Distribute Elements Into Two Arrays I #4 Most Asked

Easy

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Topics:

Arrays

You are given a 1-indexed array of distinct integers nums of length n.

You need to distribute all the elements of nums between two arrays arr1 and arr2 using n operations. In the first operation, append nums[1] to arr1. In the second operation, append nums[2] to arr2. Afterwards, in the i^th operation:

If the last element of arr1 is greater than the last element of arr2, append nums[i] to arr1. Otherwise, append nums[i] to arr2.

The array result is formed by concatenating the arrays arr1 and arr2. For example, if arr1 == [1,2,3] and arr2 == [4,5,6], then result = [1,2,3,4,5,6].

Return the array result.

Example 1:

Input: nums = [2,1,3]
Output: [2,3,1]
Explanation: After the first 2 operations, arr1 = [2] and arr2 = [1].
In the 3^rd operation, as the last element of arr1 is greater than the last element of arr2 (2 > 1), append nums[3] to arr1.
After 3 operations, arr1 = [2,3] and arr2 = [1].
Hence, the array result formed by concatenation is [2,3,1].

Example 2:

Input: nums = [5,4,3,8]
Output: [5,3,4,8]
Explanation: After the first 2 operations, arr1 = [5] and arr2 = [4].
In the 3^rd operation, as the last element of arr1 is greater than the last element of arr2 (5 > 4), append nums[3] to arr1, hence arr1 becomes [5,3].
In the 4^th operation, as the last element of arr2 is greater than the last element of arr1 (4 > 3), append nums[4] to arr2, hence arr2 becomes [4,8].
After 4 operations, arr1 = [5,3] and arr2 = [4,8].
Hence, the array result formed by concatenation is [5,3,4,8].

Constraints:

3 <= n <= 50
1 <= nums[i] <= 100
All elements in nums are distinct.

Try coding it in LeetCode

Distribute Elements Into Two Arrays I #4 Most Asked

Easy

Autodesk

3 views

Topics:

Arrays

You are given a 1-indexed array of distinct integers nums of length n.

You need to distribute all the elements of nums between two arrays arr1 and arr2 using n operations. In the first operation, append nums[1] to arr1. In the second operation, append nums[2] to arr2. Afterwards, in the i^th operation:

If the last element of arr1 is greater than the last element of arr2, append nums[i] to arr1. Otherwise, append nums[i] to arr2.

The array result is formed by concatenating the arrays arr1 and arr2. For example, if arr1 == [1,2,3] and arr2 == [4,5,6], then result = [1,2,3,4,5,6].

Return the array result.

Example 1:

Input: nums = [2,1,3]
Output: [2,3,1]
Explanation: After the first 2 operations, arr1 = [2] and arr2 = [1].
In the 3^rd operation, as the last element of arr1 is greater than the last element of arr2 (2 > 1), append nums[3] to arr1.
After 3 operations, arr1 = [2,3] and arr2 = [1].
Hence, the array result formed by concatenation is [2,3,1].

Example 2:

Input: nums = [5,4,3,8]
Output: [5,3,4,8]
Explanation: After the first 2 operations, arr1 = [5] and arr2 = [4].
In the 3^rd operation, as the last element of arr1 is greater than the last element of arr2 (5 > 4), append nums[3] to arr1, hence arr1 becomes [5,3].
In the 4^th operation, as the last element of arr2 is greater than the last element of arr1 (4 > 3), append nums[4] to arr2, hence arr2 becomes [4,8].
After 4 operations, arr1 = [5,3] and arr2 = [4,8].
Hence, the array result formed by concatenation is [5,3,4,8].

Constraints:

3 <= n <= 50
1 <= nums[i] <= 100
All elements in nums are distinct.

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 are the constraints on the range of values within the input array?
Can the input array be empty or null?
If multiple ways exist to distribute the elements satisfying the given conditions, is there a specific criteria for choosing one?
In the `insertValue` operation, if `maxA` equals `maxB`, how should I choose between inserting into `arrA` or `arrB`?
What should I return if it's impossible to distribute the elements into the two arrays according to the given procedure?

Brute Force Solution

Approach

The brute force approach involves trying out every single possible way to divide up the numbers. We will explore all combinations to find the best one that fits the specific problem requirements. This guarantees we'll find a solution, albeit inefficiently.

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

Take the first number and assign it to the first collection.
Then, take the second number and try putting it in either the first or the second collection. This creates two possibilities.
For each of those possibilities, take the third number and again try putting it in either the first or second collection. This doubles the number of possibilities we now need to consider.
Keep repeating this process for every number. Each number doubles the number of combinations we need to check.
Once all the numbers are assigned in a combination, we check if this particular arrangement is valid and meets the problem's rules. For example, the first collection has to always have all smaller values, compared to the second collection.
Keep track of all the valid arrangements you find.
After you've tried every single possible arrangement, compare all the valid ones you found and pick the 'best' arrangement according to whatever specific criteria the problem asks for (e.g., smallest sum in one collection, longest length, etc.).

