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Left and Right Sum Differences

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Arrays

You are given a 0-indexed integer array nums of size n.

Define two arrays leftSum and rightSum where:

leftSum[i] is the sum of elements to the left of the index i in the array nums. If there is no such element, leftSum[i] = 0.
rightSum[i] is the sum of elements to the right of the index i in the array nums. If there is no such element, rightSum[i] = 0.

Return an integer array answer of size n where answer[i] = |leftSum[i] - rightSum[i]|.

Example 1:

Input: nums = [10,4,8,3]
Output: [15,1,11,22]
Explanation: The array leftSum is [0,10,14,22] and the array rightSum is [15,11,3,0].
The array answer is [|0 - 15|,|10 - 11|,|14 - 3|,|22 - 0|] = [15,1,11,22].

Example 2:

Input: nums = [1]
Output: [0]
Explanation: The array leftSum is [0] and the array rightSum is [0].
The array answer is [|0 - 0|] = [0].

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁵

Try coding it in LeetCode

Left and Right Sum Differences

Easy

Asked by:

2 views

Topics:

Arrays

You are given a 0-indexed integer array nums of size n.

Define two arrays leftSum and rightSum where:

leftSum[i] is the sum of elements to the left of the index i in the array nums. If there is no such element, leftSum[i] = 0.
rightSum[i] is the sum of elements to the right of the index i in the array nums. If there is no such element, rightSum[i] = 0.

Return an integer array answer of size n where answer[i] = |leftSum[i] - rightSum[i]|.

Example 1:

Input: nums = [10,4,8,3]
Output: [15,1,11,22]
Explanation: The array leftSum is [0,10,14,22] and the array rightSum is [15,11,3,0].
The array answer is [|0 - 15|,|10 - 11|,|14 - 3|,|22 - 0|] = [15,1,11,22].

Example 2:

Input: nums = [1]
Output: [0]
Explanation: The array leftSum is [0] and the array rightSum is [0].
The array answer is [|0 - 0|] = [0].

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁵

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:

The constraints state that array elements are between 1 and 10^5. Can I assume the input will only contain positive integers, and I don't need to account for zeros or negative numbers?
Just to confirm, since the constraints specify the array length is at least 1, does this mean I don't need to handle the edge case of an empty input array?
The total sum could be up to 10^8 given the constraints. Should I be mindful of potential integer overflow and use a 64-bit integer type for my sum calculations, or will a standard 32-bit integer suffice?
Regarding space complexity, is it acceptable to use O(n) auxiliary space to store the prefix and suffix sums, or would you prefer a solution that uses O(1) constant extra space?
Should I treat the input array `nums` as read-only, or am I permitted to modify it in place if that proves useful for my implementation?

Brute Force Solution

Approach

The straightforward approach is to examine every number in the list one by one. For each number, we independently calculate the total of everything to its left and the total of everything to its right, then find the difference.

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

Create a new, empty collection to store your final answers.
For each number in the original collection, from beginning to end, perform the following actions.
First, calculate the sum of all the numbers that appear to its left. If there are no numbers to its left, this sum is zero.
Next, calculate the sum of all the numbers that appear to its right. If there are no numbers to its right, this sum is zero.
Find the absolute difference between the left sum and the right sum you just calculated.
Add this difference to your new collection of answers.
After repeating this for every single number, the new collection you built is the final result.

Code Implementation

def left_and_right_sum_differences_brute_force(input_numbers):
    # This collection will hold the final calculated differences for each position.
    answer_array = []

    # We must process each number individually to compute its unique left and right sums.
    for current_index in range(len(input_numbers)):

        # Slicing the list isolates the elements to the left and right of the current number.
        left_side_numbers = input_numbers[0:current_index]
        left_sum = sum(left_side_numbers)

        right_side_numbers = input_numbers[current_index + 1:]
        right_sum = sum(right_side_numbers)

        # The requirement is for a positive value, so we take the absolute difference.
        absolute_difference = abs(left_sum - right_sum)
        answer_array.append(absolute_difference)

    return answer_array

Big(O) Analysis

Time Complexity

O(n²) – The time complexity is driven by a process that iterates through each of the n elements in the input collection. For every single element, the algorithm independently recalculates the sum of all elements to its left and the sum of all elements to its right. Each of these summation steps can require scanning up to n-1 elements. Performing this O(n) work for each of the n elements results in a total number of operations proportional to n * n, which simplifies to O(n²).

