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Advantage Shuffle

Medium

Asked by:

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ArraysGreedy AlgorithmsTwo Pointers

You are given two integer arrays nums1 and nums2 both of the same length. The advantage of nums1 with respect to nums2 is the number of indices i for which nums1[i] > nums2[i].

Return any permutation of nums1 that maximizes its advantage with respect to nums2.

Example 1:

Input: nums1 = [2,7,11,15], nums2 = [1,10,4,11]
Output: [2,11,7,15]

Example 2:

Input: nums1 = [12,24,8,32], nums2 = [13,25,32,11]
Output: [24,32,8,12]

Constraints:

1 <= nums1.length <= 10⁵
nums2.length == nums1.length
0 <= nums1[i], nums2[i] <= 10⁹

Try coding it in LeetCode

Advantage Shuffle

Medium

Asked by:

36 views

Topics:

ArraysGreedy AlgorithmsTwo Pointers

You are given two integer arrays nums1 and nums2 both of the same length. The advantage of nums1 with respect to nums2 is the number of indices i for which nums1[i] > nums2[i].

Return any permutation of nums1 that maximizes its advantage with respect to nums2.

Example 1:

Input: nums1 = [2,7,11,15], nums2 = [1,10,4,11]
Output: [2,11,7,15]

Example 2:

Input: nums1 = [12,24,8,32], nums2 = [13,25,32,11]
Output: [24,32,8,12]

Constraints:

1 <= nums1.length <= 10⁵
nums2.length == nums1.length
0 <= nums1[i], nums2[i] <= 10⁹

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 size of `nums1` and `nums2`? What is the range of values within the arrays?
Are duplicate values allowed in `nums1` and `nums2`, and if so, how should they be handled in the advantage calculation?
If there are multiple valid permutations of `nums1` that maximize the advantage, is any one acceptable or is there a specific permutation I should aim for?
If it's not possible to assign a value from `nums1` that is greater than a value in `nums2` for a particular index, what should I do with that value in `nums1`?
Are `nums1` and `nums2` guaranteed to have the same length, and will they ever be null or empty?

Brute Force Solution

Approach

Imagine you're trying to match pairs of items between two lists to gain the most advantage. The brute force method tries every single way of pairing items from the first list with items from the second, comparing in each case if the first item is better than the second. It exhaustively tests all possible combinations.

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

Take the first item from the first list.
Compare it with every item in the second list to see if it's better.
If it's better than one of the items, mark that pairing down.
Repeat this process with the first item from the first list against all items in the second list.
Then do the same thing for the second item from the first list against all the items from the second list.
Continue this comparison for every single item in the first list against every single item in the second list, creating every possible pairing.
From all the possible pairings, look at which one yields the maximum advantage (where the item from the first list is better).
Return the arrangement that has the greatest overall advantage.

Code Implementation

import itertools

def advantage_shuffle_brute_force(first_list, second_list):
    number_of_items = len(first_list)
    best_advantage = -1
    best_permutation = []

    # Iterate through all possible permutations of first_list
    for permutation in itertools.permutations(first_list):
        current_advantage = 0
        
        # Calculate the advantage for the current permutation
        for index in range(number_of_items):
            if permutation[index] > second_list[index]:
                current_advantage += 1

        # Update the best advantage if the current one is better
        if current_advantage > best_advantage:
            best_advantage = current_advantage
            best_permutation = list(permutation)

    return best_permutation

Big(O) Analysis

Time Complexity

O(n!) – The provided solution describes a brute-force approach that explores all possible permutations of the first list (nums1) against the second list (nums2) to find the arrangement with the maximum advantage. Comparing all possible pairings of n elements from nums1 with n elements from nums2 involves generating all permutations, which grows factorially. Thus, the dominant factor determining the time complexity is the generation and evaluation of all possible orderings of nums1 against nums2. Therefore, the time complexity is O(n!).

Space Complexity

O(1) – The described brute force method iterates and compares elements without using any auxiliary data structures that scale with the input size N, where N is the number of items in each list. The algorithm only uses a fixed number of variables for indexing and comparison, independent of the input size. Therefore, the auxiliary space complexity is constant. No additional lists, dictionaries, or other data structures are created or used to store intermediate results.

Optimal Solution

Approach

The goal is to rearrange one set of numbers to 'win' against another set as often as possible. The clever trick involves strategically assigning each number from the first set to a number in the second set to guarantee a 'win' whenever feasible.

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

First, sort both sets of numbers independently, putting them in ascending order.
Imagine having two groups of cards: your cards (the first set) and your opponent's cards (the second set).
For each of your opponent's cards, try to find the smallest card in your hand that is bigger than it. If you can find one, assign that card to 'win' against the opponent's card.
If you don't have a card that can beat the opponent's current card, then sacrifice your smallest remaining card (if any) to that opponent's card. This minimizes the loss when you can't win anyway.
Repeat this process for all of the opponent's cards. You'll end up with a specific assignment of your cards to your opponent's cards.
Because of the initial sorting, the assignments created will maximize the number of wins.
Finally, arrange your numbers in the original order of the second set of numbers, so that each number is matched to the number it was strategically assigned to.

