Advantage Shuffle

Problem Statement

In this challenge, you are equipped with two integer arrays, nums1 and nums2, both containing the same number of elements. The goal is to determine a permutation of the array nums1 that maximizes its "advantage" over nums2. The "advantage" is defined as the count of indices i for which the value at nums1[i] is strictly greater than the value at nums2[i]. Your task is to devise a permutation of nums1 that, when compared index-wise to nums2, maximizes this count of favorable comparisons.

Examples

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

Approach and Intuition

To tackle this problem effectively, we need to maximize the instances where elements in nums1 are greater than corresponding ones in nums2. Here is the intuitive approach to achieving this:

1. Sort both nums1 and nums2. Sorting helps in systematic comparison starting from the smallest values up to the largest.

2. Employ a two-pointer technique to find for each element in the sorted nums2, the smallest yet bigger element from nums2. This approach attempts to just barely win over the nums2 element, thereby saving larger elements in nums1 for potentially tougher match-ups.

3. Use a greedy strategy:

   • Initialize two pointers, i for nums1 and j for nums2 both set to the start of the arrays.
   • If nums1[i] is greater than nums2[j], it is used to form the advantage pair, and both pointers are moved to the next element.
   • If nums1[i] is not greater, it means nums1[i] cannot defeat nums2[j] and is thus saved in a stack or list for possible less challenging matchups later.

4. Reassemble the final array based on the matches found and any leftover elements, ensuring to maximize the advantage at every index.

Example breakdown:

Using the above strategy on Example 1:

• Input Arrays: nums1 = [2,7,11,15], nums2 = [1,10,4,11]
• After sorting, sorted_nums1 = [2, 7, 11, 15], sorted_nums2 = [1, 4, 10, 11].
• By the two-pointer and greedy comparison we align:
  • nums1[0] which is 2 with nums2[0] which is 1 (wins),
  • look for next bigger and align accordingly leading to the output: [2, 11, 7, 15].

Through systematic and strategic comparisons, ensuring that each decision maximizes the immediate chances of scoring an advantage, this approach should efficiently yield a permutation of nums1 that attains the maximum possible advantage against nums2.

Solutions

class Solution {
    public int[] leverageAdvantage(int[] nums1, int[] nums2) {
        int[] sortedNums1 = nums1.clone();
        Arrays.sort(sortedNums1);
        int[] sortedNums2 = nums2.clone();
        Arrays.sort(sortedNums2);
    
        Map<Integer, Deque<Integer>> competitorMatchups = new HashMap();
        for (int num: nums2) competitorMatchups.put(num, new LinkedList());
    
        Deque<Integer> leftovers = new LinkedList();
    
        int k = 0;
        for (int num1: sortedNums1) {
            if (num1 > sortedNums2[k]) {
                competitorMatchups.get(sortedNums2[k++]).add(num1);
            } else {
                leftovers.add(num1);
            }
        }
    
        int[] result = new int[nums2.length];
        for (int i = 0; i < nums2.length; ++i) {
            if (competitorMatchups.get(nums2[i]).size() > 0)
                result[i] = competitorMatchups.get(nums2[i]).pop();
            else
                result[i] = leftovers.pop();
        }
        return result;
    }
}