Merge pull request #834 from 0xff-dev/1296

Add solution and test-cases for problem 1296

Author: 6boris

leetcode/1201-1300/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/README.md

@@ -1,28 +1,33 @@
 # [1296.Divide Array in Sets of K Consecutive Numbers][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
+Given an array of integers `nums` and a positive integer `k`, check whether it is possible to divide this array into sets of `k` consecutive numbers.
+
+Return `true` if it is possible. Otherwise, return `false`.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: nums = [1,2,3,3,4,4,5,6], k = 4
+Output: true
+Explanation: Array can be divided into [1,2,3,4] and [3,4,5,6].
 ```
 
-## 题意
-> ...
+**Example 2:**
 
-## 题解
-
-### 思路1
-> ...
-Divide Array in Sets of K Consecutive Numbers
-```go
+```
+Input: nums = [3,2,1,2,3,4,3,4,5,9,10,11], k = 3
+Output: true
+Explanation: Array can be divided into [1,2,3] , [2,3,4] , [3,4,5] and [9,10,11].
 ```
 
+**Example 3:**
+
+```
+Input: nums = [1,2,3,4], k = 3
+Output: false
+Explanation: Each array should be divided in subarrays of size 3.
+```
 
 ## 结语
 

leetcode/1201-1300/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/Solution.go

@@ -1,5 +1,71 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+import "sort"
+
+func Solution(nums []int, k int) bool {
+	l := len(nums)
+	if l%k != 0 {
+		return false
+	}
+	if k == 1 {
+		return true
+	}
+
+	keys := make([]int, 0)
+	keyCount := make(map[int]int)
+	for _, n := range nums {
+		keyCount[n]++
+		if keyCount[n] == 1 {
+			keys = append(keys, n)
+		}
+	}
+	sort.Ints(keys)
+
+	idx := 1
+	count := keyCount[keys[0]]
+	nextIdx := -1
+	keyCount[keys[0]] = 0
+	used := 1
+	l -= count
+
+	for idx < len(keys) {
+		if keys[idx]-keys[idx-1] != 1 {
+			return false
+		}
+		keyCount[keys[idx]] -= count
+		l -= count
+		used++
+		if keyCount[keys[idx]] < 0 {
+			return false
+		}
+		if keyCount[keys[idx]] > 0 && nextIdx == -1 {
+			nextIdx = idx
+		}
+
+		if used != k {
+			idx++
+			continue
+		}
+		if nextIdx == -1 {
+			if idx == len(keys)-1 {
+				break
+			}
+			idx++
+			count = keyCount[keys[idx]]
+			nextIdx = -1
+			keyCount[keys[idx]] = 0
+			l -= count
+			used = 1
+			idx++
+			continue
+		}
+
+		idx = nextIdx + 1
+		count = keyCount[keys[nextIdx]]
+		keyCount[keys[nextIdx]] = 0
+		nextIdx = -1
+		l -= count
+		used = 1
+	}
+	return l == 0 && used == k
 }

leetcode/1201-1300/1296.Divide-Array-in-Sets-of-K-Consecutive-Numbers/Solution_test.go

@@ -10,30 +10,31 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
+		nums   []int
+		k      int
 		expect bool
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", []int{1, 2, 3, 3, 4, 4, 5, 6}, 4, true},
+		{"TestCase2", []int{3, 2, 1, 2, 3, 4, 3, 4, 5, 9, 10, 11}, 3, true},
+		{"TestCase3", []int{1, 2, 3, 4}, 3, false},
 	}
 
 	//	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.nums, c.k)
 			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.nums, c.k)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }