Jobs

Interviews

Discussions

Sum of All Odd Length Subarrays

Easy

Asked by:

32 views

Topics:

Arrays

Given an array of positive integers arr, return the sum of all possible odd-length subarrays of arr.

A subarray is a contiguous subsequence of the array.

Example 1:

Input: arr = [1,4,2,5,3]
Output: 58
Explanation: The odd-length subarrays of arr and their sums are:
[1] = 1
[4] = 4
[2] = 2
[5] = 5
[3] = 3
[1,4,2] = 7
[4,2,5] = 11
[2,5,3] = 10
[1,4,2,5,3] = 15
If we add all these together we get 1 + 4 + 2 + 5 + 3 + 7 + 11 + 10 + 15 = 58

Example 2:

Input: arr = [1,2]
Output: 3
Explanation: There are only 2 subarrays of odd length, [1] and [2]. Their sum is 3.

Example 3:

Input: arr = [10,11,12]
Output: 66

Constraints:

1 <= arr.length <= 100
1 <= arr[i] <= 1000

Follow up:

Could you solve this problem in O(n) time complexity?

Try coding it in LeetCode

Sum of All Odd Length Subarrays

Easy

Asked by:

32 views

Topics:

Arrays

Given an array of positive integers arr, return the sum of all possible odd-length subarrays of arr.

A subarray is a contiguous subsequence of the array.

Example 1:

Input: arr = [1,4,2,5,3]
Output: 58
Explanation: The odd-length subarrays of arr and their sums are:
[1] = 1
[4] = 4
[2] = 2
[5] = 5
[3] = 3
[1,4,2] = 7
[4,2,5] = 11
[2,5,3] = 10
[1,4,2,5,3] = 15
If we add all these together we get 1 + 4 + 2 + 5 + 3 + 7 + 11 + 10 + 15 = 58

Example 2:

Input: arr = [1,2]
Output: 3
Explanation: There are only 2 subarrays of odd length, [1] and [2]. Their sum is 3.

Example 3:

Input: arr = [10,11,12]
Output: 66

Constraints:

1 <= arr.length <= 100
1 <= arr[i] <= 1000

Follow up:

Could you solve this problem in O(n) time complexity?

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 maximum size of the input array?
Can the input array contain negative numbers?
Is the input array guaranteed to be non-empty?
Should the return value be an integer or a long if the sum exceeds the integer limit?
Can you provide a small example input and its expected output to confirm my understanding?

Brute Force Solution

Approach

The brute force approach to this problem means we're going to look at every possible group of numbers we can make from the original list. We're only interested in groups that have an odd number of elements, and we will compute a running total of all these groups.

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

First, think about groups with just one number. Check each individual number from the original list to see if it counts.
Next, look at groups with three numbers. Start from the beginning of the list and take the first three numbers, and compute their sum.
Then, shift over by one position and take the next group of three numbers.
Continue this until you have checked all the groups of three numbers.
Now, repeat the process for groups of five numbers, seven numbers, and so on, until you reach the largest possible odd-sized group.
As you check each odd-sized group, add up all of the numbers in that group.
Finally, after going through all the odd-sized groups, add all of the group sums together to get the final answer.

Code Implementation

def sum_odd_length_subarrays(numbers):
    total_sum = 0

    # Iterate through possible odd subarray lengths
    for subarray_length in range(1, len(numbers) + 1, 2):

        # Iterate through starting positions for each length
        for start_index in range(len(numbers) - subarray_length + 1):

            # Calculate the end index of the current subarray
            end_index = start_index + subarray_length

            # Sum the elements in the current subarray
            subarray_sum = 0
            for index in range(start_index, end_index):
                subarray_sum += numbers[index]

            # Accumulate the sum of all odd length subarrays
            total_sum += subarray_sum

    return total_sum

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm iterates through all possible odd length subarrays of the input array. For an array of size n, we iterate through lengths 1, 3, 5, ..., up to n (or n-1 if n is even). For each odd length l, we iterate through the array to create subarrays of that length. The number of subarrays of length l is approximately n - l + 1, which is O(n). Since we consider lengths up to n, we have roughly n/2 iterations over lengths l, each costing O(n). Therefore the total time complexity approximates to n/2 * n = n^2/2, simplifying to O(n^2).

