870. Advantage Shuffle LeetCode Solution

In this guide, you will get 870. Advantage Shuffle LeetCode Solution with the best time and space complexity. The solution to Advantage Shuffle problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Advantage Shuffle solution in C++
Advantage Shuffle solution in Java
Advantage Shuffle solution in Python
Additional Resources

870. Advantage Shuffle LeetCode Solution image

Problem Statement of Advantage Shuffle

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

Complexity Analysis

Time Complexity: O(n\log n)
Space Complexity: O(n)

870. Advantage Shuffle LeetCode Solution in C++

class Solution {
 public:
  vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
    multiset<int> set{nums1.begin(), nums1.end()};

    for (int i = 0; i < nums2.size(); ++i) {
      const auto p =
          *set.rbegin() <= nums2[i] ? set.begin() : set.upper_bound(nums2[i]);
      nums1[i] = *p;
      set.erase(p);
    }

    return nums1;
  }
};
/* code provided by PROGIEZ */

870. Advantage Shuffle LeetCode Solution in Java

class Solution {
  public int[] advantageCount(int[] nums1, int[] nums2) {
    TreeMap<Integer, Integer> map = new TreeMap<>();

    for (final int num : nums1)
      map.merge(num, 1, Integer::sum);

    for (int i = 0; i < nums2.length; i++) {
      Integer key = map.higherKey(nums2[i]);
      if (key == null)
        key = map.firstKey();
      if (map.merge(key, -1, Integer::sum) == 0)
        map.remove(key);
      nums1[i] = key;
    }

    return nums1;
  }
}
// code provided by PROGIEZ

870. Advantage Shuffle LeetCode Solution in Python

from sortedcontainers import SortedList


class Solution:
  def advantageCount(self, nums1: list[int], nums2: list[int]) -> list[int]:
    sl = SortedList(nums1)

    for i, num in enumerate(nums2):
      index = 0 if sl[-1] <= num else sl.bisect_right(num)
      nums1[i] = sl[index]
      del sl[index]

    return nums1
 # code by PROGIEZ