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Split Array With Same Average

Hard

Asked by:

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Topics:

ArraysDynamic Programming

You are given an integer array nums.

You should move each element of nums into one of the two arrays A and B such that A and B are non-empty, and average(A) == average(B).

Return true if it is possible to achieve that and false otherwise.

Note that for an array arr, average(arr) is the sum of all the elements of arr over the length of arr.

Example 1:

Input: nums = [1,2,3,4,5,6,7,8]
Output: true
Explanation: We can split the array into [1,4,5,8] and [2,3,6,7], and both of them have an average of 4.5.

Example 2:

Input: nums = [3,1]
Output: false

Constraints:

1 <= nums.length <= 30
0 <= nums[i] <= 10⁴

Try coding it in LeetCode

Split Array With Same Average

Hard

Asked by:

1 view

Topics:

ArraysDynamic Programming

You are given an integer array nums.

You should move each element of nums into one of the two arrays A and B such that A and B are non-empty, and average(A) == average(B).

Return true if it is possible to achieve that and false otherwise.

Note that for an array arr, average(arr) is the sum of all the elements of arr over the length of arr.

Example 1:

Input: nums = [1,2,3,4,5,6,7,8]
Output: true
Explanation: We can split the array into [1,4,5,8] and [2,3,6,7], and both of them have an average of 4.5.

Example 2:

Input: nums = [3,1]
Output: false

Constraints:

1 <= nums.length <= 30
0 <= nums[i] <= 10⁴

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 `nums`, and what is the range of values for the integers within it?
Can the input array `nums` contain negative numbers, zero, or only positive integers?
If it's possible to split the array in multiple ways to satisfy the condition, is any valid split acceptable, or is there a specific split I should aim for?
Is an empty input array, or an array with a single element, a valid input, and if so, what should I return in those cases?
Do I need to worry about potential integer overflow issues when calculating the sum of subarrays, given the possible range of integer values and the array size?

Brute Force Solution

Approach

The brute force approach to this problem is all about trying every possible combination. We explore every possible way to divide the numbers into two groups and check if they meet our condition. It's like testing out every single possibility until we find a match.

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

First, consider creating a first group with just one number from the original set, and a second group with the remaining numbers.
Calculate the average of both groups.
Check if the average of the first group is the same as the average of the second group.
If they are the same, we have found a valid split and we're done!
If they are not the same, try a different combination.
Next, try creating a first group with two numbers, then with three numbers, and so on. For each size of the first group, consider all possible combinations of numbers that could be in it.
For each combination, the remaining numbers form the second group.
Calculate the average of both groups, and compare them.
Repeat until all possible combinations have been checked.

Code Implementation

def split_array_with_same_average_brute_force(numbers):
    total_sum = sum(numbers)
    number_count = len(numbers)

    # Iterate through all possible sizes for the first subset
    for first_subset_size in range(1, number_count):
        # Generate all combinations of indices for the first subset
        def generate_combinations(current_index, current_subset, all_combinations):
            if len(current_subset) == first_subset_size:
                all_combinations.append(current_subset[:])
                return
            if current_index == number_count:
                return

            # Include the current number in the subset
            current_subset.append(current_index)
            generate_combinations(current_index + 1, current_subset, all_combinations)
            current_subset.pop()

            # Exclude the current number from the subset
            generate_combinations(current_index + 1, current_subset, all_combinations)

        all_combinations = []
        generate_combinations(0, [], all_combinations)

        # Iterate through all combinations of the first subset
        for first_subset_indices in all_combinations:

            first_subset_sum = sum(numbers[index] for index in first_subset_indices)
            second_subset_indices = [index for index in range(number_count) if index not in first_subset_indices]
            second_subset_sum = sum(numbers[index] for index in second_subset_indices)

            # Ensure that the sum of the two subsets adds up to the total sum of the original array
            if first_subset_size * second_subset_sum == (number_count - first_subset_size) * first_subset_sum:
                return True

    # No split was found where the averages are equal
    return False

Big(O) Analysis

Time Complexity

O(2^n) – The proposed brute force approach involves examining all possible subsets of the input array of size n. For each element, we either include it in the first group or exclude it, resulting in 2 options per element. Therefore, there are 2^n possible subsets to consider. For each subset, we calculate the averages of both groups, which takes O(n) time in the worst case. Since we iterate through all possible subsets, the time complexity is O(n * 2^n). The number of subsets is the dominant factor, so the overall complexity is O(2^n).

Space Complexity

O(N) – The described brute force approach explores all possible combinations. To generate these combinations, it implicitly uses data structures to keep track of the current combination being considered, which in the worst case might hold almost all of the N input numbers. This results in an auxiliary space proportional to the number of elements in the input array. Therefore, the space complexity is O(N).

