Merge pull request #298 from sir-gon/feature/count_triplets

Feature/count triplets

Author: sir-gon

algorithm-exercises-java/src/main/java/ae/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/CountTriplets.java

@@ -8,9 +8,10 @@
  * CountTriplets.
  *
  * @link Problem definition
- *       [[docs/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/two-strings.md]]
- * @link Solution notes
- *       [[docs/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/two-strings-solution-notes.md]]
+ *       [[docs/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/count_triplets_1.md/]]
+ *
+ * @link Solution Notes
+ *       [[docs/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/count_triplets_1-solution-notes.md/]]
  */
 public class CountTriplets {
   private CountTriplets() {

algorithm-exercises-java/src/main/java/ae/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/CountTripletsBruteForce.java

@@ -0,0 +1,38 @@
+package ae.hackerrank.interview_preparation_kit.dictionaries_and_hashmaps;
+
+import java.util.List;
+
+/**
+ * CountTripletsBruteForce.
+ *
+ * @link Problem definition
+ *       [[docs/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/count_triplets_1.md/]]
+ *
+ * @see CountTriplets
+ */
+public class CountTripletsBruteForce {
+  private CountTripletsBruteForce() {
+  }
+
+  /**
+   * CountTriples.
+   */
+  static long countTriplets(List<Long> arr, long r) {
+
+    long size = arr.size();
+    long counter = 0L;
+
+    for (int i = 0; i < size - 2; i++) {
+      for (int j = i + 1; j < size - 1; j++) {
+        for (int k = j + 1; k < size; k++) {
+
+          if (r * arr.get(i) == arr.get(j) && r * arr.get(j) == arr.get(k)) {
+            counter += 1;
+          }
+        }
+      }
+    }
+
+    return counter;
+  }
+}

algorithm-exercises-java/src/test/java/ae/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/CountTripletsBruteForceTest.java

@@ -0,0 +1,52 @@
+package ae.hackerrank.interview_preparation_kit.dictionaries_and_hashmaps;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+
+import java.io.IOException;
+import java.util.List;
+import org.junit.jupiter.api.BeforeAll;
+import org.junit.jupiter.api.Test;
+import org.junit.jupiter.api.TestInstance;
+import org.junit.jupiter.api.TestInstance.Lifecycle;
+import util.JsonLoader;
+
+/**
+ * CountTripletsBruteForceTest.
+ */
+@TestInstance(Lifecycle.PER_CLASS)
+class CountTripletsBruteForceTest {
+  public static class CountTripletsBruteForceTestCase {
+    public String title;
+    public List<Long> input;
+    public Long r;
+    public Long expected;
+  }
+
+  private List<CountTripletsBruteForceTestCase> smallTestCases;
+
+  @BeforeAll
+  void setup() throws IOException {
+    String path;
+    path = String.join("/",
+        "hackerrank",
+        "interview_preparation_kit",
+        "dictionaries_and_hashmaps",
+        "count_triplets_1.small.testcases.json");
+
+    this.smallTestCases = JsonLoader.loadJson(path, CountTripletsBruteForceTestCase.class);
+  }
+
+  @Test
+  void testCountTriplets() {
+    for (CountTripletsBruteForceTestCase test : smallTestCases) {
+      Long solutionFound = CountTripletsBruteForce.countTriplets(test.input, test.r);
+
+      assertEquals(test.expected, solutionFound,
+          "%s(%s, %d) answer must be: %d".formatted(
+              "CountTripletsBruteForce.countTriplets",
+              test.input.toString(),
+              test.r,
+              test.expected));
+    }
+  }
+}

docs/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/count_triplets_1-solution-notes.md

@@ -0,0 +1,13 @@
+# [Dictionaries and Hashmaps: Count Triplets](https://www.hackerrank.com/challenges/count-triplets-1)
+
+- Difficulty:  `#medium`
+- Category: `#ProblemSolvingIntermediate`
+
+## Working solution
+
+This solution in O(N), is based on considering that each
+number analyzed is in the center of the triplet and asks
+"how many" possible cases there are on the left and
+right to calculate how many possible triplets are formed.
+
+- Source: [Hackerrank - Count Triplets Solution](https://www.thepoorcoder.com/hackerrank-count-triplets-solution/)

docs/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/count_triplets_1.md

@@ -0,0 +1,99 @@
+# [Dictionaries and Hashmaps: Count Triplets](https://www.hackerrank.com/challenges/count-triplets-1)
+
+- Difficulty:  `#medium`
+- Category: `#ProblemSolvingIntermediate`
+
+You are given an array and you need to find number of
+tripets of indices (i, j, k) such that the elements at
+those indices are in geometric progression for a given
+common ratio `r` and $ i < j < k $.
+
+## Example
+
+`arr = [1, 4, 16, 64]   r = 4`
+
+There are `[1, 4, 16]` and `[4, 16, 64]` at indices  (0, 1, 2) and (1, 2, 3).
+Return `2`.
+
+## Function Description
+
+Complete the countTriplets function in the editor below.
+
+countTriplets has the following parameter(s):
+
+- `int arr[n]`: an array of integers
+- `int r`: the common ratio
+
+## Returns
+
+- `int`: the number of triplets
+
+## Input Format
+
+The first line contains two space-separated integers `n` and `r`,
+the size of `arr` and the common ratio.
+The next line contains `n` space-seperated integers `arr[i]`.
+
+## Constraints
+
+- $ 1 \leq n \leq 10^5 $
+- $ 1 \leq r \leq 10^9 $
+- $ 1 \leq arr[i] \leq 10^9 $
+
+## Sample Input 0
+
+```text
+4 2
+1 2 2 4
+```
+
+## Sample Output 0
+
+```text
+2
+```
+
+## Explanation 0
+
+There are `2` triplets in satisfying our criteria,
+whose indices are (0, 1, 3) and (0, 2, 3)
+
+## Sample Input 1
+
+```text
+6 3
+1 3 9 9 27 81
+```
+
+## Sample Output 1
+
+```text
+6
+```
+
+## Explanation 1
+
+The triplets satisfying are index
+`(0, 1, 2)`, `(0, 1, 3)`, `(1, 2, 4)`, `(1, 3, 4)`, `(2, 4, 5)` and `(3, 4, 5)`.
+
+## Sample Input 2
+
+```text
+5 5
+1 5 5 25 125
+```
+
+## Sample Output 2
+
+```text
+4
+```
+
+## Explanation 2
+
+The triplets satisfying are index
+`(0, 1, 3)`, `(0, 2, 3)`, `(1, 2, 3)`, `(2, 3, 4)`.
+
+## Appendix
+
+[Solution notes](count_triplets_1-solution-notes.md)