How do you solve Advantage Shuffle on LeetCode?

Advantage Shuffle (LeetCode #870) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting helps in efficiently matching elements.

What is the brute force solution for Advantage Shuffle?

The brute force approach for Advantage Shuffle is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute force approach involves generating all possible permutations of nums1 and checking which permutation maximizes the advantage over nums2. This is straightforward but inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Advantage Shuffle?

Advantage Shuffle uses the following concepts: Array, Two Pointers, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting, Two Pointers.

What are the interview tips for Advantage Shuffle?

Explain your thought process clearly. Discuss the trade-offs of different approaches. Practice coding the optimal solution multiple times.

What are common mistakes when solving Advantage Shuffle?

Not considering the original indices when sorting. Failing to handle cases where no elements can beat the corresponding elements.

#870

Advantage Shuffle

Medium

ArrayTwo PointersGreedySortingGreedySortingTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n log n)
Space	O(n)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution uses sorting and a greedy approach. By sorting both arrays, we can efficiently match elements in nums1 to maximize the count of indices where nums1[i] > nums2[i]. This allows us to use a two-pointer technique to find the best matches.

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Algorithm

3 steps

1Step 1: Sort nums1 and nums2 while keeping track of the original indices of nums2.
2Step 2: Use two pointers: one for nums1 and one for nums2.
3Step 3: For each element in nums1, if it can beat the current element in nums2, assign it and move both pointers. Otherwise, assign the smallest remaining element in nums1.

solution.py13 lines

1def advantageCount(nums1, nums2):
2    sorted_nums1 = sorted(nums1)
3    sorted_indices = sorted(range(len(nums2)), key=lambda i: nums2[i])
4    result = [0] * len(nums1)
5    j = 0
6    for num in sorted_nums1:
7        if num > nums2[sorted_indices[j]]:
8            result[sorted_indices[j]] = num
9            j += 1
10        else:
11            result[sorted_indices[-1]] = num
12            sorted_indices.pop()
13    return result

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting of nums1 and the indices. The space complexity is O(n) for storing the result and indices.

1Sorting helps in efficiently matching elements.
2Using indices allows us to keep track of original positions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.