Merge pull request #1241 from 0xff-dev/2138

Add solution and test-cases for problem 2138

Author: 6boris

leetcode/2101-2200/2138.Divide-a-String-Into-Groups-of-Size-k/README.md

@@ -1,28 +1,39 @@
 # [2138.Divide a String Into Groups of Size k][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
 
+A string `s` can be partitioned into groups of size `k` using the following procedure:
+
+- The first group consists of the first `k` characters of the string, the second group consists of the next k characters of the string, and so on. Each element can be a part of **exactly one** group.
+- For the last group, if the string **does not** have `k` characters remaining, a character `fill` is used to complete the group.
+
+Note that the partition is done so that after removing the `fill` character from the last group (if it exists) and concatenating all the groups in order, the resultant string should be `s`.
+
+Given the string `s`, the size of each group `k` and the character `fill`, return a string array denoting the **composition of every group** `s` has been divided into, using the above procedure.
+
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: s = "abcdefghi", k = 3, fill = "x"
+Output: ["abc","def","ghi"]
+Explanation:
+The first 3 characters "abc" form the first group.
+The next 3 characters "def" form the second group.
+The last 3 characters "ghi" form the third group.
+Since all groups can be completely filled by characters from the string, we do not need to use fill.
+Thus, the groups formed are "abc", "def", and "ghi".
 ```
 
-## 题意
-> ...
-
-## 题解
+**Example 2:**
 
-### 思路1
-> ...
-Divide a String Into Groups of Size k
-```go
 ```
-
+Input: s = "abcdefghij", k = 3, fill = "x"
+Output: ["abc","def","ghi","jxx"]
+Explanation:
+Similar to the previous example, we are forming the first three groups "abc", "def", and "ghi".
+For the last group, we can only use the character 'j' from the string. To complete this group, we add 'x' twice.
+Thus, the 4 groups formed are "abc", "def", "ghi", and "jxx".
+```
 
 ## 结语
 

leetcode/2101-2200/2138.Divide-a-String-Into-Groups-of-Size-k/Solution.go

@@ -1,5 +1,17 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(s string, k int, fill byte) []string {
+	l := len(s)
+	bs := []byte(s)
+	mod := l % k
+	if mod != 0 {
+		for i := 0; i < k-mod; i++ {
+			bs = append(bs, fill)
+		}
+	}
+	var ans []string
+	for i := 0; i < len(bs); i += k {
+		ans = append(ans, string(bs[i:i+k]))
+	}
+	return ans
 }

leetcode/2101-2200/2138.Divide-a-String-Into-Groups-of-Size-k/Solution_test.go

@@ -10,30 +10,31 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		inputs string
+		k      int
+		fill   byte
+		expect []string
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", "abcdefghi", 3, 'x', []string{"abc", "def", "ghi"}},
+		{"TestCase2", "abcdefghij", 3, 'x', []string{"abc", "def", "ghi", "jxx"}},
 	}
 
 	//	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.inputs, c.k, c.fill)
 			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 %v",
+					c.expect, got, c.inputs, c.k, c.fill)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }