[Hacker Rank] Interview Preparation Kit: Greedy Algorithms: Luck Balance. Solved ✅.

Author: Gonzalo Diaz

docs/hackerrank/interview_preparation_kit/greedy_algorithms/luck-balance.md

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+# [Greedy Algorithms: Luck Balance](https://www.hackerrank.com/challenges/luck-balance)
+
+- Difficulty:  `#easy`
+- Category: `#ProblemSolvingBasic`
+
+Lena is preparing for an important coding competition that is preceded
+by a number of sequential preliminary contests.
+Initially, her luck balance is 0.
+She believes in "saving luck", and wants to check her theory.
+Each contest is described by two integers, `L[i]` and `T[i]`:
+
+- `L[i]` is the amount of luck associated with a contest.
+If Lena wins the contest, her luck balance will decrease by `L[i]`;
+if she loses it, her luck balance will increase by `L[i]`.
+
+- `T[i]` denotes the contest's importance rating.
+It's equal to `1` if the contest is important, and it's equal to `0` if it's unimportant.
+
+If Lena loses no more than `k` important contests, what is the maximum amount
+of luck she can have after competing in all the preliminary contests?
+This value may be negative.
+
+## Example
+
+```text
+Contest     L[i]    T[i]
+1           5       1
+2           1       1
+3           4       0
+```
+
+If Lena loses all of the contests, her will be `5 + 1 +4 = 10`.
+Since she is allowed to lose  important contests,
+and there are only `2` important contests,
+she can lose all three contests to maximize her luck at `10`.
+
+If `k = 1`, she has to win at least  of the  important contests.
+She would choose to win the lowest value important contest worth `1`.
+Her final luck will be `5 + 4 - 1 = 8`.
+
+## Function Description
+
+Complete the luckBalance function in the editor below.
+
+luckBalance has the following parameter(s):
+
+- `int k`: the number of important contests Lena can lose
+- `int contests[n][2]`: a 2D array of integers where each `contests[i]`
+contains two integers that represent the luck balance and importance of the  contest
+
+## Returns
+
+- `int`: the maximum luck balance achievable
+
+## Input Format
+
+The first line contains two space-separated integers  `n` and `k`,
+the number of preliminary contests and the maximum number
+of important contests Lena can lose.
+
+Each of the next  lines contains two space-separated integers,
+`L[i]` and `T[i]`, the contest's luck balance and its importance rating.
+
+## Constraints
+
+- $ 1 \leq n \leq 100 $
+- $ 0 \leq k \leq N $
+- $ 1 \leq L[i] \leq 10^4 $
+- $ T[i] \isin \{0,1\} $
+
+## Sample Input
+
+```text
+STDIN       Function
+-----       --------
+6 3         n = 6, k = 3
+5 1         contests = [[5, 1], [2, 1], [1, 1], [8, 1], [10, 0], [5, 0]]
+2 1
+1 1
+8 1
+10 0
+5 0
+```
+
+## Sample Output
+
+```text
+29
+```
+
+## Explanation
+
+There are `n = 6` contests. Of these contests, `4` are important
+and she cannot lose more than  of them.
+Lena maximizes her luck if she wins the $ 3^{rd} $ important contest
+(where `L[i] = 1`) and loses all of the other five contests for a total
+luck balance of `5 + 2 + 8 + 10 + 5 - 1 = 29`.

src/hackerrank/interview_preparation_kit/greedy_algorithms/luck-balance.js

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+/**
+ * @link Problem definition [[docs/hackerrank/interview_preparation_kit/greedy_algorithms/luck-balance.md]]
+ */
+
+function dynamicSort(array, sortBy) {
+  const initialValue = 0;
+  return array.sort((a, b) =>
+    sortBy.reduce((result, current) => {
+      const { property, order = 'asc' } = current;
+
+      let comparison = 0;
+      if (a[property] > b[property]) {
+        comparison = 1;
+      } else if (a[property] < b[property]) {
+        comparison = -1;
+      }
+
+      if (order === 'desc') {
+        comparison *= -1;
+      }
+
+      return comparison;
+    }, initialValue)
+  );
+}
+
+class Contest {
+  constructor(luck, important) {
+    this.luck = luck;
+    this.important = important;
+  }
+}
+
+export function luckBalance(k, contests) {
+  // Write your code here
+  let importantContests = [];
+  const nonimportantContests = [];
+
+  contests.forEach((contest) => {
+    const [luck, important] = [...contest];
+    if (important === 1) {
+      importantContests.push(new Contest(luck, important));
+    } else {
+      nonimportantContests.push(new Contest(luck, important));
+    }
+  });
+
+  importantContests = dynamicSort(importantContests, [
+    { property: 'important', order: 'desc' },
+    { property: 'luck', order: 'desc' }
+  ]);
+
+  let total = 0;
+  const size = importantContests.length;
+  const cut = Math.min(k, size);
+
+  for (let i = 0; i < cut; i++) {
+    total += importantContests[i].luck;
+  }
+
+  for (let i = cut; i < size; i++) {
+    total -= importantContests[i].luck;
+  }
+
+  nonimportantContests.forEach((contest) => {
+    total += contest.luck;
+  });
+
+  return total;
+}
+
+export default { luckBalance };

src/hackerrank/interview_preparation_kit/greedy_algorithms/luck-balance.test.js

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+import { describe, expect, it } from '@jest/globals';
+import { logger as console } from '../../../logger.js';
+
+import { luckBalance } from './luck-balance.js';
+
+const TEST_CASES = [
+  {
+    'title': 'Sample Test case 0',
+    'k': 3,
+    'contests': [
+      [5, 1],
+      [2, 1],
+      [1, 1],
+      [8, 1],
+      [10, 0],
+      [5, 0]
+    ],
+    'expected': 29
+  },
+  {
+    'title': 'Sample Test case 1',
+    'k': 5,
+    'contests': [
+      [13, 1],
+      [10, 1],
+      [9, 1],
+      [8, 1],
+      [13, 1],
+      [12, 1],
+      [18, 1],
+      [13, 1]
+    ],
+    'expected': 42
+  },
+  {
+    'title': 'Sample Test case 1',
+    'k': 2,
+    'contests': [
+      [5, 1],
+      [4, 0],
+      [6, 1],
+      [2, 1],
+      [8, 0]
+    ],
+    'expected': 21
+  }
+];
+
+describe('luck-balance', () => {
+  it('luckBalance test cases', () => {
+    expect.assertions(3);
+
+    TEST_CASES.forEach((test) => {
+      const answer = luckBalance(test.k, test.contests);
+
+      console.debug(
+        `luckBalance(${test.k}, ${test.contests}) solution found: ${answer}`
+      );
+
+      expect(answer).toStrictEqual(test.expected);
+    });
+  });
+});