Merge pull request 6boris#361 from 0xff-dev/1657

Add solution and test-cases for problem  1657

Author: 6boris

leetcode/1601-1700/1657.Determine-if-Two-Strings-Are-Close/README.md

@@ -1,28 +1,46 @@
 # [1657.Determine if Two Strings Are Close][title]
 
-> [!WARNING|style:flat]
-> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
-
 ## Description
+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: a = "11", b = "1"
-Output: "100"
+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"
 ```
 
-## 题意
-> ...
-
-## 题解
+**Example 2:**
 
-### 思路1
-> ...
-Determine if Two Strings Are Close
-```go
 ```
+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"
+```
 
 ## 结语
 

leetcode/1601-1700/1657.Determine-if-Two-Strings-Are-Close/Solution.go

@@ -1,5 +1,29 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+import "sort"
+
+func Solution(word1 string, word2 string) bool {
+	if len(word1) != len(word2) {
+		return false
+	}
+	bucket1, bucket2 := make([]int, 26), make([]int, 26)
+	for idx := 0; idx < len(word1); idx++ {
+		bucket1[word1[idx]-'a']++
+		bucket2[word2[idx]-'a']++
+	}
+
+	for i := 0; i < 26; i++ {
+		if bucket2[i] != 0 && bucket1[i] == 0 || bucket1[i] != 0 && bucket2[i] == 0 {
+			return false
+		}
+	}
+	sort.Ints(bucket1)
+	sort.Ints(bucket2)
+
+	for i := 0; i < 26; i++ {
+		if bucket1[i] != bucket2[i] {
+			return false
+		}
+	}
+	return true
 }

leetcode/1601-1700/1657.Determine-if-Two-Strings-Are-Close/Solution_test.go

@@ -10,30 +10,30 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
+		s1, s2 string
 		expect bool
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", "abc", "bca", true},
+		{"TestCase2", "a", "aa", false},
+		{"TestCase3", "cabbba", "abbccc", true},
 	}
 
 	//	开始测试
 	for i, c := range cases {
 		t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
-			got := Solution(c.inputs)
+			got := Solution(c.s1, c.s2)
 			if !reflect.DeepEqual(got, c.expect) {
-				t.Fatalf("expected: %v, but got: %v, with inputs: %v",
-					c.expect, got, c.inputs)
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v %v",
+					c.expect, got, c.s1, c.s2)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }