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* LAMBDA PROGRAMMING LABORATORY
*
* For each exercise, develop a solution using Java SE 8 Lambda/Streams
* and remove the @Ignore tag. Then run the tests.
*
* In NetBeans, Ctrl-F6 will run the project's tests, which default to
* the unsolved exercises (as opposed to the solutions). Alt-F6 [PC] or
* or Cmd-F6 [Mac] will run just the tests in the currently selected file.
*
* Several of the exercises read data from a text file. The field named
* "reader" is a BufferedReader which will be opened for you on the text file.
* In any exercise that refers to reading from the text file, you can simply
* use the "reader" variable without worry about opening or closing it.
* This is setup by JUnit using the @Before and @After methods at the bottom of
* this file. The text file is "SonnetI.txt" (Shakespeare's first sonnet) which
* is located at the root of this NetBeans project.
import java.util.AbstractMap;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.OptionalInt;
import java.util.Random;
import java.util.Set;
import java.util.stream.Collector;
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import java.util.stream.Collectors;
import java.util.stream.IntStream;
import java.util.stream.Stream;
import org.junit.After;
import org.junit.Before;
import org.junit.Ignore;
import org.junit.Test;
import static org.junit.Assert.assertEquals;
import static org.junit.Assert.assertFalse;
import static org.junit.Assert.assertTrue;
public class G_Challenges {
/**
* Denormalize this map. The input is a map whose keys are the number of legs of an animal
* and whose values are lists of names of animals. Run through the map and generate a
* "denormalized" list of strings describing the animal, with the animal's name separated
* by a colon from the number of legs it has. The ordering in the output list is not
* considered significant.
*
* Input is Map<Integer, List<String>>:
* { 4=["ibex", "hedgehog", "wombat"],
* 6=["ant", "beetle", "cricket"],
* ...
* }
*
* Output should be a List<String>:
* [ "ibex:4",
* "hedgehog:4",
* "wombat:4",
* "ant:6",
* "beetle:6",
* "cricket:6",
* ...
* ]
public void ex26_denormalizeMap() {
Map<Integer, List<String>> input = new HashMap<>();
input.put(4, Arrays.asList("ibex", "hedgehog", "wombat"));
input.put(6, Arrays.asList("ant", "beetle", "cricket"));
input.put(8, Arrays.asList("octopus", "spider", "squid"));
input.put(10, Arrays.asList("crab", "lobster", "scorpion"));
input.put(750, Arrays.asList("millipede"));
List<String> result = null; // TODO
assertTrue(result.contains("ibex:4"));
assertTrue(result.contains("hedgehog:4"));
assertTrue(result.contains("wombat:4"));
assertTrue(result.contains("ant:6"));
assertTrue(result.contains("beetle:6"));
assertTrue(result.contains("cricket:6"));
assertTrue(result.contains("octopus:8"));
assertTrue(result.contains("spider:8"));
assertTrue(result.contains("squid:8"));
assertTrue(result.contains("crab:10"));
assertTrue(result.contains("lobster:10"));
assertTrue(result.contains("scorpion:10"));
assertTrue(result.contains("millipede:750"));
}
// Hint 1:
// <editor-fold defaultstate="collapsed">
// There are several ways to approach this. You could use a stream of map keys,
// a stream of map entries, or nested forEach() methods.
// </editor-fold>
// Hint 2:
// <editor-fold defaultstate="collapsed">
// If you use streams, consider using Stream.flatMap().
// </editor-fold>
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// ========================================================
// NEW FOR 2016
// ========================================================
/**
* Invert a "multi-map". (P. Sandoz)
*
* Given a Map<X, Set<Y>>, convert it to Map<Y, Set<X>>.
*/
@Test @Ignore
public void ex27_invertMultiMap() {
Map<String, List<Integer>> input = new HashMap<>();
input.put("a", Arrays.asList(1, 2));
input.put("b", Arrays.asList(2, 3));
input.put("c", Arrays.asList(1, 3));
input.put("d", Arrays.asList(1, 4));
input.put("e", Arrays.asList(2, 4));
input.put("f", Arrays.asList(3, 4));
Map<Integer, List<String>> result = null; // TODO
assertEquals(new HashSet<>(Arrays.asList("a", "c", "d")), result.get(1));
assertEquals(new HashSet<>(Arrays.asList("a", "b", "e")), result.get(2));
assertEquals(new HashSet<>(Arrays.asList("b", "c", "f")), result.get(3));
assertEquals(new HashSet<>(Arrays.asList("d", "e", "f")), result.get(4));
assertEquals(4, result.size());
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}
/**
* Select from the input list each word that longer than the immediately
* preceding word. Include the first word, since it is longer than the
* (nonexistent) word that precedes it.
*
* XXX - compare to ex11
*/
@Test @Ignore
public void ex28_selectLongerThanPreceding() {
List<String> input = Arrays.asList(
"alfa", "bravo", "charlie", "delta", "echo", "foxtrot", "golf", "hotel");
List<String> result = null; // TODO
assertEquals("[alfa, bravo, charlie, foxtrot, hotel]", result.toString());
}
// Hint:
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// <editor-fold defaultstate="collapsed">
// Instead of a stream of words (Strings), run an IntStream of positions.
