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Maximum Size Subarray Sum Equals k

Medium

Google

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Topics:

Arrays

Given an array of integers nums and a target value k, find the maximum length of a subarray that sums to k.

For example:

nums = [1, -1, 5, -2, 3], k = 3. The answer is 4. (Because the subarray [1, -1, 5, -2] sums to 3.)
nums = [ -2, -1, 2, 1 ], k = 1. The answer is 2. (Because the subarray [-1, 2] sums to 1.)
nums = [1, 2, 3, 4, 5], k = 0. The answer is 0. (There is no subarray that sums to 0.)
nums = [], k = 0. The answer is 0.

Explain your approach and provide the time and space complexity analysis.

Maximum Size Subarray Sum Equals k

Medium

Google

2 views

Topics:

Arrays

Given an array of integers nums and a target value k, find the maximum length of a subarray that sums to k.

For example:

nums = [1, -1, 5, -2, 3], k = 3. The answer is 4. (Because the subarray [1, -1, 5, -2] sums to 3.)
nums = [ -2, -1, 2, 1 ], k = 1. The answer is 2. (Because the subarray [-1, 2] sums to 1.)
nums = [1, 2, 3, 4, 5], k = 0. The answer is 0. (There is no subarray that sums to 0.)
nums = [], k = 0. The answer is 0.

Explain your approach and provide the time and space complexity analysis.

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 are the possible values for the elements within the input array, including negative numbers, zero, or positive numbers?
What should be returned if no subarray sums to `k`? Should I return null, an empty array, or some other indicator?
What is the data type of `k`? Is it guaranteed to be an integer, or could it be a floating-point number?
Is the input array guaranteed to be non-null and non-empty? What should I do if it is null or empty?
If multiple subarrays have a sum equal to `k`, should I return the one with the maximum size or any valid subarray?

Brute Force Solution

Approach

The brute force strategy involves checking every possible piece of the given list to see if it meets our desired sum. We will examine every possible continuous section of the list, calculate its sum, and see if it matches the target value. The goal is to find the longest section that adds up to the target.

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

Start by considering the first number in the list as the beginning of a potential section.
Then, include the second number, and check if the sum of the first two numbers equals the target. If it does then store the size of this piece.
Continue adding numbers one by one to this section from the beginning, each time checking if the total sum matches the target and saving the size if applicable.
Once you've checked all possible sections starting from the first number, move to the second number in the list.
Repeat the same process, creating sections starting from the second number, then the third, and so on, until you've checked every possible starting point.
At each step, remember to compare the length of any matching sections with previously found sections, always storing the longest section that matches the target sum.

Code Implementation

def max_size_subarray_sum_equals_k_brute_force(numbers, target):
    maximum_size = 0

    for start_index in range(len(numbers)): # Iterate through possible start indices

        current_sum = 0

        for end_index in range(start_index, len(numbers)): # Iterate through possible end indices

            current_sum += numbers[end_index]

            #Check if the current subarray sum is equal to the target
            if current_sum == target:

                subarray_size = end_index - start_index + 1

                # We want to maximize the size of the subarray
                if subarray_size > maximum_size:

                    maximum_size = subarray_size

    return maximum_size

Big(O) Analysis

Time Complexity

O(n²) – The provided brute force solution iterates through all possible subarrays of the input array. The outer loop iterates from the first element to the last element (n iterations), defining the starting point of the subarray. The inner loop iterates from the starting point to the end of the array, calculating the sum of the current subarray. In the worst-case scenario, for each starting element, we potentially iterate through all remaining elements. Thus, the total number of operations is proportional to n + (n-1) + (n-2) + ... + 1, which is equivalent to n(n+1)/2. This simplifies to O(n²).

Space Complexity

O(1) – The provided brute force solution iterates through subarrays of the input list. It only uses a few variables to keep track of the start and end indices of the subarrays being considered, as well as the maximum length found so far. These variables consume a constant amount of space, irrespective of the input list's size, N. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The goal is to find the longest continuous section within a list of numbers that adds up to a specific target number. Instead of checking every possible section, we use a trick to quickly identify potential sections that meet the criteria.

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

Keep track of the running total as you go through the list of numbers.
Store each running total along with its position in the list in a special lookup table.
For each running total, check if there's a previous running total in the table that, when subtracted from the current running total, equals the target number.
If such a previous running total exists, it means the numbers between those two positions add up to the target number.
Calculate the length of that section and remember the longest one found so far.
If a running total is not already in the lookup table, add it along with its position. Do this only if it's not already there to ensure we find the earliest, and therefore the longest, matching subarray.
After going through the entire list, you'll have the length of the longest section that adds up to the target number.

