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added ShortestRangeInKSortedLists question
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‎README.md‎

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@@ -61,7 +61,7 @@ Maven build script is provided.
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- RangeMerge (RangeMergeTest) - sorted array of ints. Encode/Archive/Zip numbers into ranges. Optimized solution
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- TrappingRainWater (TrappingRainWaterTest) - array of landscape. Calculate how many units of water will remain after the rain inside.
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- KClosestPoints (KClosestPointsTest) - find k closest 2d points to particular one
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- ShortestRangeInKSortedLists (ShortestRangeInKSortedListsTest) - take one element from each list - make the range of taken numbers the shortest possible
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# Yandex tasks
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- Bencode (BencodeTest) - implementation of Bencode protocol for encoding/decoding data. Widely used in torrents. More info: https://en.wikipedia.org/wiki/Bencode
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‎src/main/java/iurii/job/interview/amazon/ShortestRangeInKSortedLists.java‎

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package iurii.job.interview.amazon;
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import java.util.*;
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/**
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* Problem statement in detail:
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* Given k sorted lists of integers of size n each,
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* find the smallest range that includes at least element from each of the k lists.
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* If more than one smallest ranges are found, print any one of them.
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* <p>
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* Problem is similar to {@link iurii.job.interview.facebook.MinIteratorOfSortedAscendingLists}
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* <p>
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* https://www.youtube.com/watch?v=zplklOy7ENo
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*/
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public class ShortestRangeInKSortedLists {
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public Range findRange(List<List<Integer>> lists) {
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// checks if needed
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if (lists == null || lists.isEmpty()) {
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return null;
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}
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if (lists.stream().anyMatch(List::isEmpty)) {
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return null;
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}
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long totalTransitionSize = lists.stream().mapToLong(list -> list.size() - 1).sum();
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// pointers
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int listsCount = lists.size();
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List<Integer> pointers = new ArrayList<>(Collections.nCopies(listsCount, 0));
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// initial value - will be changed, because min > max
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Range result = new Range();
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result.min = Integer.MIN_VALUE;
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result.max = Integer.MAX_VALUE;
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// find
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while (totalTransitionSize-- > 0) { // check all transitions
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long min = Integer.MAX_VALUE;
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long max = Integer.MIN_VALUE;
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int minFromExisting = Integer.MAX_VALUE;
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int minPointer = -1;
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for (int i = 0; i < listsCount; i++) {
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int pointer = pointers.get(i);
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max = Math.max(lists.get(i).get(pointer), max);
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min = Math.min(lists.get(i).get(pointer), min);
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// find min from lists that have one more element
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if (lists.get(i).get(pointer) < minFromExisting && pointer + 1 < lists.get(i).size()) {
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minFromExisting = Math.min(lists.get(i).get(pointer), minFromExisting);
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minPointer = i;
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}
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}
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pointers.set(minPointer, pointers.get(minPointer) + 1);
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if (max - min < result.max - result.min) {
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result.min = min;
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result.max = max;
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}
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}
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return result;
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}
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// long not to have overflow in case wide range of integers
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public static class Range {
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long min;
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long max;
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}
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public Range findRangeWithPriorityQueue(List<List<Integer>> lists) {
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// checks if needed
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if (lists == null || lists.isEmpty()) {
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return null;
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}
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if (lists.stream().anyMatch(List::isEmpty)) {
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return null;
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}
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Range result = new Range();
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result.min = Integer.MIN_VALUE;
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result.max = Integer.MAX_VALUE;
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MixedListsIterator iterator = new MixedListsIterator(lists);
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while (iterator.hasNext()) {
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Range range = iterator.next();
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if (range.max - range.min < result.max - result.min) {
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result = range;
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}
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}
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return result;
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}
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interface MinIteratorOfSortedAscendingLists<E> extends Iterator<E> {
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}
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private class MixedListsIterator implements MinIteratorOfSortedAscendingLists<Range> {
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private final List<List<Integer>> lists;
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private final PriorityQueue<ListIterator> minQueue;
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private final PriorityQueue<Integer> maxQueue;
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private long totalTransitionSize;
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public MixedListsIterator(List<List<Integer>> lists) {
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this.lists = lists;
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minQueue = new PriorityQueue<>(lists.size());
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maxQueue = new PriorityQueue<>(lists.size(), Comparator.reverseOrder());
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for (List<Integer> list : lists) {
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minQueue.add(new ListIterator(list.iterator()));
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maxQueue.add(list.get(0));
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}
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totalTransitionSize = lists.stream().mapToLong(list -> list.size() - 1).sum();
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}
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@Override
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public boolean hasNext() {
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return totalTransitionSize != 0;
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}
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@Override
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public Range next() {
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if (hasNext()) {
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ListIterator listIterator = minQueue.poll();
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Range result = new Range();
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if (listIterator.hasNext()) {
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int min = listIterator.next();
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int max = maxQueue.peek();
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result.min = min;
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result.max = max;
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maxQueue.remove(min);
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minQueue.add(listIterator);
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if (listIterator.hasNext()) {
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maxQueue.add(listIterator.value);
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}
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}
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return result;
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}
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throw new NoSuchElementException();
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}
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}
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public static class ListIterator implements Comparable<ListIterator>, Iterator<Integer> {
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private Iterator<Integer> iterator;
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private Integer value;
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public ListIterator(Iterator<Integer> iterator) {
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this.iterator = iterator;
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if (iterator.hasNext()) {
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value = iterator.next();
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}
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}
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@Override
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public int compareTo(ListIterator o) {
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if (iterator.hasNext() ^ o.iterator.hasNext()) {
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return Integer.compare(value, o.value);
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} else {
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return iterator.hasNext() ? -1 : 1;
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}
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}
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@Override
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public boolean hasNext() {
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return value != null;
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}
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@Override
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public Integer next() {
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int result = value;
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if (iterator.hasNext()) {
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value = iterator.next();
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} else {
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value = null;
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}
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return result;
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}
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}
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}

‎src/test/java/iurii/job/interview/amazon/ShortestRangeInKSortedListsTest.java‎

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package iurii.job.interview.amazon;
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import org.junit.Test;
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import java.util.Arrays;
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import java.util.List;
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import static org.assertj.core.api.Assertions.assertThat;
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public class ShortestRangeInKSortedListsTest {
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@Test
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public void test() {
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List<Integer> list1 = Arrays.asList(1, 4, 7);
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List<Integer> list2 = Arrays.asList(2, 5, 8);
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List<Integer> list3 = Arrays.asList(3, 4, 6);
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List<List<Integer>> lists = Arrays.asList(list1, list2, list3);
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ShortestRangeInKSortedLists shortestRangeInKSortedLists = new ShortestRangeInKSortedLists();
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ShortestRangeInKSortedLists.Range range = shortestRangeInKSortedLists.findRange(lists);
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assertThat(range.min).isEqualTo(4);
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assertThat(range.max).isEqualTo(5);
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}
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@Test
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public void testWithPriorityQueue() {
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List<Integer> list1 = Arrays.asList(1, 4, 7);
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List<Integer> list2 = Arrays.asList(2, 5, 8);
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List<Integer> list3 = Arrays.asList(3, 4, 6);
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List<List<Integer>> lists = Arrays.asList(list1, list2, list3);
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ShortestRangeInKSortedLists shortestRangeInKSortedLists = new ShortestRangeInKSortedLists();
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ShortestRangeInKSortedLists.Range range = shortestRangeInKSortedLists.findRangeWithPriorityQueue(lists);
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assertThat(range.min).isEqualTo(4);
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assertThat(range.max).isEqualTo(5);
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}
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}

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