Minimum Interval to Include Each Query LeetCode solution in Java

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class Solution {
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  public int[] minInterval(int[][] intervals, int[] queries) {
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    // sort intervals
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    Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
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    // add [index,query]
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    int[][] q = new int[queries.length][2];
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    for (int i = 0; i < queries.length; i++) {
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      q[i][0] = i;
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      q[i][1] = queries[i];
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    }
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    // sort the queries by query val
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    Arrays.sort(q, (a, b) -> a[1] - b[1]);
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    // store the minimum intervals in the priority queue, min heap
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    Queue<int[]> pq = new PriorityQueue<>((a, b) -> (a[1] - a[0]) - (b[1] - b[0]));
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    int[] result = new int[queries.length];
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    int j = 0;
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    for (int i = 0; i < q.length; i++) {
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      int index = q[i][0];
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      int val = q[i][1];
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      // if start is less than query val, then add to pq
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      while (j < intervals.length && intervals[j][0] <= val) pq.offer(intervals[j++]);
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      // anything which has the end lesser than the val then remove it
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      while (!pq.isEmpty() && pq.peek()[1] < val) pq.poll();
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      // add difference to the result
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      result[index] = pq.isEmpty() ? -1 : pq.peek()[1] - pq.peek()[0] + 1;
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    }
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    return result;
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  }
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}