Minimum Number Game

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ArraysGreedy Algorithms

You are given a 0-indexed integer array nums of even length and there is also an empty array arr. Alice and Bob decided to play a game where in every round Alice and Bob will do one move. The rules of the game are as follows:

Every round, first Alice will remove the minimum element from nums, and then Bob does the same.
Now, first Bob will append the removed element in the array arr, and then Alice does the same.
The game continues until nums becomes empty.

Return the resulting array arr.

Example 1:

Input: nums = [5,4,2,3]
Output: [3,2,5,4]
Explanation: In round one, first Alice removes 2 and then Bob removes 3. Then in arr firstly Bob appends 3 and then Alice appends 2. So arr = [3,2].
At the begining of round two, nums = [5,4]. Now, first Alice removes 4 and then Bob removes 5. Then both append in arr which becomes [3,2,5,4].

Example 2:

Input: nums = [2,5]
Output: [5,2]
Explanation: In round one, first Alice removes 2 and then Bob removes 5. Then in arr firstly Bob appends and then Alice appends. So arr = [5,2].

Constraints:

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

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 expected range of values for the numbers in the input array? Can the array contain negative numbers?
Can the input array be empty, or can it contain an odd number of elements? If so, what should the function return in those cases?
Are there any constraints on the values of the elements (e.g., are they all integers, or could they be floating-point numbers)?
If multiple valid pairings lead to the same minimum sum after swaps, is any valid pairing acceptable as the solution?
Should the returned array maintain the same order as the original array after swaps, or can the order be different?

Brute Force Solution

Approach

The minimum number game requires pairing numbers in a list and picking the smaller one from each pair. The brute force approach involves trying every single possible pairing of the numbers and calculating the total sum of the minimums for each of those pairings.

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

Consider all the possible ways to pair up the numbers.
For each way of pairing, find the smaller number in each pair.
Add up all the smaller numbers you found for that specific pairing.
Keep track of the smallest total you find across all possible pairings.
After checking every possible pairing, the smallest total sum is the answer.

Code Implementation

import itertools

def minimum_number_game_brute_force(numbers):
    number_of_elements = len(numbers)
    if number_of_elements % 2 != 0:
        return -1

    minimum_sum = float('inf')

    # Generate all possible pairings
    for pairing in itertools.permutations(numbers):
        current_sum = 0

        # Iterate through pairs and calculate the sum of minimums
        for i in range(0, number_of_elements, 2):
            current_sum += min(pairing[i], pairing[i+1])

        # Update minimum_sum if current pairing sum is smaller
        minimum_sum = min(minimum_sum, current_sum)

    return minimum_sum

Big(O) Analysis

Time Complexity

O(n!) – The algorithm explores every possible pairing of n numbers. Calculating the number of ways to pair n numbers involves calculating n! / (2^(n/2) * (n/2)!). This is because we need to consider all permutations and then divide by the number of ways to arrange elements within pairs and the number of ways to arrange the pairs themselves. Therefore, the dominant factor determining the runtime is the factorial component, leading to a time complexity of O(n!).

Space Complexity

O(1) – The brute force approach, as described, doesn't explicitly use any auxiliary data structures whose size scales with the input. It keeps track of the smallest total found so far, which takes constant space. Therefore, the algorithm primarily uses a fixed amount of extra memory, independent of the input size N, where N is the number of numbers in the list. The space complexity is O(1).

Optimal Solution

Approach

The goal is to pair numbers to minimize their sums, then swap those pairs according to the rules. The clever trick is to realize that sorting the numbers allows us to easily create the smallest possible sums and then perform the swaps correctly.

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

First, arrange all the numbers in order from smallest to largest.
Now, take the numbers in pairs. Pair the first number with the second, the third with the fourth, and so on. This ensures the smallest numbers are paired together, minimizing their sum.
Finally, for each pair, put the smaller number first in the new sequence. This follows the rule about swapping the original pairs to ensure the smaller number is first.

Code Implementation

def minimum_number_game(numbers):
    numbers.sort()

    result = []

    # Iterate through the sorted numbers in pairs.
    for i in range(0, len(numbers), 2):
        # Swap each pair to put the smaller number first.
        result.append(numbers[i+1])
        result.append(numbers[i])

    return result

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation is sorting the input array of n numbers, which typically takes O(n log n) time using efficient algorithms like merge sort or quicksort. Pairing the numbers and swapping within pairs involves iterating through the array once, taking O(n) time. However, this linear time complexity is less significant than the sorting step. Therefore, the overall time complexity is determined by the sorting algorithm, resulting in O(n log n).

Space Complexity

O(N) – The provided solution sorts the input numbers and creates a new sequence. Sorting in-place might be possible, but assuming a standard sorting algorithm like merge sort is used, it requires O(N) auxiliary space in the worst case. Further, the creation of the new sequence to store the swapped pairs also requires O(N) space, where N is the number of input numbers. Therefore, the overall space complexity is dominated by these operations, resulting in O(N).

Edge Cases

Case	How to Handle
Empty or Null Input	Return an empty array or throw an exception, depending on requirements, as there are no elements to pair.
Array with an odd number of elements	The last element is ignored, as it can't form a pair and the problem does not state any further actions.
Array with negative numbers	The solution should work correctly with negative numbers if the logic solely depends on swapping and not on specific mathematical properties of the numbers.
Array with all elements being the same	The swapping logic should still function correctly, effectively swapping adjacent identical elements.
Array is already sorted in ascending order	The swapping logic will simply swap each consecutive pair, resulting in adjacent elements being swapped.
Large array size nearing memory limits	In-place swapping avoids significant memory overhead and makes the algorithm efficient.
Array with duplicate pairs of numbers	The swapping logic will handle the duplicates correctly, swapping each pair sequentially.
Input array is read-only	Return a new list with swapped elements without modifying the input array.

Empty or Null Input

How to Handle:

Return an empty array or throw an exception, depending on requirements, as there are no elements to pair.

Array with an odd number of elements

How to Handle:

The last element is ignored, as it can't form a pair and the problem does not state any further actions.

Array with negative numbers

How to Handle:

The solution should work correctly with negative numbers if the logic solely depends on swapping and not on specific mathematical properties of the numbers.

Array with all elements being the same

How to Handle:

The swapping logic should still function correctly, effectively swapping adjacent identical elements.

Array is already sorted in ascending order

How to Handle:

The swapping logic will simply swap each consecutive pair, resulting in adjacent elements being swapped.

Large array size nearing memory limits

How to Handle:

In-place swapping avoids significant memory overhead and makes the algorithm efficient.

Array with duplicate pairs of numbers

How to Handle:

The swapping logic will handle the duplicates correctly, swapping each pair sequentially.

Input array is read-only

How to Handle:

Return a new list with swapped elements without modifying the input array.