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Split Array Largest Sum

Hard

Split Array Largest Sum

Hard

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 value ranges for the integers within the input array?
Is it possible for the input array to be empty or null?
If there are multiple ways to split the array that result in the same largest sum of subarrays, can I return any of them, or is there a preference?
Is the number of subarrays, `k`, always a valid number, meaning will it always be within the bounds of 1 and the length of the input array?
Can the input value `k` be equal to 1 or the length of the array? If so, what is the expected output?

Brute Force Solution

Approach

The brute force approach for this problem is all about trying every possible way to split the group of numbers into subgroups. We want to find the split where the largest sum of any of these subgroups is as small as possible. Essentially, we check everything to find the best arrangement.

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

Consider splitting the numbers into one group first and calculate the sum of that group.
Then, try splitting the numbers into two groups in every possible way. For each split, find the larger of the two group sums.
Next, try splitting the numbers into three groups, again considering all possibilities. Find the largest group sum for each split.
Continue this process until you've tried splitting the numbers into the maximum allowed number of groups.
For each possible splitting arrangement, remember the largest sum among the subgroups.
Finally, compare all the 'largest sums' you've recorded, and choose the smallest one. That smallest value is the answer.

Code Implementation

def split_array_largest_sum_brute_force(numbers, group_count):
    minimum_largest_sum = float('inf')

    # Iterate through all possible splits
    for i in range(1 << (len(numbers) - 1)):
        number_of_groups = 1
        current_sum = 0
        largest_sum_so_far = 0
        group_sums = []
        current_group = []

        # Simulate the splitting process based on the bitmask
        for j in range(len(numbers)):
            current_sum += numbers[j]
            current_group.append(numbers[j])

            # Check if we need to split into a new group
            if j < len(numbers) - 1 and (i >> j) & 1:
                group_sums.append(current_sum)
                largest_sum_so_far = max(largest_sum_so_far, current_sum)
                current_sum = 0
                number_of_groups += 1
                current_group = []

        # Handle the last group
        group_sums.append(current_sum)
        largest_sum_so_far = max(largest_sum_so_far, current_sum)

        # Filter for valid splits based on group count
        if number_of_groups == group_count:
            # Only consider splits with the correct group count
            minimum_largest_sum = min(minimum_largest_sum, largest_sum_so_far)

    return minimum_largest_sum

Big(O) Analysis

Time Complexity

O(n^k) – The brute force approach explores all possible ways to split an array of size n into k subarrays. To split an array of n elements into k subarrays, we need to choose k-1 split points among the n-1 possible positions. The number of ways to do this is given by the binomial coefficient (n-1 choose k-1), which can be represented as (n-1)! / ((k-1)! * (n-k)!). In the worst-case scenario, we have to iterate through a significant portion of these combinations and calculate the sum of each subarray for each combination. Therefore, the time complexity is roughly proportional to a combination of n and k, making it O(n^k) where the specific value of k influences the polynomial order but remains highly dependent on the number of subarrays. As k grows, the complexity increases dramatically.

Space Complexity

O(1) – The brute force approach, as described, exhaustively explores splitting the array into subgroups and keeps track of the minimum largest sum found so far. It does not explicitly mention the use of auxiliary data structures like arrays, lists, or hash maps to store intermediate splits or sums. The algorithm primarily involves calculating sums on the fly and comparing them, requiring only a few variables to store the current split's sum and the overall minimum largest sum. Hence, the auxiliary space is constant, independent of the input array size N.

Optimal Solution

Approach

The best way to solve this problem is to use a strategy called binary search. Instead of checking every possible way to split the array, we'll cleverly narrow down the range of possible largest sums until we find the smallest one that works.