Space Complexity

O(1) – The brute force approach described calculates the sum of odd-length subarrays directly without using any auxiliary data structures like temporary arrays or hash maps to store intermediate subarray sums or indices. It only requires a few integer variables to keep track of the starting position and the length of the current subarray. Therefore, the space complexity remains constant irrespective of the size of the input array, which we denote as N. This results in a space complexity of O(1).

Optimal Solution

Approach

The goal is to add up numbers from smaller chunks of a list, but only the chunks that have an odd number of elements. Instead of checking every possible chunk, we will calculate how many times each individual number in the list will be part of an odd-sized chunk and then add it up only that many times.

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

Consider each number in the list one by one.
For each number, figure out how many total chunks (both odd and even length) it will be part of. This depends on its position in the list.
Then, figure out how many of those chunks have an odd length.
Multiply the number by the count of odd-length chunks it appears in.
Add up all these products to get the final answer. This way, you only touch each number a few times instead of repeatedly adding it for every chunk.

Code Implementation

def sum_odd_length_subarrays(array_of_numbers):
    total_sum = 0
    array_length = len(array_of_numbers)

    for index, number in enumerate(array_of_numbers):
        # Calculate total subarrays the number is in
        total_subarrays = (index + 1) * (array_length - index)

        # Calculate odd length subarrays it is in
        odd_length_subarrays = (total_subarrays + 1) // 2

        # Add the contribution of this number to the total sum
        total_sum += number * odd_length_subarrays

    return total_sum

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each of the n elements in the input array once. For each element, it performs a constant number of arithmetic operations to calculate how many odd-length subarrays it appears in. Since the number of operations per element is constant and independent of n, the overall time complexity is directly proportional to the input size n. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The provided explanation describes calculating the number of odd-length subarrays a single element contributes to and accumulating the product of this count and the element's value. No auxiliary data structures like arrays, lists, or hash maps are explicitly created. Only a few variables (counters and an accumulator for the final sum) are used to perform calculations. Therefore, the space complexity is constant, independent of the input array size N.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as there are no subarrays.
Array with a single element	Return the value of that single element as it's the only odd-length subarray.
Array with all elements being zero	The standard algorithm should correctly sum the zeros for each odd-length subarray.
Array with all identical non-zero elements	The result should be a sum of that element multiplied by the counts of odd length subarrays.
Array containing negative numbers	The core logic should correctly add negative numbers to the sum.
Array with a large number of elements (performance)	Optimize the solution to avoid redundant calculations and aim for O(n) or O(n log n) complexity to handle large inputs efficiently.
Integer overflow when calculating the sum	Use a data type with a larger range (e.g., long) to store the sum to prevent overflow.
Very large individual array elements	Handle large individual elements by preventing overflow during intermediate calculations by casting to a larger type when necessary.

Null or empty input array

How to Handle:

Return 0 immediately as there are no subarrays.

Array with a single element

How to Handle:

Return the value of that single element as it's the only odd-length subarray.

Array with all elements being zero

How to Handle:

The standard algorithm should correctly sum the zeros for each odd-length subarray.

Array with all identical non-zero elements

How to Handle:

The result should be a sum of that element multiplied by the counts of odd length subarrays.

Array containing negative numbers

How to Handle:

The core logic should correctly add negative numbers to the sum.

Array with a large number of elements (performance)

How to Handle:

Optimize the solution to avoid redundant calculations and aim for O(n) or O(n log n) complexity to handle large inputs efficiently.

Integer overflow when calculating the sum

How to Handle:

Use a data type with a larger range (e.g., long) to store the sum to prevent overflow.

Very large individual array elements

How to Handle:

Handle large individual elements by preventing overflow during intermediate calculations by casting to a larger type when necessary.