Space Complexity

O(N) – The primary driver of space complexity is the "new, empty collection" created to store the results. For an input collection of size N, the algorithm calculates and stores one difference value for each element, resulting in an answer collection of size N. While simple variables are used to temporarily hold the left and right sums during calculations, their space requirement is constant. Thus, the auxiliary space used by the algorithm is directly proportional to the input size.

Optimal Solution

Approach

The most efficient way to solve this is to avoid recalculating sums from scratch for every single number. Instead, we can cleverly build up the sums for the left and right sides by making two passes over the collection of numbers, one from each direction.

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

First, we will figure out the sum of all numbers to the left of every position. We do this by sweeping through the numbers from beginning to end.
We'll keep a running total, which starts at zero. For each number we visit, its 'left sum' is simply our current running total.
We record this left sum for the current position. Then, we add the number at this position to our running total before moving to the next one.
After this first pass is complete, we will have a record of the 'left sum' for every single position.
Next, we'll do a similar process for the right side. We start a new running total, also at zero, but this time we sweep from the end of the collection back to the beginning.
For each number we visit on this backward pass, its 'right sum' is our current running total for this direction.
Now we have both the left sum (from our first pass) and the right sum for this position. We find the absolute difference between these two sums, which is our final answer for this spot.
Finally, we add the number at the current position to our backward running total before moving on to the next spot.
By repeating this for the entire backward pass, we efficiently calculate the final answer for every position without any repetitive work.

Code Implementation

def left_and_right_sum_differences(input_numbers: list[int]) -> list[int]:
    number_of_elements = len(input_numbers)
    answer_array = [0] * number_of_elements

    # This first pass populates our temporary list with the sum of all numbers to the left.

    left_running_sum = 0
    for index in range(number_of_elements):
        answer_array[index] = left_running_sum
        left_running_sum += input_numbers[index]

    # The second pass calculates right-side sums while simultaneously finding the final difference.

    right_running_sum = 0
    for index in range(number_of_elements - 1, -1, -1):
        # We can now calculate the final value because we have the left sum and the right sum.

        answer_array[index] = abs(answer_array[index] - right_running_sum)
        right_running_sum += input_numbers[index]

    return answer_array

Big(O) Analysis

Time Complexity

O(n) – The time complexity is determined by the two separate passes made over the input array of size n, where n is the number of elements. The first pass iterates from beginning to end to calculate the cumulative sum from the left for each position, which involves n operations. The second pass iterates from the end to the beginning to calculate the right sums and compute the final difference, also requiring n operations. Since these two passes are sequential and not nested, the total number of operations is approximately n + n, or 2n, which simplifies to a time complexity of O(n).

Space Complexity

O(N) – The algorithm's space complexity is primarily driven by the storage of intermediate results from the first pass. The explanation states it creates "a record of the 'left sum' for every single position," which requires an auxiliary array of size N, where N is the number of input elements. This array is essential because the second pass, which moves from right to left, needs these stored values to compute the final absolute difference at each position. While the running totals for each pass use constant space, this array's size scales directly with the input, resulting in an O(N) auxiliary space complexity.