Code Implementation

def distribute_elements_brute_force(numbers):
    all_possible_distributions = []

    def generate_distributions(index, first_collection, second_collection):
        if index == len(numbers):
            # We have reached the end of our number list
            all_possible_distributions.append((first_collection.copy(), second_collection.copy()))
            return

        number = numbers[index]

        # Try adding the current number to the first collection.
        first_collection.append(number)
        generate_distributions(index + 1, first_collection, second_collection)

        # Backtrack. Remove added element.
        first_collection.pop()

        # Try adding the current number to the second collection.
        second_collection.append(number)
        generate_distributions(index + 1, first_collection, second_collection)

        # Backtrack. Remove added element.
        second_collection.pop()

    generate_distributions(0, [], [])
    return all_possible_distributions

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores all possible combinations of placing each of the n elements into two arrays. For each element, there are two choices: put it in the first array or the second array. Therefore, the total number of combinations is 2 * 2 * ... * 2 (n times), which equals 2^n. For each of these 2^n combinations, we need to perform some constant amount of work to validate if it's a valid arrangement, and possibly compare it to the current best arrangement. This constant amount of work does not affect the overall Big O complexity. Therefore, the overall time complexity is O(2^n).

Space Complexity

O(2^N) – The brute force approach generates all possible combinations of assigning each of the N numbers into two arrays. In the worst case, it needs to store all valid arrangements. Because each number has two choices, this leads to a maximum of 2^N combinations. Although the plain english mentions keeping track of the valid arrangements and picking the best one later, the storage of all the combinations will dominate the space usage. Therefore, the space complexity is O(2^N).

Optimal Solution

Approach

The best way to solve this puzzle is to cleverly sort each number into one of two groups while making sure their running sums stay in order. We use a system of comparisons to guide our decisions, ensuring we always place each number in the correct group without backtracking.

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

Imagine you have two empty boxes labeled A and B. These boxes will hold your numbers.
Start with the first number in your list. Compare it to the running total of box A and to the running total of box B.
If the number is larger than both running totals, put it in the box whose running total is smaller. This ensures that the totals always stay in the correct order.
If the number is smaller than or equal to one of the running totals, you *must* place it in the other box to maintain the order.
Keep repeating this process for each number in the list.
By consistently comparing each number and carefully choosing its box, you will end up with two groups that meet the specified requirements.

Code Implementation

def distribute_elements(number_list):
    array_a = []
    array_b = []
    running_total_a = 0
    running_total_b = 0

    for number in number_list:
        # Compare the number to the running totals.
        if number > running_total_a and number > running_total_b:
            # Place in array with smaller sum to maintain order.
            if running_total_a <= running_total_b:
                array_a.append(number)
                running_total_a += number

            else:
                array_b.append(number)
                running_total_b += number

        else:
            # Ensure that the sums of elements stay in order.
            if running_total_a >= running_total_b:
                array_a.append(number)
                running_total_a += number

            else:
                array_b.append(number)
                running_total_b += number

    return array_a, array_b

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each of the n elements in the input array once. Inside the loop, it performs a fixed number of constant-time comparisons (comparing the current element with the running sums of two arrays A and B). Since there are no nested loops or recursive calls, the dominant operation is the single pass through the input array. Therefore, the time complexity is directly proportional to the input size n, resulting in O(n).

Space Complexity

O(N) – The algorithm describes placing each number from the input list into one of two implied arrays (box A and box B). In the worst-case scenario, one of these boxes could contain all N elements of the input list. Therefore, the auxiliary space required to store these distributed elements scales linearly with the input size N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty array or throw an IllegalArgumentException as appropriate for the language to signal invalid input.
Input array with a single element	Return an empty array or a pre-defined error array since it cannot be split into two arrays.
Array contains only identical elements	The solution should correctly distribute identical elements between the two arrays based on the defined comparison criteria.
Array contains negative numbers	The comparison logic should handle negative numbers correctly, without assuming non-negativity.
Array contains zero values	The comparison logic needs to handle zero values correctly, as zeros could lead to unexpected comparison outcomes.
Integer overflow during comparisons	Use appropriate data types (e.g., long in Java) or careful comparison strategies to avoid integer overflow issues.
Large input array (performance)	Optimize the solution to have a time complexity of O(n log n) or better to handle large arrays efficiently.
Array already sorted in ascending or descending order	The solution should still generate valid arrays even if the input array is already sorted in some way.

Null or empty input array

How to Handle:

Return an empty array or throw an IllegalArgumentException as appropriate for the language to signal invalid input.

Input array with a single element

How to Handle:

Return an empty array or a pre-defined error array since it cannot be split into two arrays.

Array contains only identical elements

How to Handle:

The solution should correctly distribute identical elements between the two arrays based on the defined comparison criteria.

Array contains negative numbers

How to Handle:

The comparison logic should handle negative numbers correctly, without assuming non-negativity.

Array contains zero values

How to Handle:

The comparison logic needs to handle zero values correctly, as zeros could lead to unexpected comparison outcomes.

Integer overflow during comparisons

How to Handle:

Use appropriate data types (e.g., long in Java) or careful comparison strategies to avoid integer overflow issues.

Large input array (performance)

How to Handle:

Optimize the solution to have a time complexity of O(n log n) or better to handle large arrays efficiently.

Array already sorted in ascending or descending order

How to Handle:

The solution should still generate valid arrays even if the input array is already sorted in some way.