Code Implementation

def advantage_count(first_list, second_list): 
    list_size = len(first_list)
    sorted_first_list = sorted(first_list)
    indexed_second_list = sorted((value, index) for index, value in enumerate(second_list))

    result_list = [0] * list_size
    assigned_indices = set()
    left_index = 0
    right_index = list_size - 1

    # Assign largest possible number in first_list to each number in second_list
    for second_list_value, second_list_index in indexed_second_list:
        if sorted_first_list[right_index] > second_list_value:
            result_list[second_list_index] = sorted_first_list[right_index]
            assigned_indices.add(right_index)
            right_index -= 1
        else:
            #If no number is larger, assign smallest number
            result_list[second_list_index] = sorted_first_list[left_index]
            assigned_indices.add(left_index)
            left_index += 1

    return result_list

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operations are the sorting steps performed on both input arrays, each of size n. Sorting algorithms like merge sort or quicksort have a time complexity of O(n log n). While there is a loop iterating through the sorted second array, the operations within this loop (finding a suitable element or assigning a 'loss') take at most O(n) time in total, but it's dominated by the O(n log n) sort. Therefore, the overall time complexity is O(n log n).

Space Complexity

O(N) – The algorithm sorts both input arrays, but the space complexity analysis focuses on auxiliary space. The primary auxiliary space usage comes from storing the indices of the second array (nums2) after sorting to reconstruct the output in the original order. This requires creating an array of size N to hold these indices, where N is the length of the input arrays. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty or null nums1 or nums2	Return an empty list or handle appropriately, such as by throwing an IllegalArgumentException if null input is not allowed.
nums1 and nums2 have different lengths	Throw an IllegalArgumentException or return an empty list as the problem states arrays must be of the same length.
nums1 and nums2 contain only equal values	Assign the smallest possible values of nums1 to nums2 to minimize disadvantage, resulting in zero advantage in this case.
nums1 is sorted in descending order and nums2 is sorted in ascending order	Greedily assign each element in nums1 to the smallest element in nums2 that it can beat to maximize advantage.
nums1 contains all large values and nums2 contains all small values	Each value in nums1 should be matched to the smallest unmatched value in nums2.
nums1 contains all small values and nums2 contains all large values	We are forced to minimize the disadvantage by assigning the smallest possible values of nums1 to nums2, leading to a worst-case permutation.
Integer overflow when comparing very large numbers	Ensure the comparison does not lead to integer overflow; consider using long or appropriate handling to avoid overflow.
Arrays with extreme boundary values (e.g., Integer.MAX_VALUE, Integer.MIN_VALUE)	The solution should work correctly even with extreme integer values without causing any exceptions or incorrect results.

Empty or null nums1 or nums2

How to Handle:

Return an empty list or handle appropriately, such as by throwing an IllegalArgumentException if null input is not allowed.

nums1 and nums2 have different lengths

How to Handle:

Throw an IllegalArgumentException or return an empty list as the problem states arrays must be of the same length.

nums1 and nums2 contain only equal values

How to Handle:

Assign the smallest possible values of nums1 to nums2 to minimize disadvantage, resulting in zero advantage in this case.

nums1 is sorted in descending order and nums2 is sorted in ascending order

How to Handle:

Greedily assign each element in nums1 to the smallest element in nums2 that it can beat to maximize advantage.

nums1 contains all large values and nums2 contains all small values

How to Handle:

Each value in nums1 should be matched to the smallest unmatched value in nums2.

nums1 contains all small values and nums2 contains all large values

How to Handle:

We are forced to minimize the disadvantage by assigning the smallest possible values of nums1 to nums2, leading to a worst-case permutation.

Integer overflow when comparing very large numbers

How to Handle:

Ensure the comparison does not lead to integer overflow; consider using long or appropriate handling to avoid overflow.

Arrays with extreme boundary values (e.g., Integer.MAX_VALUE, Integer.MIN_VALUE)

How to Handle:

The solution should work correctly even with extreme integer values without causing any exceptions or incorrect results.

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 size of `nums1` and `nums2`? What is the range of values within the arrays?
Are duplicate values allowed in `nums1` and `nums2`, and if so, how should they be handled in the advantage calculation?
If there are multiple valid permutations of `nums1` that maximize the advantage, is any one acceptable or is there a specific permutation I should aim for?
If it's not possible to assign a value from `nums1` that is greater than a value in `nums2` for a particular index, what should I do with that value in `nums1`?
Are `nums1` and `nums2` guaranteed to have the same length, and will they ever be null or empty?