Code Implementation

def max_size_subarray_sum_equals_k(nums, target):
    running_sum = 0
    max_subarray_length = 0
    sum_index_map = {}

    # Store initial running sum of 0 at index -1
    sum_index_map[0] = -1

    for i, num in enumerate(nums):
        running_sum += num

        # Check if running_sum - target exists in the map
        if (running_sum - target) in sum_index_map:
            # Update the max length
            subarray_length = i - sum_index_map[running_sum - target]
            max_subarray_length = max(max_subarray_length, subarray_length)

        # Store the running sum and index
        # Only if it's not already present. Ensure longest subarray.
        if running_sum not in sum_index_map:
            sum_index_map[running_sum] = i

    return max_subarray_length

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n once to compute the running sum. Inside the loop, a constant-time operation (lookup and insertion in a hash map) is performed. Therefore, the time complexity is dominated by the single pass through the array, resulting in O(n) time complexity.

Space Complexity

O(N) – The space complexity is determined by the lookup table, which is used to store the running total and its index. In the worst-case scenario, where no two prefix sums are equal, we may end up storing all N running totals along with their corresponding indices in the lookup table. Thus, the auxiliary space used grows linearly with the input size N. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
Empty array input	Return 0 immediately as there is no subarray.
Null array input	Throw an IllegalArgumentException or return 0, depending on requirements.
Array with a single element that equals k	Return 1 as the subarray of size 1 satisfies the condition.
Array with a single element that does not equal k	Return 0 as no subarray sums to k.
Array with all elements equal to zero and k is zero	Return the length of the array, as the entire array sums to zero.
Array with all positive numbers and k is negative	Return 0 as no subarray can have a negative sum.
Large array with values causing integer overflow during summation	Use long data type to store the prefix sums to prevent integer overflow.
No subarray sums to k	Return 0 after iterating through the entire array without finding a suitable subarray.

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 are the possible values for the elements within the input array, including negative numbers, zero, or positive numbers?
What should be returned if no subarray sums to `k`? Should I return null, an empty array, or some other indicator?
What is the data type of `k`? Is it guaranteed to be an integer, or could it be a floating-point number?
Is the input array guaranteed to be non-null and non-empty? What should I do if it is null or empty?
If multiple subarrays have a sum equal to `k`, should I return the one with the maximum size or any valid subarray?

Brute Force Solution

Approach

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

Start by considering the first number in the list as the beginning of a potential section.
Then, include the second number, and check if the sum of the first two numbers equals the target. If it does then store the size of this piece.
Continue adding numbers one by one to this section from the beginning, each time checking if the total sum matches the target and saving the size if applicable.
Once you've checked all possible sections starting from the first number, move to the second number in the list.
Repeat the same process, creating sections starting from the second number, then the third, and so on, until you've checked every possible starting point.
At each step, remember to compare the length of any matching sections with previously found sections, always storing the longest section that matches the target sum.

Code Implementation

def max_size_subarray_sum_equals_k_brute_force(numbers, target):
    maximum_size = 0

    for start_index in range(len(numbers)): # Iterate through possible start indices

        current_sum = 0

        for end_index in range(start_index, len(numbers)): # Iterate through possible end indices

            current_sum += numbers[end_index]

            #Check if the current subarray sum is equal to the target
            if current_sum == target:

                subarray_size = end_index - start_index + 1

                # We want to maximize the size of the subarray
                if subarray_size > maximum_size:

                    maximum_size = subarray_size

    return maximum_size

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Keep track of the running total as you go through the list of numbers.
Store each running total along with its position in the list in a special lookup table.
For each running total, check if there's a previous running total in the table that, when subtracted from the current running total, equals the target number.
If such a previous running total exists, it means the numbers between those two positions add up to the target number.
Calculate the length of that section and remember the longest one found so far.
If a running total is not already in the lookup table, add it along with its position. Do this only if it's not already there to ensure we find the earliest, and therefore the longest, matching subarray.
After going through the entire list, you'll have the length of the longest section that adds up to the target number.

Code Implementation

def max_size_subarray_sum_equals_k(nums, target):
    running_sum = 0
    max_subarray_length = 0
    sum_index_map = {}

    # Store initial running sum of 0 at index -1
    sum_index_map[0] = -1

    for i, num in enumerate(nums):
        running_sum += num

        # Check if running_sum - target exists in the map
        if (running_sum - target) in sum_index_map:
            # Update the max length
            subarray_length = i - sum_index_map[running_sum - target]
            max_subarray_length = max(max_subarray_length, subarray_length)

        # Store the running sum and index
        # Only if it's not already present. Ensure longest subarray.
        if running_sum not in sum_index_map:
            sum_index_map[running_sum] = i

    return max_subarray_length

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty array input	Return 0 immediately as there is no subarray.
Null array input	Throw an IllegalArgumentException or return 0, depending on requirements.
Array with a single element that equals k	Return 1 as the subarray of size 1 satisfies the condition.
Array with a single element that does not equal k	Return 0 as no subarray sums to k.
Array with all elements equal to zero and k is zero	Return the length of the array, as the entire array sums to zero.
Array with all positive numbers and k is negative	Return 0 as no subarray can have a negative sum.
Large array with values causing integer overflow during summation	Use long data type to store the prefix sums to prevent integer overflow.
No subarray sums to k	Return 0 after iterating through the entire array without finding a suitable subarray.