Merge pull request #472 from sir-gon/feature/count-triplets-1

[Hacker Rank] Interview Preparation Kit: Dictionaries and Hashmaps: C…

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

@@ -0,0 +1,113 @@
+# [Dictionaries and Hashmaps: Count Triplets](https://www.hackerrank.com/challenges/count-triplets-1)
+
+- Difficulty:  `#medium`
+- Category: `#ProblemSolvingIntermediate`
+
+## Failed tries
+
+### Introducing optimizations and cache
+
+This solution is based on the idea of traversing each possible case,
+similar to brute force, but introducing a cache mechanism
+for each case found and cutting the number of times it would have to be traversed.
+
+This has following problems:
+
+1) The problem is that it preserves the complexity of O(n^3) (same as brute force)
+
+2) "Cognitive Complexity" (measured by sonarlint) it is very high.
+This due to the number of nesting.
+
+3) Initially, the innermost calculation assumes that the values in the array
+"should" be ordered, that is, it expects that the values to the right of the
+currently evaluated value would be larger.
+With this idea in mind, an initial ordering of values was introduced.
+But the problem requires that the values be in positions i < j < k.
+That is why initially an ordering of the input was executed,
+but such ordering introduces unexpected possibilities. In an unordered list,
+it can produce overcounts of favorable cases.
+
+```python
+def count_triplets_with_cache_and_cuts(arr: list[int], r: int) -> int:
+
+    # arrc = arr[:]
+    # arrc = sorted(arrc)
+    size = len(arr)
+
+    cache: dict[tuple[int, int, int], bool] = {}
+    counter = 0
+
+    i_resume = True
+    i = 0
+    while i < size - 2 and i_resume:
+        j_resume = True
+        j = i + 1
+        while j < size - 1 and j_resume:
+            if arr[j] > r * arr[i]:
+                j_resume = False
+
+            k_resume = True
+            k = j + 1
+            while k < size and k_resume:
+                if arr[k] > r * arr[j]:
+                    k_resume = False
+
+                triplet = (arr[i], arr[j], arr[k])
+
+                if triplet in cache and cache[triplet]:
+                    counter += 1
+                else:
+                    if r * arr[i] == arr[j] and r * arr[j] == arr[k]:
+                        cache[triplet] = True
+                        counter += 1
+                    else:
+                        cache[triplet] = False
+
+                k += 1
+            j += 1
+        i += 1
+
+    return counter
+```
+
+### Reducing complexity
+
+This solutions reduce complexity to O(N), but has the same problem
+with unordered lists as the previous case.
+
+```python
+def count_triplets_doenst_work_with_unsorted_input(arr: list[int], r: int) -> int:
+    count: int = 0
+
+    if len(arr) < 3:
+        return count
+
+    counter = Counter(arr)
+    size = len(counter)
+    limit = size - 2
+
+    i = 0
+    for k, v in counter.items():
+        knext = k * r
+        knext_next = k * r * r
+
+        if i < limit and knext in counter and knext_next in counter:
+            next_elem_cnt = counter[knext]
+            next_next_elem_cnt = counter[knext_next]
+            count += v * (next_elem_cnt * next_next_elem_cnt)
+        elif r == 1:
+            count += math.factorial(v) // (math.factorial(3) * math.factorial(v - 3))
+
+        i += 1
+
+    return count
+```
+
+## 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)

src/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/count_triplets_1_bruteforce.js

@@ -0,0 +1,26 @@
+/**
+ * @link Problem definition [[docs/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/count_triplets_1.md]]
+ * @see Solution Notes: [[docs/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/count_triplets_1-solution-notes.md]]
+ */
+import { logger as console } from '../../../logger.js';
+
+export function countTriplets(arr, ratio) {
+  const size = arr.length;
+  let counter = 0;
+
+  for (let i = 0; i < size - 2; i++) {
+    for (let j = i + 1; j < size - 1; j++) {
+      for (let k = j + 1; k < size; k++) {
+        console.debug(`${arr[i]}, ${arr[j]}, ${arr[k]}`);
+
+        if (ratio * arr[i] === arr[j] && ratio * arr[j] === arr[k]) {
+          counter += 1;
+        }
+      }
+    }
+  }
+
+  return counter;
+}
+
+export default { countTriplets };

src/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/count_triplets_1_bruteforce.test.js