Optimal Solution

Approach

The problem asks us to determine if we can split a group of numbers into two smaller groups that have the exact same average value. Instead of trying every single combination, we use a mathematical trick to reframe the problem and check for the existence of a specific sum. This allows us to quickly rule out impossible scenarios and significantly reduce the number of cases we need to examine.

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

First, calculate the total sum of all the numbers in the original group.
Recognize that if we can split the original group, it implies a smaller group exists with a specific sum that satisfies a mathematical condition derived from the equal average requirement. That is, find out if there exists a group of numbers where the total sum of the numbers in the group, divided by the number of elements in the group, is equal to the total sum of the numbers in the original group, divided by the total number of numbers in the original group.
To avoid dealing with division and fractions, rearrange the equation such that it only involves sums and multiplications. This significantly simplifies the computations.
Iterate through all possible sizes for the smaller group. The size must be between 1 and the number of elements in the original group, minus one.
For each possible size, check if a smaller group with the calculated sum exists within the original group. This can be done by checking all possible subsets of that size, or by applying Dynamic Programming.
If a smaller group with the required sum is found, it means that it is possible to split the original group into two groups with the same average. If we exhaust all possibilities, and don't find such a group, we cannot split the group into two groups with the same average.

Code Implementation

def can_split_array_same_average(number_group):
    total_sum = sum(number_group)
    number_count = len(number_group)

    # Iterate through possible sizes of the first sub-group
    for subgroup_size in range(1, number_count):
        # Check if the required sub-group sum is an integer.
        if (total_sum * subgroup_size) % number_count != 0:
            continue

        required_subgroup_sum = (total_sum * subgroup_size) // number_count

        # Creating a set to track achievable sums.
        possible_sums = {0}

        # Iterate through numbers, building possible sums.
        for number in number_group:
            new_sums = set()
            for possible_sum in possible_sums:
                new_sums.add(possible_sum + number)
            possible_sums.update(new_sums)

        # Check if the required sum is achievable.
        if required_subgroup_sum in possible_sums:
            #If achievable, return True immediately
            return True

    # No valid split found.
    return False

Big(O) Analysis

Time Complexity

O(n * C(n, n/2)) or O(n * 2^n) in the worst case – The algorithm iterates through all possible sizes of the smaller group, which takes O(n) time, where n is the size of the original group. For each size, it checks if a subset with a specific sum exists. The number of subsets of size k from a set of size n is given by the binomial coefficient C(n, k), where C(n, k) = n! / (k! * (n-k)!). In the worst-case scenario, where we might need to examine a significant portion of all possible subsets, the time complexity for subset checking can approach C(n, n/2), which is approximately proportional to 2^n in the worst case. Combining the iteration through sizes and the subset checking, the overall time complexity can be expressed as O(n * C(n, n/2)) or O(n * 2^n) in the worst case if no memoization is performed.

Space Complexity

O(N*M) – The dominant space complexity stems from the use of Dynamic Programming (or checking all possible subsets). Specifically, to check for a subset of size 'm' that sums to 's', where 'm' ranges from 1 to N-1, and 's' is a potentially large value, we might use a table (or memoization). If using dynamic programming, where N is the number of elements in the input array, and M is the target sum we are searching for in the subsets, a table of dimensions approximately (N x M) would be needed. This leads to an auxiliary space complexity of O(N*M) in the worst case, where M can be proportional to the sum of all elements, which could be much larger than N in some cases.

Edge Cases

Case	How to Handle
Null or empty input array	Return false immediately as a split is impossible.
Input array with only one element	Return false as a split into two non-empty subarrays is impossible.
Input array with two elements having the same value	Check if the values are equal; if so, return true since the average of each subarray will be the same.
Input array with two elements having different values	Return false since any split will result in different averages.
Array with all elements being zero	Return true if the array has more than one element, as the average of any subarray will be zero.
Array with large numbers leading to potential integer overflow during sum calculation	Use long data type to store the sum to prevent integer overflow.
Array with negative numbers	The algorithm should handle negative numbers correctly as they will affect the average calculation.
Large array size impacting time complexity	Optimize the solution to avoid unnecessary computations and ensure it runs within acceptable time limits, possibly using dynamic programming to avoid redundant calculations.

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 `nums`, and what is the range of values for the integers within it?
Can the input array `nums` contain negative numbers, zero, or only positive integers?
If it's possible to split the array in multiple ways to satisfy the condition, is any valid split acceptable, or is there a specific split I should aim for?
Is an empty input array, or an array with a single element, a valid input, and if so, what should I return in those cases?
Do I need to worry about potential integer overflow issues when calculating the sum of subarrays, given the possible range of integer values and the array size?