Edge Cases

Case	How to Handle
Input array with a single element	The solution correctly produces an output array [0] as both the left and right sums are zero by definition.
Calculation for the first element at index 0	The algorithm correctly handles this boundary by ensuring the leftSum for the first element is always zero.
Calculation for the last element at index n-1	The algorithm correctly handles this boundary by ensuring the rightSum for the last element is always zero.
Maximum size input (n=1000) affecting performance	An efficient O(n) linear-time solution is required to avoid a timeout that a naive O(n^2) approach would cause.
Potential for integer overflow when calculating sums	The solution must confirm that the maximum possible sum, around 10^8, safely fits within a standard 32-bit signed integer.
Input array containing all identical values	The solution's logic is independent of the input value distribution and correctly processes arrays with uniform numbers.
Input values being strictly positive integers	The solution can rely on the problem's constraint that numbers are positive, which means prefix sums will be monotonically increasing.
Space complexity for storing intermediate sums	An optimal solution uses O(1) extra space by calculating sums on-the-fly and reusing the output array, avoiding explicit `leftSum` and `rightSum` arrays.

Input array with a single element

How to Handle:

The solution correctly produces an output array [0] as both the left and right sums are zero by definition.

Calculation for the first element at index 0

How to Handle:

The algorithm correctly handles this boundary by ensuring the leftSum for the first element is always zero.

Calculation for the last element at index n-1

How to Handle:

The algorithm correctly handles this boundary by ensuring the rightSum for the last element is always zero.

Maximum size input (n=1000) affecting performance

How to Handle:

An efficient O(n) linear-time solution is required to avoid a timeout that a naive O(n^2) approach would cause.

Potential for integer overflow when calculating sums

How to Handle:

The solution must confirm that the maximum possible sum, around 10^8, safely fits within a standard 32-bit signed integer.

Input array containing all identical values

How to Handle:

The solution's logic is independent of the input value distribution and correctly processes arrays with uniform numbers.

Input values being strictly positive integers

How to Handle:

The solution can rely on the problem's constraint that numbers are positive, which means prefix sums will be monotonically increasing.

Space complexity for storing intermediate sums

How to Handle:

An optimal solution uses O(1) extra space by calculating sums on-the-fly and reusing the output array, avoiding explicit `leftSum` and `rightSum` arrays.

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:

The constraints state that array elements are between 1 and 10^5. Can I assume the input will only contain positive integers, and I don't need to account for zeros or negative numbers?
Just to confirm, since the constraints specify the array length is at least 1, does this mean I don't need to handle the edge case of an empty input array?
The total sum could be up to 10^8 given the constraints. Should I be mindful of potential integer overflow and use a 64-bit integer type for my sum calculations, or will a standard 32-bit integer suffice?
Regarding space complexity, is it acceptable to use O(n) auxiliary space to store the prefix and suffix sums, or would you prefer a solution that uses O(1) constant extra space?
Should I treat the input array `nums` as read-only, or am I permitted to modify it in place if that proves useful for my implementation?

Brute Force Solution

Approach

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

Create a new, empty collection to store your final answers.
For each number in the original collection, from beginning to end, perform the following actions.
First, calculate the sum of all the numbers that appear to its left. If there are no numbers to its left, this sum is zero.
Next, calculate the sum of all the numbers that appear to its right. If there are no numbers to its right, this sum is zero.
Find the absolute difference between the left sum and the right sum you just calculated.
Add this difference to your new collection of answers.
After repeating this for every single number, the new collection you built is the final result.

Code Implementation

def left_and_right_sum_differences_brute_force(input_numbers):
    # This collection will hold the final calculated differences for each position.
    answer_array = []

    # We must process each number individually to compute its unique left and right sums.
    for current_index in range(len(input_numbers)):

        # Slicing the list isolates the elements to the left and right of the current number.
        left_side_numbers = input_numbers[0:current_index]
        left_sum = sum(left_side_numbers)

        right_side_numbers = input_numbers[current_index + 1:]
        right_sum = sum(right_side_numbers)