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

First, figure out the smallest and largest possible values for the largest sum. The smallest it could be is the biggest single number in the array. The largest it could be is the sum of all the numbers.
Now, guess a value in the middle of that range. Let's call it our 'target sum'.
Check if you can split the array into the required number of parts, where the sum of each part is no more than the 'target sum'. You can do this by going through the numbers one by one, adding them to the current part, and starting a new part whenever the current part's sum goes over the 'target sum'.
If you can split the array into the right number of parts or fewer with that 'target sum', it means your guess was too high. So, adjust the 'largest possible value' to be one less than the target sum, and try again with a new middle value.
If you can't split the array into the right number of parts, it means your guess was too low. So, adjust the 'smallest possible value' to be one more than the target sum, and try again with a new middle value.
Keep repeating this guessing and adjusting process until the smallest and largest possible values are right next to each other. That final value is the smallest possible largest sum you can achieve.

Code Implementation

def split_array(nums, k):
    smallest_possible_value = max(nums)
    largest_possible_value = sum(nums)

    while smallest_possible_value <= largest_possible_value:
        middle_value = (smallest_possible_value + largest_possible_value) // 2
        
        # Check if it's possible to split the array with the middle value
        if can_split(nums, k, middle_value):
            # If possible, reduce the largest possible value.
            largest_possible_value = middle_value - 1
        else:
            # If not possible, increase the smallest possible value.
            smallest_possible_value = middle_value + 1

    return smallest_possible_value

def can_split(nums, k, target_sum):
    number_of_subarrays = 1
    current_subarray_sum = 0

    for num in nums:
        if current_subarray_sum + num <= target_sum:
            current_subarray_sum += num
        else:
            number_of_subarrays += 1

            # Reset with the current number.
            current_subarray_sum = num

            # If num > target_sum, it means target_sum is too small.
            if current_subarray_sum > target_sum:
                return False

    # Check if the split is possible with given k.
    return number_of_subarrays <= k

Big(O) Analysis

Time Complexity

O(n log S) – The algorithm uses binary search to find the minimum largest sum. The binary search space is between the maximum element in the array and the sum of all elements, which we denote as S. Therefore, the binary search itself takes O(log S) time. In each step of the binary search, we need to check if a given 'target sum' is feasible, which involves iterating through the array of n elements once to count the number of subarrays required. This feasibility check takes O(n) time. Thus, the overall time complexity is O(n log S).

Space Complexity

O(1) – The algorithm primarily uses variables to store the current sum of a part, the number of parts created, and the range for the binary search (smallest and largest possible sums). These variables consume a constant amount of space regardless of the input array's size (N). No auxiliary data structures like arrays or hash maps are used; therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0, as there's no sum to split.
k is 1 (split into one subarray)	Return the sum of the entire array.
k is equal to the length of nums (split into n subarrays)	Return the maximum element in the array.
Array contains very large numbers that could lead to integer overflow when summed	Use long data type for sum calculations to prevent overflow.
All elements in the array are 0	The binary search should still converge to 0, representing the largest sum possible with zeroed subarrays.
Array contains only one element	Return the single element if k = 1, or the element itself if k >= array length, or handle it with base cases
Array contains extremely large positive numbers and very small negative numbers	The binary search range might need adjustments to handle the negative impact on overall sums.
k is greater than the array's length	Treat k as the array's length since we cannot split it into more subarrays than elements.

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 value ranges for the integers within the input array?
Is it possible for the input array to be empty or null?
If there are multiple ways to split the array that result in the same largest sum of subarrays, can I return any of them, or is there a preference?
Is the number of subarrays, `k`, always a valid number, meaning will it always be within the bounds of 1 and the length of the input array?
Can the input value `k` be equal to 1 or the length of the array? If so, what is the expected output?

Brute Force Solution

Approach

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

Consider splitting the numbers into one group first and calculate the sum of that group.
Then, try splitting the numbers into two groups in every possible way. For each split, find the larger of the two group sums.
Next, try splitting the numbers into three groups, again considering all possibilities. Find the largest group sum for each split.
Continue this process until you've tried splitting the numbers into the maximum allowed number of groups.
For each possible splitting arrangement, remember the largest sum among the subgroups.
Finally, compare all the 'largest sums' you've recorded, and choose the smallest one. That smallest value is the answer.