Brute Force Solution

Approach

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

First, consider creating a first group with just one number from the original set, and a second group with the remaining numbers.
Calculate the average of both groups.
Check if the average of the first group is the same as the average of the second group.
If they are the same, we have found a valid split and we're done!
If they are not the same, try a different combination.
Next, try creating a first group with two numbers, then with three numbers, and so on. For each size of the first group, consider all possible combinations of numbers that could be in it.
For each combination, the remaining numbers form the second group.
Calculate the average of both groups, and compare them.
Repeat until all possible combinations have been checked.

Code Implementation

def split_array_with_same_average_brute_force(numbers):
    total_sum = sum(numbers)
    number_count = len(numbers)

    # Iterate through all possible sizes for the first subset
    for first_subset_size in range(1, number_count):
        # Generate all combinations of indices for the first subset
        def generate_combinations(current_index, current_subset, all_combinations):
            if len(current_subset) == first_subset_size:
                all_combinations.append(current_subset[:])
                return
            if current_index == number_count:
                return

            # Include the current number in the subset
            current_subset.append(current_index)
            generate_combinations(current_index + 1, current_subset, all_combinations)
            current_subset.pop()

            # Exclude the current number from the subset
            generate_combinations(current_index + 1, current_subset, all_combinations)

        all_combinations = []
        generate_combinations(0, [], all_combinations)

        # Iterate through all combinations of the first subset
        for first_subset_indices in all_combinations:

            first_subset_sum = sum(numbers[index] for index in first_subset_indices)
            second_subset_indices = [index for index in range(number_count) if index not in first_subset_indices]
            second_subset_sum = sum(numbers[index] for index in second_subset_indices)

            # Ensure that the sum of the two subsets adds up to the total sum of the original array
            if first_subset_size * second_subset_sum == (number_count - first_subset_size) * first_subset_sum:
                return True

    # No split was found where the averages are equal
    return False

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, calculate the total sum of all the numbers in the original group.
Recognize that if we can split the original group, it implies a smaller group exists with a specific sum that satisfies a mathematical condition derived from the equal average requirement. That is, find out if there exists a group of numbers where the total sum of the numbers in the group, divided by the number of elements in the group, is equal to the total sum of the numbers in the original group, divided by the total number of numbers in the original group.
To avoid dealing with division and fractions, rearrange the equation such that it only involves sums and multiplications. This significantly simplifies the computations.
Iterate through all possible sizes for the smaller group. The size must be between 1 and the number of elements in the original group, minus one.
For each possible size, check if a smaller group with the calculated sum exists within the original group. This can be done by checking all possible subsets of that size, or by applying Dynamic Programming.
If a smaller group with the required sum is found, it means that it is possible to split the original group into two groups with the same average. If we exhaust all possibilities, and don't find such a group, we cannot split the group into two groups with the same average.

Code Implementation

def can_split_array_same_average(number_group):
    total_sum = sum(number_group)
    number_count = len(number_group)

    # Iterate through possible sizes of the first sub-group
    for subgroup_size in range(1, number_count):
        # Check if the required sub-group sum is an integer.
        if (total_sum * subgroup_size) % number_count != 0:
            continue

        required_subgroup_sum = (total_sum * subgroup_size) // number_count

        # Creating a set to track achievable sums.
        possible_sums = {0}

        # Iterate through numbers, building possible sums.
        for number in number_group:
            new_sums = set()
            for possible_sum in possible_sums:
                new_sums.add(possible_sum + number)
            possible_sums.update(new_sums)

        # Check if the required sum is achievable.
        if required_subgroup_sum in possible_sums:
            #If achievable, return True immediately
            return True

    # No valid split found.
    return False

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return false immediately as a split is impossible.
Input array with only one element	Return false as a split into two non-empty subarrays is impossible.
Input array with two elements having the same value	Check if the values are equal; if so, return true since the average of each subarray will be the same.
Input array with two elements having different values	Return false since any split will result in different averages.
Array with all elements being zero	Return true if the array has more than one element, as the average of any subarray will be zero.
Array with large numbers leading to potential integer overflow during sum calculation	Use long data type to store the sum to prevent integer overflow.
Array with negative numbers	The algorithm should handle negative numbers correctly as they will affect the average calculation.
Large array size impacting time complexity	Optimize the solution to avoid unnecessary computations and ensure it runs within acceptable time limits, possibly using dynamic programming to avoid redundant calculations.