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 maximum size of the input array?
Can the input array contain negative numbers?
Is the input array guaranteed to be non-empty?
Should the return value be an integer or a long if the sum exceeds the integer limit?
Can you provide a small example input and its expected output to confirm my understanding?

Brute Force Solution

Approach

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

First, think about groups with just one number. Check each individual number from the original list to see if it counts.
Next, look at groups with three numbers. Start from the beginning of the list and take the first three numbers, and compute their sum.
Then, shift over by one position and take the next group of three numbers.
Continue this until you have checked all the groups of three numbers.
Now, repeat the process for groups of five numbers, seven numbers, and so on, until you reach the largest possible odd-sized group.
As you check each odd-sized group, add up all of the numbers in that group.
Finally, after going through all the odd-sized groups, add all of the group sums together to get the final answer.

Code Implementation

def sum_odd_length_subarrays(numbers):
    total_sum = 0

    # Iterate through possible odd subarray lengths
    for subarray_length in range(1, len(numbers) + 1, 2):

        # Iterate through starting positions for each length
        for start_index in range(len(numbers) - subarray_length + 1):

            # Calculate the end index of the current subarray
            end_index = start_index + subarray_length

            # Sum the elements in the current subarray
            subarray_sum = 0
            for index in range(start_index, end_index):
                subarray_sum += numbers[index]

            # Accumulate the sum of all odd length subarrays
            total_sum += subarray_sum

    return total_sum

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Consider each number in the list one by one.
For each number, figure out how many total chunks (both odd and even length) it will be part of. This depends on its position in the list.
Then, figure out how many of those chunks have an odd length.
Multiply the number by the count of odd-length chunks it appears in.
Add up all these products to get the final answer. This way, you only touch each number a few times instead of repeatedly adding it for every chunk.

Code Implementation

def sum_odd_length_subarrays(array_of_numbers):
    total_sum = 0
    array_length = len(array_of_numbers)

    for index, number in enumerate(array_of_numbers):
        # Calculate total subarrays the number is in
        total_subarrays = (index + 1) * (array_length - index)

        # Calculate odd length subarrays it is in
        odd_length_subarrays = (total_subarrays + 1) // 2

        # Add the contribution of this number to the total sum
        total_sum += number * odd_length_subarrays

    return total_sum

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as there are no subarrays.
Array with a single element	Return the value of that single element as it's the only odd-length subarray.
Array with all elements being zero	The standard algorithm should correctly sum the zeros for each odd-length subarray.
Array with all identical non-zero elements	The result should be a sum of that element multiplied by the counts of odd length subarrays.
Array containing negative numbers	The core logic should correctly add negative numbers to the sum.
Array with a large number of elements (performance)	Optimize the solution to avoid redundant calculations and aim for O(n) or O(n log n) complexity to handle large inputs efficiently.
Integer overflow when calculating the sum	Use a data type with a larger range (e.g., long) to store the sum to prevent overflow.
Very large individual array elements	Handle large individual elements by preventing overflow during intermediate calculations by casting to a larger type when necessary.

Null or empty input array

How to Handle:

Return 0 immediately as there are no subarrays.

Array with a single element

How to Handle:

Return the value of that single element as it's the only odd-length subarray.

Array with all elements being zero

How to Handle:

The standard algorithm should correctly sum the zeros for each odd-length subarray.

Array with all identical non-zero elements

How to Handle:

The result should be a sum of that element multiplied by the counts of odd length subarrays.

Array containing negative numbers

How to Handle:

The core logic should correctly add negative numbers to the sum.

Array with a large number of elements (performance)

How to Handle:

Optimize the solution to avoid redundant calculations and aim for O(n) or O(n log n) complexity to handle large inputs efficiently.

Integer overflow when calculating the sum

How to Handle:

Use a data type with a larger range (e.g., long) to store the sum to prevent overflow.

Very large individual array elements

How to Handle:

Handle large individual elements by preventing overflow during intermediate calculations by casting to a larger type when necessary.