Code Implementation

def split_array_largest_sum_brute_force(numbers, group_count):
    minimum_largest_sum = float('inf')

    # Iterate through all possible splits
    for i in range(1 << (len(numbers) - 1)):
        number_of_groups = 1
        current_sum = 0
        largest_sum_so_far = 0
        group_sums = []
        current_group = []

        # Simulate the splitting process based on the bitmask
        for j in range(len(numbers)):
            current_sum += numbers[j]
            current_group.append(numbers[j])

            # Check if we need to split into a new group
            if j < len(numbers) - 1 and (i >> j) & 1:
                group_sums.append(current_sum)
                largest_sum_so_far = max(largest_sum_so_far, current_sum)
                current_sum = 0
                number_of_groups += 1
                current_group = []

        # Handle the last group
        group_sums.append(current_sum)
        largest_sum_so_far = max(largest_sum_so_far, current_sum)

        # Filter for valid splits based on group count
        if number_of_groups == group_count:
            # Only consider splits with the correct group count
            minimum_largest_sum = min(minimum_largest_sum, largest_sum_so_far)

    return minimum_largest_sum

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, figure out the smallest and largest possible values for the largest sum. The smallest it could be is the biggest single number in the array. The largest it could be is the sum of all the numbers.
Now, guess a value in the middle of that range. Let's call it our 'target sum'.
Check if you can split the array into the required number of parts, where the sum of each part is no more than the 'target sum'. You can do this by going through the numbers one by one, adding them to the current part, and starting a new part whenever the current part's sum goes over the 'target sum'.
If you can split the array into the right number of parts or fewer with that 'target sum', it means your guess was too high. So, adjust the 'largest possible value' to be one less than the target sum, and try again with a new middle value.
If you can't split the array into the right number of parts, it means your guess was too low. So, adjust the 'smallest possible value' to be one more than the target sum, and try again with a new middle value.
Keep repeating this guessing and adjusting process until the smallest and largest possible values are right next to each other. That final value is the smallest possible largest sum you can achieve.

Code Implementation

def split_array(nums, k):
    smallest_possible_value = max(nums)
    largest_possible_value = sum(nums)

    while smallest_possible_value <= largest_possible_value:
        middle_value = (smallest_possible_value + largest_possible_value) // 2
        
        # Check if it's possible to split the array with the middle value
        if can_split(nums, k, middle_value):
            # If possible, reduce the largest possible value.
            largest_possible_value = middle_value - 1
        else:
            # If not possible, increase the smallest possible value.
            smallest_possible_value = middle_value + 1

    return smallest_possible_value

def can_split(nums, k, target_sum):
    number_of_subarrays = 1
    current_subarray_sum = 0

    for num in nums:
        if current_subarray_sum + num <= target_sum:
            current_subarray_sum += num
        else:
            number_of_subarrays += 1

            # Reset with the current number.
            current_subarray_sum = num

            # If num > target_sum, it means target_sum is too small.
            if current_subarray_sum > target_sum:
                return False

    # Check if the split is possible with given k.
    return number_of_subarrays <= k

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0, as there's no sum to split.
k is 1 (split into one subarray)	Return the sum of the entire array.
k is equal to the length of nums (split into n subarrays)	Return the maximum element in the array.
Array contains very large numbers that could lead to integer overflow when summed	Use long data type for sum calculations to prevent overflow.
All elements in the array are 0	The binary search should still converge to 0, representing the largest sum possible with zeroed subarrays.
Array contains only one element	Return the single element if k = 1, or the element itself if k >= array length, or handle it with base cases
Array contains extremely large positive numbers and very small negative numbers	The binary search range might need adjustments to handle the negative impact on overall sums.
k is greater than the array's length	Treat k as the array's length since we cannot split it into more subarrays than elements.