870. Advantage Shuffle 

Description

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

Solutions

Solution 1

Python3

class Solution:
    def advantageCount(self, nums1: List[int], nums2: List[int]) -> List[int]:
        nums1.sort()
        t = sorted((v, i) for i, v in enumerate(nums2))
        n = len(nums2)
        ans = [0] * n
        i, j = 0, n - 1
        for v in nums1:
            if v <= t[i][0]:
                ans[t[j][1]] = v
                j -= 1
            else:
                ans[t[i][1]] = v
                i += 1
        return ans

Java

class Solution {
    public int[] advantageCount(int[] nums1, int[] nums2) {
        int n = nums1.length;
        int[][] t = new int[n][2];
        for (int i = 0; i < n; ++i) {
            t[i] = new int[] {nums2[i], i};
        }
        Arrays.sort(t, (a, b) -> a[0] - b[0]);
        Arrays.sort(nums1);
        int[] ans = new int[n];
        int i = 0, j = n - 1;
        for (int v : nums1) {
            if (v <= t[i][0]) {
                ans[t[j--][1]] = v;
            } else {
                ans[t[i++][1]] = v;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
        int n = nums1.size();
        vector<pair<int, int>> t;
        for (int i = 0; i < n; ++i) t.push_back({nums2[i], i});
        sort(t.begin(), t.end());
        sort(nums1.begin(), nums1.end());
        int i = 0, j = n - 1;
        vector<int> ans(n);
        for (int v : nums1) {
            if (v <= t[i].first)
                ans[t[j--].second] = v;
            else
                ans[t[i++].second] = v;
        }
        return ans;
    }
};

Go

func advantageCount(nums1 []int, nums2 []int) []int {
	n := len(nums1)
	t := make([][]int, n)
	for i, v := range nums2 {
		t[i] = []int{v, i}
	}
	sort.Slice(t, func(i, j int) bool {
		return t[i][0] < t[j][0]
	})
	sort.Ints(nums1)
	ans := make([]int, n)
	i, j := 0, n-1
	for _, v := range nums1 {
		if v <= t[i][0] {
			ans[t[j][1]] = v
			j--
		} else {
			ans[t[i][1]] = v
			i++
		}
	}
	return ans
}

TypeScript

function advantageCount(nums1: number[], nums2: number[]): number[] {
    const n = nums1.length;
    const idx = Array.from({ length: n }, (_, i) => i);
    idx.sort((i, j) => nums2[i] - nums2[j]);
    nums1.sort((a, b) => a - b);

    const ans = new Array(n).fill(0);
    let left = 0;
    let right = n - 1;
    for (let i = 0; i < n; i++) {
        if (nums1[i] > nums2[idx[left]]) {
            ans[idx[left]] = nums1[i];
            left++;
        } else {
            ans[idx[right]] = nums1[i];
            right--;
        }
    }
    return ans;
}

Rust

impl Solution {
    pub fn advantage_count(mut nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
        let n = nums1.len();
        let mut idx = (0..n).collect::<Vec<usize>>();
        idx.sort_by(|&i, &j| nums2[i].cmp(&nums2[j]));
        nums1.sort();
        let mut res = vec![0; n];
        let mut left = 0;
        let mut right = n - 1;
        for &num in nums1.iter() {
            if num > nums2[idx[left]] {
                res[idx[left]] = num;
                left += 1;
            } else {
                res[idx[right]] = num;
                right -= 1;
            }
        }
        res
    }
}