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 <= 105nums2.length == nums1.length0 <= nums1[i], nums2[i] <= 109This 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].nums1 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.C++
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.
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.nums1 for each element in nums2. Efficient for matching large sets due to logarithmic lookup complexity.C#
Time Complexity: O(n log n), due to sorting and map operations.
Space Complexity: O(n), as we store frequencies in a map.
| Approach | Complexity |
|---|---|
| Approach 1: Greedy with Sorting and Two Pointers | 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. |
| Approach 2: Optimal Utilization with Map and Binary Search | Time Complexity: O(n log n), due to sorting and map operations. Space Complexity: O(n), as we store frequencies in a map. |
Shuffle the Array (Constant Space) - Leetcode 1470 - Python • NeetCodeIO • 19,611 views views
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