        # The requirement is for a positive value, so we take the absolute difference.
        absolute_difference = abs(left_sum - right_sum)
        answer_array.append(absolute_difference)

    return answer_array

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, we will figure out the sum of all numbers to the left of every position. We do this by sweeping through the numbers from beginning to end.
We'll keep a running total, which starts at zero. For each number we visit, its 'left sum' is simply our current running total.
We record this left sum for the current position. Then, we add the number at this position to our running total before moving to the next one.
After this first pass is complete, we will have a record of the 'left sum' for every single position.
Next, we'll do a similar process for the right side. We start a new running total, also at zero, but this time we sweep from the end of the collection back to the beginning.
For each number we visit on this backward pass, its 'right sum' is our current running total for this direction.
Now we have both the left sum (from our first pass) and the right sum for this position. We find the absolute difference between these two sums, which is our final answer for this spot.
Finally, we add the number at the current position to our backward running total before moving on to the next spot.
By repeating this for the entire backward pass, we efficiently calculate the final answer for every position without any repetitive work.

Code Implementation

def left_and_right_sum_differences(input_numbers: list[int]) -> list[int]:
    number_of_elements = len(input_numbers)
    answer_array = [0] * number_of_elements

    # This first pass populates our temporary list with the sum of all numbers to the left.

    left_running_sum = 0
    for index in range(number_of_elements):
        answer_array[index] = left_running_sum
        left_running_sum += input_numbers[index]

    # The second pass calculates right-side sums while simultaneously finding the final difference.

    right_running_sum = 0
    for index in range(number_of_elements - 1, -1, -1):
        # We can now calculate the final value because we have the left sum and the right sum.

        answer_array[index] = abs(answer_array[index] - right_running_sum)
        right_running_sum += input_numbers[index]

    return answer_array

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input array with a single element	The solution correctly produces an output array [0] as both the left and right sums are zero by definition.
Calculation for the first element at index 0	The algorithm correctly handles this boundary by ensuring the leftSum for the first element is always zero.
Calculation for the last element at index n-1	The algorithm correctly handles this boundary by ensuring the rightSum for the last element is always zero.
Maximum size input (n=1000) affecting performance	An efficient O(n) linear-time solution is required to avoid a timeout that a naive O(n^2) approach would cause.
Potential for integer overflow when calculating sums	The solution must confirm that the maximum possible sum, around 10^8, safely fits within a standard 32-bit signed integer.
Input array containing all identical values	The solution's logic is independent of the input value distribution and correctly processes arrays with uniform numbers.
Input values being strictly positive integers	The solution can rely on the problem's constraint that numbers are positive, which means prefix sums will be monotonically increasing.
Space complexity for storing intermediate sums	An optimal solution uses O(1) extra space by calculating sums on-the-fly and reusing the output array, avoiding explicit `leftSum` and `rightSum` arrays.

Input array with a single element

How to Handle:

The solution correctly produces an output array [0] as both the left and right sums are zero by definition.

Calculation for the first element at index 0

How to Handle:

The algorithm correctly handles this boundary by ensuring the leftSum for the first element is always zero.

Calculation for the last element at index n-1

How to Handle:

The algorithm correctly handles this boundary by ensuring the rightSum for the last element is always zero.

Maximum size input (n=1000) affecting performance

How to Handle:

An efficient O(n) linear-time solution is required to avoid a timeout that a naive O(n^2) approach would cause.

Potential for integer overflow when calculating sums

How to Handle:

The solution must confirm that the maximum possible sum, around 10^8, safely fits within a standard 32-bit signed integer.

Input array containing all identical values

How to Handle:

The solution's logic is independent of the input value distribution and correctly processes arrays with uniform numbers.

Input values being strictly positive integers

How to Handle:

The solution can rely on the problem's constraint that numbers are positive, which means prefix sums will be monotonically increasing.

Space complexity for storing intermediate sums

How to Handle:

An optimal solution uses O(1) extra space by calculating sums on-the-fly and reusing the output array, avoiding explicit `leftSum` and `rightSum` arrays.