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Split the Array

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Arrays

You are given an integer array nums of even length. You have to split the array into two parts nums1 and nums2 such that:

nums1.length == nums2.length == nums.length / 2.
nums1 should contain distinct elements.
nums2 should also contain distinct elements.

Return true if it is possible to split the array, and false otherwise.

Example 1:

Input: nums = [1,1,2,2,3,4]
Output: true
Explanation: One of the possible ways to split nums is nums1 = [1,2,3] and nums2 = [1,2,4].

Example 2:

Input: nums = [1,1,1,1]
Output: false
Explanation: The only possible way to split nums is nums1 = [1,1] and nums2 = [1,1]. Both nums1 and nums2 do not contain distinct elements. Therefore, we return false.

Constraints:

1 <= nums.length <= 100
nums.length % 2 == 0
1 <= nums[i] <= 100

Split the Array

Easy

Visa

0 views

Topics:

Arrays

You are given an integer array nums of even length. You have to split the array into two parts nums1 and nums2 such that:

nums1.length == nums2.length == nums.length / 2.
nums1 should contain distinct elements.
nums2 should also contain distinct elements.

Return true if it is possible to split the array, and false otherwise.

Example 1:

Input: nums = [1,1,2,2,3,4]
Output: true
Explanation: One of the possible ways to split nums is nums1 = [1,2,3] and nums2 = [1,2,4].

Example 2:

Input: nums = [1,1,1,1]
Output: false
Explanation: The only possible way to split nums is nums1 = [1,1] and nums2 = [1,1]. Both nums1 and nums2 do not contain distinct elements. Therefore, we return false.

Constraints:

1 <= nums.length <= 100
nums.length % 2 == 0
1 <= nums[i] <= 100

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 range of values for the integers within the input array `nums`? Can they be negative, zero, or only positive?
What should I return if the input array `nums` is null or empty?
Are there any constraints on the size of the input array `nums`?
If multiple valid splits are possible, is it sufficient to return `true` as soon as I find one, or is there a preference for a specific split (e.g., the split with the smallest left subarray)?
Can I assume the input array `nums` only contains integers, or might there be floating-point numbers or other data types?

Brute Force Solution

Approach

The brute force approach means trying every possible way to split a group of numbers into two smaller groups. We check each split to see if it meets our desired conditions. This ensures we explore all options to find a valid solution, if one exists.

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

First, consider splitting the numbers so that the first group has only the first number.
Then, check if the two groups formed by this split satisfy the condition.
If not, try splitting the numbers so that the first group has the first two numbers.
Check again if the two groups now satisfy the condition.
Keep repeating this process, each time adding one more number to the first group and checking the condition.
Continue until the first group contains all but the last number.
If none of the splits satisfied the condition, then there is no solution.

Code Implementation

def split_the_array_brute_force(data):
    number_of_elements = len(data)
    best_difference = float('inf')
    best_split = None

    # Iterate through all possible subsets of the data
    for i in range(1 << number_of_elements):
        group_one = []
        group_two = []

        # Assign elements to groups based on bitmask
        for j in range(number_of_elements):
            if (i >> j) & 1:
                group_one.append(data[j])
            else:
                group_two.append(data[j])

        # Handle edge case where either group is empty
        if not group_one or not group_two:
            continue

        sum_group_one = sum(group_one)
        sum_group_two = sum(group_two)
        current_difference = abs(sum_group_one - sum_group_two)

        # Find the split with the minimum difference
        if current_difference < best_difference:

            # Keep track of the difference for comparison
            best_difference = current_difference

            # Store current division as the best split
            best_split = (group_one, group_two)

    # Return the best split found
    return best_split

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n^2) – The described brute force approach iterates through almost every possible split point in the array, considering each split from one element in the first group to all but one element. For each such split, the algorithm must, at a minimum, iterate through each of the two subgroups to perform a condition check. This checking process involves looking at the 'n' elements of the array, potentially performing a constant time operation on each, approximately 'n' times for each split. Since the outer loop for splitting the array runs approximately 'n' times and the inner operation for condition check runs 'n' times, the total number of operations becomes proportional to n * n. Thus, the time complexity is O(n^2).

Space Complexity

O(1) – The provided brute force approach iteratively splits the input array of size N into two groups. It doesn't explicitly create any new arrays or data structures to store these groups. The algorithm only needs to keep track of the splitting point, which can be represented by a single index. Thus, the auxiliary space used is constant and does not depend on the size of the input array N.

Optimal Solution

Approach

The goal is to divide the given set of numbers into two groups so that the difference between the sums of the two groups is as small as possible. The optimal approach involves making small adjustments to the sum of the smaller group to inch closer to the optimal value.

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

First, calculate the total sum of all the numbers.
Figure out the target sum for one of the groups, which would be half of the total sum.
Think about building one group at a time. Start with an empty group.
Now, go through the numbers one at a time and consider adding each to the group. If adding a number makes the group's sum closer to the target sum (without going over), add it.
Keep doing this until you've considered all the numbers.
At the end, the numbers in your group will sum to a value as close to the target as possible. The remaining numbers form the second group.
The difference between the sums of these two groups will be the smallest possible difference you can achieve.

Code Implementation

def split_the_array(numbers):
    total_sum = sum(numbers)
    target_sum = total_sum / 2

    closest_sum = 0

    # Use dynamic programming to find closest subset sum.
    subset_sums = {0}

    for number in numbers:
        new_subset_sums = set()
        for current_sum in subset_sums:
            new_sum = current_sum + number
            new_subset_sums.add(new_sum)

        subset_sums.update(new_subset_sums)

    # Find closest sum to the target.
    for current_sum in subset_sums:
        if abs(current_sum - target_sum) < abs(closest_sum - target_sum):
            closest_sum = current_sum

    # Calculate the two group sums
    group_one_sum = closest_sum

    # The other group's sum is whatever remains.
    group_two_sum = total_sum - closest_sum

    # Return the absolute difference.
    return abs(group_one_sum - group_two_sum)

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n once. Inside the loop, a constant-time comparison is performed to determine whether to include the current number in the first group. Since the primary operation involves a single pass through the array, the time complexity is directly proportional to the number of elements, n. Therefore, the overall time complexity is O(n).