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// </editor-fold>
/**
* Generate a list of words that is the concatenation of each adjacent
* pair of words in the input list. For example, if the input is
* [x, y, z]
* the output should be
* [xy, yz]
*
* XXX - compare to ex11
*/
@Test @Ignore
public void ex29_concatenateAdjacent() {
List<String> input = Arrays.asList(
"alfa", "bravo", "charlie", "delta", "echo", "foxtrot");
List<String> result = null; // TODO
assertEquals("[alfabravo, bravocharlie, charliedelta, deltaecho, echofoxtrot]",
result.toString());
}
// Hint:
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// <editor-fold defaultstate="collapsed">
// Instead of a stream of words (Strings), run an IntStream of positions.
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// </editor-fold>
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/**
* Select the longest words from the input list. That is, select the words
* whose lengths are equal to the maximum word length. For this exercise,
* it's easiest to perform two passes over the input list.
*
* XXX - compare to ex09 and ex10
*/
@Test @Ignore
public void ex30_selectLongestWords() {
List<String> input = Arrays.asList(
"alfa", "bravo", "charlie", "delta", "echo", "foxtrot", "golf", "hotel");
List<String> result = null; // TODO
assertEquals("[charlie, foxtrot]", result.toString());
}
/**
* Select the longest words from the input stream. That is, select the words
* whose lengths are equal to the maximum word length. For this exercise,
* you must compute the result in a single pass over the input stream.
*
* XXX - compare to ex30
*/
@Test @Ignore
public void ex31_selectLongestWordsOnePass() {
Stream<String> input = Stream.of(
"alfa", "bravo", "charlie", "delta", "echo", "foxtrot", "golf", "hotel");
List<String> result = null; // TODO
assertEquals("[charlie, foxtrot]", result.toString());
}
/**
* Create a list of non-overlapping sublists of the input list, where each
* sublist (except for the first one) starts with a word whose first character is a ":".
* For example, given the input list
* [w, x, :y, z]
* the output should be
* [[w, x], [:y, z]]
*/
@Test @Ignore
public void ex32_partitionIntoSublists() {
List<String> input = Arrays.asList(
"alfa", "bravo", ":charlie", "delta", ":echo", ":foxtrot", "golf", "hotel");
List<List<String>> result = null; // TODO
assertEquals("[[alfa, bravo], [:charlie, delta], [:echo], [:foxtrot, golf, hotel]]",
result.toString());
}
/**
* Given a stream of integers, compute separate sums of the even and odd values
* in this stream. Since the input is a stream, this necessitates making a single
* pass over the input.
*/
@Test @Ignore
public void ex33_separateOddEvenSums() {
IntStream input = new Random(987523).ints(20, 0, 100);
int sumEvens = 0; // TODO
int sumOdds = 0; // TODO
assertEquals(516, sumEvens);
assertEquals(614, sumOdds);
}
// Hint:
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// <editor-fold defaultstate="collapsed">
// Use Collectors.partitioningBy().
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// </editor-fold>
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/**
* Given a string, split it into a list of strings consisting of
* consecutive characters from the original string. Note: this is
* similar to Python's itertools.groupby function, but it differs
* from Java's Collectors.groupingBy() collector.
*
* XXX - compare to ex32
*/
@Test @Ignore
public void ex34_splitCharacterRuns() {
String input = "aaaaabbccccdeeeeeeaaafff";
List<String> result = null; // TODO
assertEquals("[aaaaa, bb, cccc, d, eeeeee, aaa, fff]", result.toString());
}
/**
* Given a string, find the substring containing the longest run of consecutive,
* equal characters.
*
* XXX - compare to ex34
*/
@Test @Ignore
public void ex35_longestCharacterRuns() {
String input = "aaaaabbccccdeeeeeeaaafff";
String result = null; // TODO
assertEquals("eeeeee", result);
}
/**
* Given a parallel stream of strings, collect them into a collection in reverse order.
* Since the stream is parallel, you MUST write a proper combiner function in order to get
* the correct result.
*/
@Test @Ignore
public void ex36_reversingCollector() {
Stream<String> input =
IntStream.range(0, 100).mapToObj(String::valueOf).parallel();
Collection<String> result =
input.collect(Collector.of(null, null, null));
// TODO fill in collector functions above
assertEquals(
IntStream.range(0, 100)
.map(i -> 99 - i)
.mapToObj(String::valueOf)
.collect(Collectors.toList()),
new ArrayList<>(result));
}
/**
* Given an array of int, find the int value that occurs a majority
* of times in the array (that is, strictly more than half of the
* elements are that value), and return it in an OptionalInt. If there
* is no majority value, return an empty OptionalInt.
*/
OptionalInt majority(int[] array) {
return null; // TODO
}
@Test @Ignore
public void ex37_majority() {
int[] array1 = { 3, 3, 4, 2, 4, 4, 2, 4, 4 };
int[] array2 = { 3, 3, 4, 2, 4, 4, 2, 4 };
OptionalInt result1 = majority(array1);
OptionalInt result2 = majority(array2);
assertEquals(OptionalInt.of(4), result1);
assertFalse(result2.isPresent());
}