Advantage shuffle

Array Two Pointers Greedy Sorting

54.3% acceptance

Feb 22, 2026

1691

101

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.

Solution

Rust

Time O(n log n)

Space O(n)

LeetCode

solution.rs

impl Solution {
  pub fn advantage_count(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
    let n = nums1.len();
    let mut sorted1 = nums1.clone();
    sorted1.sort_unstable();
    let mut deq: std::collections::VecDeque<i32> = sorted1.into_iter().collect();
    let mut idx2: Vec<usize> = (0..n).collect();
    idx2.sort_unstable_by(|&a, &b| nums2[b].cmp(&nums2[a]));
    let mut ans = vec![0; n];
    for i in idx2 {
      if *deq.back().unwrap() > nums2[i] {
        ans[i] = deq.pop_back().unwrap();
      } else {
        ans[i] = deq.pop_front().unwrap();
      }
    }
    ans
  }
}