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This approach involves sorting the arrays and using a two-pointer technique to maximize the advantage. By sorting, we can easily find which elements can pair to maximize the advantage.
Steps:
nums1 and create an array of indices of nums2 sorted based on values.nums1 to the sorted indices of nums2, ensuring we maximize comparisons where nums1[i] > nums2[i].Time Complexity: O(n log n), primarily due to sorting of arrays.
Space Complexity: O(n), due to the storage of results and sorted indices.
1def advantage_count(nums1, nums2):
2 nums1.sort()
3 sorted_indices = sorted(range(len(nums2)), key=lambda i: nums2[i])
4 result = [-1] * len(nums1)
5 left, right = 0, len(nums1) - 1
6 for num in nums1:
7 if num > nums2[sorted_indices[left]]:
8 result[sorted_indices[left]] = num
9 left += 1
10 else:
11 result[sorted_indices[right]] = num
12 right -= 1
13 return result
14nums1 and uses two pointers to optimally distribute its elements to maximize advantage. Unsuited elements replace the largest unmatched element from nums2 which minimizes their impact.This method involves using a map to store elements of nums1 and using binary search to quickly find the smallest usable element to maximize the advantage.
Steps:
nums1 in a map.nums2, find the smallest greater element from the map to maximize the advantage, using binary search capabilities.Time Complexity: O(n log n), due to sorting and map operations.
Space Complexity: O(n), as we store frequencies in a map.
1import java.util.*;
2
nums1 for each element in nums2. Efficient for matching large sets due to logarithmic lookup complexity.