Space Complexity

O(1) – The described approach primarily involves iterative processing and does not create any auxiliary data structures that scale with the input size N (the number of elements). The algorithm calculates the total sum and target sum, which are stored in constant space. There is no mention of creating temporary lists, hashmaps or recursion, therefore it only utilizes a few variables to keep track of the current sum and potentially the difference. This results in constant extra space usage, independent of N.

Edge Cases

Case	How to Handle
Null or empty input array	Return false immediately, as there are no subarrays to split.
Array with only one element	Return false, as the subarrays must be non-empty and splitting requires at least two elements.
Array with two elements where the sum of both sides equals the target	Return false, as both sides must contain at least one element.
Array where the total sum is odd	Return false, as an odd sum cannot be divided into two equal integer sums.
Array with all elements equal to zero	Return true, as any split will result in two subarrays with sum zero.
Array containing large positive numbers that could lead to integer overflow when summing	Use a larger data type (e.g., long) to store the sums to prevent overflow.
Array with both very large positive and very large negative numbers, potentially causing overflow/underflow during intermediate sum calculations	Employ long data types for calculations and consider using a stable summation algorithm if necessary to mitigate precision issues.
Array where no such split exists	The algorithm should iterate through all possible split points and return false if no valid split is found after exhausting all options.

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 range of values for the integers within the input array `nums`? Can they be negative, zero, or only positive?
What should I return if the input array `nums` is null or empty?
Are there any constraints on the size of the input array `nums`?
If multiple valid splits are possible, is it sufficient to return `true` as soon as I find one, or is there a preference for a specific split (e.g., the split with the smallest left subarray)?
Can I assume the input array `nums` only contains integers, or might there be floating-point numbers or other data types?

Brute Force Solution

Approach

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

First, consider splitting the numbers so that the first group has only the first number.
Then, check if the two groups formed by this split satisfy the condition.
If not, try splitting the numbers so that the first group has the first two numbers.
Check again if the two groups now satisfy the condition.
Keep repeating this process, each time adding one more number to the first group and checking the condition.
Continue until the first group contains all but the last number.
If none of the splits satisfied the condition, then there is no solution.

Code Implementation

def split_the_array_brute_force(data):
    number_of_elements = len(data)
    best_difference = float('inf')
    best_split = None

    # Iterate through all possible subsets of the data
    for i in range(1 << number_of_elements):
        group_one = []
        group_two = []

        # Assign elements to groups based on bitmask
        for j in range(number_of_elements):
            if (i >> j) & 1:
                group_one.append(data[j])
            else:
                group_two.append(data[j])

        # Handle edge case where either group is empty
        if not group_one or not group_two:
            continue

        sum_group_one = sum(group_one)
        sum_group_two = sum(group_two)
        current_difference = abs(sum_group_one - sum_group_two)

        # Find the split with the minimum difference
        if current_difference < best_difference:

            # Keep track of the difference for comparison
            best_difference = current_difference

            # Store current division as the best split
            best_split = (group_one, group_two)

    # Return the best split found
    return best_split

Try it yourself on LeetCode

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.
Figure out the target sum for one of the groups, which would be half of the total sum.
Think about building one group at a time. Start with an empty group.
Now, go through the numbers one at a time and consider adding each to the group. If adding a number makes the group's sum closer to the target sum (without going over), add it.
Keep doing this until you've considered all the numbers.
At the end, the numbers in your group will sum to a value as close to the target as possible. The remaining numbers form the second group.
The difference between the sums of these two groups will be the smallest possible difference you can achieve.

Code Implementation

def split_the_array(numbers):
    total_sum = sum(numbers)
    target_sum = total_sum / 2

    closest_sum = 0

    # Use dynamic programming to find closest subset sum.
    subset_sums = {0}

    for number in numbers:
        new_subset_sums = set()
        for current_sum in subset_sums:
            new_sum = current_sum + number
            new_subset_sums.add(new_sum)

        subset_sums.update(new_subset_sums)

    # Find closest sum to the target.
    for current_sum in subset_sums:
        if abs(current_sum - target_sum) < abs(closest_sum - target_sum):
            closest_sum = current_sum

    # Calculate the two group sums
    group_one_sum = closest_sum

    # The other group's sum is whatever remains.
    group_two_sum = total_sum - closest_sum

    # Return the absolute difference.
    return abs(group_one_sum - group_two_sum)

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return false immediately, as there are no subarrays to split.
Array with only one element	Return false, as the subarrays must be non-empty and splitting requires at least two elements.
Array with two elements where the sum of both sides equals the target	Return false, as both sides must contain at least one element.
Array where the total sum is odd	Return false, as an odd sum cannot be divided into two equal integer sums.
Array with all elements equal to zero	Return true, as any split will result in two subarrays with sum zero.
Array containing large positive numbers that could lead to integer overflow when summing	Use a larger data type (e.g., long) to store the sums to prevent overflow.
Array with both very large positive and very large negative numbers, potentially causing overflow/underflow during intermediate sum calculations	Employ long data types for calculations and consider using a stable summation algorithm if necessary to mitigate precision issues.
Array where no such split exists	The algorithm should iterate through all possible split points and return false if no valid split is found after exhausting all options.