[Hacker Rank] Interview Preparation Kit: String Manipulation: Making …

…Anagrams. Solved ✅.

docs/hackerrank/interview_preparation_kit/string_manipulation/ctci-making-anagrams.md

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+# [Strings: Making Anagrams](https://www.hackerrank.com/challenges/ctci-making-anagrams)
+
+How many characters should one delete to make two given strings anagrams
+of each other?
+
+- Difficulty:  `#easy`
+- Category: `#ProblemSolvingBasic` `#strings`
+
+A student is taking a cryptography class and has found anagrams to be very useful.
+Two strings are anagrams of each other if the first
+string's letters can be rearranged to form the second string.
+In other words, both strings must contain the same
+exact letters in the same exact frequency.
+For example, `bacdc` and `dcbac` are anagrams,
+but `bacdc` and `dcbad` are not.
+
+The student decides on an encryption scheme that involves two large strings.
+The encryption is dependent on the minimum number of character deletions
+required to make the two strings anagrams. Determine this number.
+
+Given two strings, `a` and `b`, that may or may not be of the same length,
+determine the minimum number of character deletions required
+to make `a` and `b` anagrams.
+Any characters can be deleted from either of the strings.
+
+## Example
+
+```python
+a = 'cde'
+b = 'dcf'
+```
+
+Delete `e` from `a` and `f` from `b` so that the remaining strings
+are `cd` and `dc` which are anagrams. This takes `2`  character deletions.
+
+## Function Description
+
+Complete the makeAnagram function in the editor below.
+
+makeAnagram has the following parameter(s):
+
+- `string a`: a string
+- `string b`: another string
+
+## Returns
+
+- `int`: the minimum total characters that must be deleted
+
+## Input Format
+
+The first line contains a single string, `a`.
+The second line contains a single string, `b`.
+
+## Constraints
+
+- $ 1 \leq |a|, |b| \leq 10^4 $
+- The strings `a` and `b` consist of lowercase English alphabetic letters, ascii[a-z].
+
+## Sample Input
+
+```text
+cde
+abc
+```
+
+## Sample Output
+
+```text
+4
+```
+
+## Explanation
+
+Delete the following characters from the strings make them anagrams:
+
+1. Remove `d` and `e` from `cde` to get `c`.
+2. Remove `a` and `b` from `abc` to get `c`.
+
+It takes `4` deletions to make both strings anagrams.

src/hackerrank/interview_preparation_kit/string_manipulation/ctci_making_anagrams.test.ts

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+import { describe, expect, it } from '@jest/globals';
+
+import { makeAnagram } from './ctci_making_anagrams';
+import TEST_CASES from './ctci_making_anagrams.testcases.json';
+
+describe('makeAnagram', () => {
+  it('makeAnagram test cases', () => {
+    expect.assertions(3);
+
+    TEST_CASES.forEach((test) => {
+      const result = makeAnagram(test.a, test.b);
+
+      expect(result).toStrictEqual(test.expected);
+    });
+  });
+});

src/hackerrank/interview_preparation_kit/string_manipulation/ctci_making_anagrams.testcases.json

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+[
+  {
+    "title": "Sample Test case 0",
+    "a": "cde",
+    "b": "abc",
+    "expected": 4
+  },
+  {
+    "title": "Sample Test case 1",
+    "a": "fcrxzwscanmligyxyvym",
+    "b": "jxwtrhvujlmrpdoqbisbwhmgpmeoke",
+    "expected": 30
+  },
+  {
+    "title": "Sample Test case 2",
+    "a": "showman",
+    "b": "woman",
+    "expected": 2
+  }
+]

src/hackerrank/interview_preparation_kit/string_manipulation/ctci_making_anagrams.ts

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+/**
+ * @link Problem definition [[docs/hackerrank/interview_preparation_kit/string_manipulation/m]]
+ */
+
+function charToDicMap(word: string): Record<string, number> {
+  const output: Record<string, number> = {};
+
+  for (const letter of word.split('')) {
+    output[letter] = output?.[letter] ? output[letter] + 1 : 1;
+  }
+
+  return output;
+}
+
+function sum(values: number[]): number {
+  return values.reduce(
+    (previousValue: number, currentValue: number) =>
+      previousValue + currentValue,
+    0
+  );
+}
+
+export function makeAnagram(a: string, b: string): number {
+  const aMap = charToDicMap(a);
+  const bMap = charToDicMap(b);
+
+  for (const key of Object.keys(aMap)) {
+    if (bMap?.[key]) {
+      aMap[key] = Math.abs(aMap[key] - bMap[key]);
+      bMap[key] = 0;
+    }
+  }
+
+  return sum(Object.values(aMap)) + sum(Object.values(bMap));
+}
+
+export default { makeAnagram };