@@ -0,0 +1,47 @@
+import { describe, expect, it } from '@jest/globals';
+import { logger as console } from '../../../logger.js';
+
+import { countTriplets } from './count_triplets_1_bruteforce.js';
+
+const SMALL_TEST_CASES = [
+  {
+    title: 'Sample Test Case 0',
+    input: [1, 2, 2, 4],
+    r: 2,
+    expected: 2
+  },
+  {
+    title: 'Sample Test Case 1',
+    input: [1, 3, 9, 9, 27, 81],
+    r: 3,
+    expected: 6
+  },
+  {
+    title: 'Sample Test Case 1 (unsorted)',
+    input: [9, 3, 1, 81, 9, 27],
+    r: 3,
+    expected: 1
+  },
+  {
+    title: 'Sample Test Case 12',
+    input: [1, 5, 5, 25, 125],
+    r: 5,
+    expected: 4
+  }
+];
+
+describe('count_triplets_1', () => {
+  it('countTriplets test cases', () => {
+    expect.assertions(4);
+
+    SMALL_TEST_CASES.forEach((test) => {
+      const answer = countTriplets(test.input, test.r);
+
+      console.debug(
+        `countTriplets(${test.input}, ${test.r}) solution found: ${answer}`
+      );
+
+      expect(answer).toStrictEqual(test.expected);
+    });
+  });
+});

src/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/count_triplets_1_optimized.test.js

@@ -0,0 +1,49 @@
+import { describe, expect, it } from '@jest/globals';
+import { logger as console } from '../../../logger.js';
+
+import { countTriplets } from './count_triplets_1_optmized.js';
+import SMALL_TEST_CASES from './count_triplets_1_testcases.json';
+
+const BIG_TEST_CASES = [
+  {
+    title: 'Sample Test Case 2',
+    input: [
+      1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
+      1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
+      1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
+      1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
+    ],
+    r: 1,
+    expected: 161700
+  }
+];
+
+describe('count_triplets_1 (optimized)', () => {
+  it('countTriplets small test cases', () => {
+    expect.assertions(4);
+
+    SMALL_TEST_CASES.forEach((test) => {
+      const answer = countTriplets(test.input, test.r);
+
+      console.debug(
+        `countTriplets(${test.input}, ${test.r}) solution found: ${answer}`
+      );
+
+      expect(answer).toStrictEqual(test.expected);
+    });
+  });
+
+  it('countTriplets big test cases', () => {
+    expect.assertions(1);
+
+    BIG_TEST_CASES.forEach((test) => {
+      const answer = countTriplets(test.input, test.r);
+
+      console.debug(
+        `countTriplets(${test.input}, ${test.r}) solution found: ${answer}`
+      );
+
+      expect(answer).toStrictEqual(test.expected);
+    });
+  });
+});

src/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/count_triplets_1_optmized.js

@@ -0,0 +1,38 @@
+/**
+ * @link Problem definition [[docs/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/count_triplets_1.md]]
+ * @see Solution Notes: [[docs/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/count_triplets_1-solution-notes.md]]
+ */
+
+export function countTriplets(arr, ratio) {
+  let triplets = 0;
+
+  const aCounter = arr.reduce((accumulator, entry) => {
+    if (entry in accumulator) {
+      accumulator[entry] += 1;
+    } else {
+      accumulator[entry] = 1;
+    }
+    return accumulator;
+  }, {});
+
+  const bCounter = {};
+
+  arr.forEach((x) => {
+    const j = Math.floor(x / ratio);
+    const k = x * ratio;
+    aCounter[x] -= 1;
+    if (bCounter[j] && aCounter[k] && x % ratio === 0) {
+      triplets += bCounter[j] * aCounter[k];
+    }
+
+    if (x in bCounter) {
+      bCounter[x] += 1;
+    } else {
+      bCounter[x] = 1;
+    }
+  });
+
+  return triplets;
+}
+
+export default { countTriplets };