problem: new problem solution - 1657 . Determine if Two Strings Are C…

…lose

README.md

@@ -162,6 +162,7 @@ My approach to solving LeetCode problems typically involves the following steps:
 | 334  | [Increasing Triplet Subsequence](https://leetcode.com/problems/increasing-triplet-subsequence/)                                                             | Algorithms | [TypeScript](./problems/algorithms/increasingTripletSubsequence/IncreasingTripletSubsequence.ts)                                                 | Medium     |
 | 1347 | [Minimum Number of Steps to Make Two Strings Anagram](https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram/)                   | Algorithms | [TypeScript](./problems/algorithms/minimumNumberOfStepsToMakeTwoStringsAnagram/MinimumNumberOfStepsToMakeTwoStringsAnagram.ts)                   | Medium     |
 | 3005 | [Count Elements With Maximum Frequency](https://leetcode.com/problems/count-elements-with-maximum-frequency/)                                               | Algorithms | [TypeScript](./problems/algorithms/countElementsWithMaximumFrequency/CountElementsWithMaximumFrequency.ts)                                       | Easy       |
+| 1657 | [Determine if Two Strings Are Close](https://leetcode.com/problems/determine-if-two-strings-are-close/)                                                     | Algorithms | [TypeScript](./problems/algorithms/determineIfTwoStringsAreClose/DetermineIfTwoStringsAreClose.ts)                                               | Medium     |
 | ...  | ...                                                                                                                                                         | ...        | ...                                                                                                                                              | ...        |
 
 In this table:

problems/algorithms/determineIfTwoStringsAreClose/DetermineIfTwoStringsAreClose.test.ts

@@ -0,0 +1,49 @@
+// Source : https://leetcode.com/problems/determine-if-two-strings-are-close/
+// Author : francisco
+// Date   : 2024-01-14
+
+import { closeStrings } from "./DetermineIfTwoStringsAreClose";
+
+describe("determine if two strings are close", () => {
+  test("example 1", () => {
+    const result: boolean = closeStrings("abc", "bca");
+
+    expect(result).toBe(true);
+  });
+
+  test("example 2", () => {
+    const result: boolean = closeStrings("a", "aa");
+
+    expect(result).toBe(false);
+  });
+
+  test("example 3", () => {
+    const result: boolean = closeStrings("cabbba", "abbccc");
+
+    expect(result).toBe(true);
+  });
+
+  test("additional testcase 4", () => {
+    const result: boolean = closeStrings("sadjkfasjfkg", "sadjkfasjfk");
+
+    expect(result).toBe(false);
+  });
+
+  test("additional testcase 5", () => {
+    const result: boolean = closeStrings("abbccc", "addccc");
+
+    expect(result).toBe(false);
+  });
+
+  test("failed submission - 117 / 153 testcases passed", () => {
+    const result: boolean = closeStrings("abbbzcf", "babzzcz");
+
+    expect(result).toBe(false);
+  });
+
+  test("failed submission - 132 / 153 testcases passed", () => {
+    const result: boolean = closeStrings("abbzzca", "babzzcz");
+
+    expect(result).toBe(false);
+  });
+});

problems/algorithms/determineIfTwoStringsAreClose/DetermineIfTwoStringsAreClose.ts

@@ -0,0 +1,111 @@
+// Source : https://leetcode.com/problems/determine-if-two-strings-are-close/
+// Author : francisco
+// Date   : 2024-01-14
+
+/*****************************************************************************************************
+ *
+ * Two strings are considered close if you can attain one from the other using the following
+ * operations:
+ *
+ * 	Operation 1: Swap any two existing characters.
+ *
+ * 		For example, abcde -> aecdb
+ *
+ * 	Operation 2: Transform every occurrence of one existing character into another existing
+ * character, and do the same with the other character.
+ *
+ * 		For example, aacabb -> bbcbaa (all a's turn into b's, and all b's turn into a's)
+ *
+ * You can use the operations on either string as many times as necessary.
+ *
+ * Given two strings, word1 and word2, return true if word1 and word2 are close, and false otherwise.
+ *
+ * Example 1:
+ *
+ * Input: word1 = "abc", word2 = "bca"
+ * Output: true
+ * Explanation: You can attain word2 from word1 in 2 operations.
+ * Apply Operation 1: "abc" -> "acb"
+ * Apply Operation 1: "acb" -> "bca"
+ *
+ * Example 2:
+ *
+ * Input: word1 = "a", word2 = "aa"
+ * Output: false
+ * Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of
+ * operations.
+ *
+ * Example 3:
+ *
+ * Input: word1 = "cabbba", word2 = "abbccc"
+ * Output: true
+ * Explanation: You can attain word2 from word1 in 3 operations.
+ * Apply Operation 1: "cabbba" -> "caabbb"
+ * Apply Operation 2: "caabbb" -> "baaccc"
+ * Apply Operation 2: "baaccc" -> "abbccc"
+ *
+ * Constraints:
+ *
+ * 	1 <= word1.length, word2.length <= 10^5
+ * 	word1 and word2 contain only lowercase English letters.
+ ******************************************************************************************************/
+
+/**
+ * @param {string} word1
+ * @param {string} word2
+ * @returns {boolean}
+ * return true if word1 and word2 are close, and false otherwise
+ * NOTE: close means - same length && same characters && same counts
+
+ * Tests:
+ * I: word1 = "abc", word2 = "bca" -> O: true
+ * I: word1 = "a", word2 = "aa" -> O: false
+ * I: word1 = "cabbba", word2 = "abbccc" -> O: true
+ * I: word1 = "sadjkfasjfkg", word2 = "sadjkfasjfk" -> O: false
+ * I: word1 = "abbccc", word2 = "addccc" -> O: false
+ * I: word1 = "abbbzcf", word2 = "babzzcz" -> O: false
+ * I: word1 = "abbzzca", word2 = "babzzcz" -> O: false
+
+ * Template:
+ * if the lengths are different they cannot be close
+ * iterate word 1 and get its characters and character counts, counts1
+ * iterate word 2: check if any of the characters in word 2 is not in counts1
+ * add word 2's counts to its own variable
+ * iterate each word's counts: hashtable for the frequency of certain count
+ * for word 1's counts add 1 to the frequency, for word 2's counts sub 1 from frequency
+ * if every frequency is zero it means there is the same number of counts for each word
+
+ * Time Complexity: O(n + max(m, k)), where n is word1.length;
+    m is values1.length and k is values2.length
+ * Space Complexity: O(m + k)
+ */
+export function closeStrings(word1: string, word2: string): boolean {
+  // same length
+  if (word1.length !== word2.length) return false;
+
+  const counts1: Record<string, number> = {};
+  const counts2: Record<string, number> = {};
+
+  for (let i: number = 0; i < word1.length; i++) {
+    counts1[word1[i] as string] = (counts1[word1[i] as string] ?? 0) + 1;
+  }
+
+  for (let i: number = 0; i < word2.length; i++) {
+    // same characters
+    if (counts1[word2[i] as string] === undefined) return false;
+    counts2[word2[i] as string] = (counts2[word2[i] as string] ?? 0) + 1;
+  }
+
+  const values1: number[] = Object.values(counts1);
+  const values2: number[] = Object.values(counts2);
+
+  const counts: Record<number, number> = {};
+
+  for (let i: number = 0; i < values1.length; i++) {
+    counts[values1[i] as number] = (counts[values1[i] as number] ?? 0) + 1;
+    counts[values2[i] as number] = (counts[values2[i] as number] ?? 0) - 1;
+  }
+
+  // same counts
+  return Object.values(counts).every((count: number) => count === 0);
+}