Brute Force Solution

Approach

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

Take the first item from the first list.
Compare it with every item in the second list to see if it's better.
If it's better than one of the items, mark that pairing down.
Repeat this process with the first item from the first list against all items in the second list.
Then do the same thing for the second item from the first list against all the items from the second list.
Continue this comparison for every single item in the first list against every single item in the second list, creating every possible pairing.
From all the possible pairings, look at which one yields the maximum advantage (where the item from the first list is better).
Return the arrangement that has the greatest overall advantage.

Code Implementation

import itertools

def advantage_shuffle_brute_force(first_list, second_list):
    number_of_items = len(first_list)
    best_advantage = -1
    best_permutation = []

    # Iterate through all possible permutations of first_list
    for permutation in itertools.permutations(first_list):
        current_advantage = 0
        
        # Calculate the advantage for the current permutation
        for index in range(number_of_items):
            if permutation[index] > second_list[index]:
                current_advantage += 1

        # Update the best advantage if the current one is better
        if current_advantage > best_advantage:
            best_advantage = current_advantage
            best_permutation = list(permutation)

    return best_permutation

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, sort both sets of numbers independently, putting them in ascending order.
Imagine having two groups of cards: your cards (the first set) and your opponent's cards (the second set).
For each of your opponent's cards, try to find the smallest card in your hand that is bigger than it. If you can find one, assign that card to 'win' against the opponent's card.
If you don't have a card that can beat the opponent's current card, then sacrifice your smallest remaining card (if any) to that opponent's card. This minimizes the loss when you can't win anyway.
Repeat this process for all of the opponent's cards. You'll end up with a specific assignment of your cards to your opponent's cards.
Because of the initial sorting, the assignments created will maximize the number of wins.
Finally, arrange your numbers in the original order of the second set of numbers, so that each number is matched to the number it was strategically assigned to.

Code Implementation

def advantage_count(first_list, second_list): 
    list_size = len(first_list)
    sorted_first_list = sorted(first_list)
    indexed_second_list = sorted((value, index) for index, value in enumerate(second_list))

    result_list = [0] * list_size
    assigned_indices = set()
    left_index = 0
    right_index = list_size - 1

    # Assign largest possible number in first_list to each number in second_list
    for second_list_value, second_list_index in indexed_second_list:
        if sorted_first_list[right_index] > second_list_value:
            result_list[second_list_index] = sorted_first_list[right_index]
            assigned_indices.add(right_index)
            right_index -= 1
        else:
            #If no number is larger, assign smallest number
            result_list[second_list_index] = sorted_first_list[left_index]
            assigned_indices.add(left_index)
            left_index += 1

    return result_list

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty or null nums1 or nums2	Return an empty list or handle appropriately, such as by throwing an IllegalArgumentException if null input is not allowed.
nums1 and nums2 have different lengths	Throw an IllegalArgumentException or return an empty list as the problem states arrays must be of the same length.
nums1 and nums2 contain only equal values	Assign the smallest possible values of nums1 to nums2 to minimize disadvantage, resulting in zero advantage in this case.
nums1 is sorted in descending order and nums2 is sorted in ascending order	Greedily assign each element in nums1 to the smallest element in nums2 that it can beat to maximize advantage.
nums1 contains all large values and nums2 contains all small values	Each value in nums1 should be matched to the smallest unmatched value in nums2.
nums1 contains all small values and nums2 contains all large values	We are forced to minimize the disadvantage by assigning the smallest possible values of nums1 to nums2, leading to a worst-case permutation.
Integer overflow when comparing very large numbers	Ensure the comparison does not lead to integer overflow; consider using long or appropriate handling to avoid overflow.
Arrays with extreme boundary values (e.g., Integer.MAX_VALUE, Integer.MIN_VALUE)	The solution should work correctly even with extreme integer values without causing any exceptions or incorrect results.

Empty or null nums1 or nums2

How to Handle:

Return an empty list or handle appropriately, such as by throwing an IllegalArgumentException if null input is not allowed.

nums1 and nums2 have different lengths

How to Handle:

Throw an IllegalArgumentException or return an empty list as the problem states arrays must be of the same length.

nums1 and nums2 contain only equal values

How to Handle:

Assign the smallest possible values of nums1 to nums2 to minimize disadvantage, resulting in zero advantage in this case.

nums1 is sorted in descending order and nums2 is sorted in ascending order

How to Handle:

Greedily assign each element in nums1 to the smallest element in nums2 that it can beat to maximize advantage.

nums1 contains all large values and nums2 contains all small values

How to Handle:

Each value in nums1 should be matched to the smallest unmatched value in nums2.

nums1 contains all small values and nums2 contains all large values

How to Handle:

We are forced to minimize the disadvantage by assigning the smallest possible values of nums1 to nums2, leading to a worst-case permutation.

Integer overflow when comparing very large numbers

How to Handle:

Ensure the comparison does not lead to integer overflow; consider using long or appropriate handling to avoid overflow.

Arrays with extreme boundary values (e.g., Integer.MAX_VALUE, Integer.MIN_VALUE)

How to Handle:

The solution should work correctly even with extreme integer values without causing any exceptions or incorrect results.