problem: new problem solution - 1660 . Find Words That Can Be Formed …

…by Characters

README.md

@@ -116,6 +116,7 @@ My approach to solving LeetCode problems typically involves the following steps:
 | 563  | [Binary Tree Tilt](https://leetcode.com/problems/binary-tree-tilt/)                                                                               | Algorithms | [TypeScript](./problems/algorithms/binaryTreeTilt/BinaryTreeTilt.ts)                                                                     | Easy       |
 | 572  | [Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)                                                                 | Algorithms | [TypeScript](./problems/algorithms/subtreeOfAnotherTree/SubtreeOfAnotherTree.ts)                                                         | Easy       |
 | 1662 | [Check If Two String Arrays are Equivalent](https://leetcode.com/problems/check-if-two-string-arrays-are-equivalent/)                             | Algorithms | [TypeScript](./problems/algorithms/checkIfTwoStringArraysAreEquivalent/CheckIfTwoStringArraysAreEquivalent.ts)                           | Easy       |
+| 1160 | [Find Words That Can Be Formed by Characters](https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/)                         | Algorithms | [TypeScript](./problems/algorithms/findWordsThatCanBeFormedByCharacters/FindWordsThatCanBeFormedByCharacters.ts)                         | Easy       |
 | ...  | ...                                                                                                                                               | ...        | ...                                                                                                                                      | ...        |
 
 In this table:

problems/algorithms/findWordsThatCanBeFormedByCharacters/FindWordsThatCanBeFormedByCharacters.test.ts

@@ -0,0 +1,23 @@
+// Source : https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/
+// Author : Francisco Tomas
+// Date   : 2023-12-02
+
+import { countCharacters } from "./FindWordsThatCanBeFormedByCharacters";
+
+describe("find words tha can be formed by characters", () => {
+  test("example 1", () => {
+    const result: number = countCharacters(
+      ["cat", "bt", "hat", "tree"],
+      "atach",
+    );
+    expect(result).toBe(6);
+  });
+
+  test("example 2", () => {
+    const result: number = countCharacters(
+      ["hello", "world", "leetcode"],
+      "welldonehoneyr",
+    );
+    expect(result).toBe(10);
+  });
+});

problems/algorithms/findWordsThatCanBeFormedByCharacters/FindWordsThatCanBeFormedByCharacters.ts

@@ -0,0 +1,123 @@
+// Source : https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/
+// Author : Francisco Tomas
+// Date   : 2023-12-02
+
+/*****************************************************************************************************
+ *
+ * You are given an array of strings words and a string chars.
+ *
+ * A string is good if it can be formed by characters from chars (each character can only be used
+ * once).
+ *
+ * Return the sum of lengths of all good strings in words.
+ *
+ * Example 1:
+ *
+ * Input: words = ["cat","bt","hat","tree"], chars = "atach"
+ * Output: 6
+ * Explanation: The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6.
+ *
+ * Example 2:
+ *
+ * Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr"
+ * Output: 10
+ * Explanation: The strings that can be formed are "hello" and "world" so the answer is 5 + 5 = 10.
+ *
+ * Constraints:
+ *
+ * 	1 <= words.length <= 1000
+ * 	1 <= words[i].length, chars.length <= 100
+ * 	words[i] and chars consist of lowercase English letters.
+ ******************************************************************************************************/
+
+/**
+ * string[] string -> number
+ * given an array of strings, words, and a string, chars, return the sum of lengths of all good strins in words
+ * NOTE: a string is good if it can be formed by characters from chars (each character can only be used once)
+
+ * Stub:
+ function countCharacters(words: string[], chars: string): number {return 0}
+
+ * Tests:
+ * I: words = ["cat","bt","hat","tree"], chars = "atach" -> O: 6
+ * I: words = ["hello","world","leetcode"], chars = "welldonehoneyr" -> O: 10
+
+ * Constraints:
+ * - 1 <= words.length <= 1000
+ * - 1 <= words[i].length, chars.length <= 100
+ * - words[i] and chars consist of lowercase English letters.
+ */
+
+/**
+ * Time Complexity: O(n + m * k), where n is the length of chars, m is the length of words, and k is the average length of each word in words
+ * Space Complexity: O(1), charsFrequency and wordFrequency size never exceeds 26
+
+ * Runtime: 178ms (13.89%)
+ * Memory: 50.92MB (80.56%)
+ */
+export function countCharactersV1(words: string[], chars: string): number {
+  const charsFrequency: Record<string, number> = {};
+  let ans: number = 0;
+
+  for (const char of chars) {
+    // eslint-disable-next-line @typescript-eslint/strict-boolean-expressions
+    charsFrequency[char] = (charsFrequency[char] ?? 0) + 1;
+  }
+
+  for (const word of words) {
+    const wordFrequency = { ...charsFrequency };
+    let indicator: boolean = true;
+
+    for (const char of word) {
+      if (wordFrequency[char] === undefined) {
+        indicator = false;
+        break;
+      }
+      wordFrequency[char] -= 1;
+    }
+
+    if (indicator) {
+      ans += word.length;
+    }
+  }
+
+  return ans;
+}
+
+/**
+ * Time Complexity: O(n + m * k), where n is the length of chars, m is the length of words, and k is the average length of each word in words
+ * Space Complexity: O(1), charsFrequency and c both have a fixed length of 26
+
+ * Runtime: 71ms (94.44%)
+ * Memory: 52.20MB (22.22%)
+ */
+export function countCharacters(words: string[], chars: string): number {
+  const charsFrequency: number[] = new Array(26).fill(0);
+  for (const char of chars) {
+    charsFrequency[char.charCodeAt(0) - "a".charCodeAt(0)]++;
+  }
+
+  let ans: number = 0;
+  for (const word of words) {
+    if (canFormWord(word, charsFrequency)) {
+      ans += word.length;
+    }
+  }
+
+  return ans;
+}
+
+function canFormWord(word: string, chars: number[]): boolean {
+  const c: number[] = new Array(26).fill(0);
+
+  for (const char of word) {
+    const ch = char.charCodeAt(0) - "a".charCodeAt(0);
+    c[ch]++;
+    // eslint-disable-next-line @typescript-eslint/no-non-null-assertion
+    if (c[ch]! > chars[ch]!) {
+      return false;
+    }
+  }
+
+  return true;
+}