Advantage Shuffle LeetCode solution in Java

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class Solution {
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  public int[] advantageCount(int[] nums1, int[] nums2) {
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    Arrays.sort(nums1);
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    PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> (b[0] - a[0]));
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    for (int i = 0; i < nums2.length; i++) pq.add(new int[] {nums2[i], i});
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    int left = 0, right = nums1.length - 1;
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    while (left <= right) {
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      int[] get = pq.poll();
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      int pos = get[1], max = get[0];
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      if (nums1[right] > max) {
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        nums2[pos] = nums1[right];
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        right--;
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      } else {
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        nums2[pos] = nums1[left];
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        left++;
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      }
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    }
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    return nums2;
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  }
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}