Amazon Interview Question for Solutions Engineers

Country: United States
Interview Type: Phone Interview

Comment hidden because of low score. Click to expand.
of 2 vote

public static void merge(List<Stream> inputStreams, Stream outputStream) {
  PriorityQueue<PQKeyVal> q = new PriorityQueue<PQKeyVal>();
  for (Stream s: inputStreams) {
    q.add(new PQKeyVal(s));
  for (;;)
    PQKeyVal top = q.poll();
    q.offer(new PQKeyVal(;

where PQKeyVal defined like that:

class PQKeyVal implements Comparable<PQKeyVal> {
  public final int key;
  public final Stream stream;
  public PQKeyVal(Stream stream) { = stream;
    this.key = stream.peek();

  public int compareTo(PQKeyVal that) {
    return this.key - that.key;

- Anonymous October 06, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
of 0 vote

We can use a priority queue where the key is the head value and the value is a pointer to the corresponding stream. We just need to repeatedly pop the min element, output the key, advance the stream to the next element and push it back to the queue using the new head value.

- adr October 06, 2019 | Flag Reply

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