting_number = starting_number - 1
            test_current_sum = test_starting_number
            test_is_valid = True

            for number in numbers:
                test_current_sum += number

                # Check if decrementing the number is valid or not
                if test_current_sum <= 0:
                    test_is_valid = False
                    break

            if test_is_valid:
                starting_number = test_starting_number
            else:
                # If we can't decrement, return the current valid value.
                return starting_number
        else:
            # If the starting number is invalid, increment it and try again.
            starting_number += 1

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Keep a running total as you go through the numbers one by one.
At each step, update the running total by adding the current number.
Keep track of the lowest running total you've seen so far.
The minimum starting value we need is the opposite of that lowest running total, plus one. This makes sure we are always positive.

Code Implementation

def minimum_value_to_get_positive_step_by_step_sum(numbers):
    running_total = 0
    minimum_running_total = 0

    for number in numbers:
        running_total += number

        # Track the lowest running total seen so far.
        minimum_running_total = min(minimum_running_total, running_total)

    # Need to start with at least this value.
    minimum_start_value = abs(minimum_running_total) + 1

    return minimum_start_value

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 1 since an empty array requires only a starting value of 1 to maintain a positive sum.
Array with all positive numbers	The minimum startValue will be 1, as the sum will always be positive.
Array with all negative numbers	The minimum startValue will need to offset the cumulative sum of the negative numbers.
Array with mixed positive and negative numbers	Track the minimum cumulative sum and calculate the required startValue to keep the sum above 0.
Array containing zero values	Zero values do not impact the cumulative sum, so they should be processed normally without special handling.
Integer overflow when calculating cumulative sum	Use long data type to store the cumulative sum to prevent overflow.
Array with extremely large negative numbers such that the required startValue also overflows	Consider adding a check for cases where even a long data type can't hold the negative sums, and either return an error or throw an exception.
Very large array size impacting performance	The solution iterates through the array once, achieving O(n) time complexity which should be efficient enough even for large inputs.