feat: add solutions to lc problem: No.0683

No.0683.K Empty Slots

Author: yanglbme

solution/0600-0699/0683.K Empty Slots/README.md

@@ -54,22 +54,206 @@ k: 1
 
 <!-- 这里可写通用的实现逻辑 -->
 
+树状数组。
+
+树状数组，也称作“二叉索引树”（Binary Indexed Tree）或 Fenwick 树。 它可以高效地实现如下两个操作：
+
+1. **单点更新** `update(x, delta)`： 把序列 x 位置的数加上一个值 delta；
+1. **前缀和查询** `query(x)`：查询序列 `[1,...x]` 区间的区间和，即位置 x 的前缀和。
+
+这两个操作的时间复杂度均为 `O(log n)`。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
+class BinaryIndexedTree:
+    def __init__(self, n):
+        self.n = n
+        self.c = [0] * (n + 1)
 
+    @staticmethod
+    def lowbit(x):
+        return x & -x
+
+    def update(self, x, delta):
+        while x <= self.n:
+            self.c[x] += delta
+            x += BinaryIndexedTree.lowbit(x)
+    
+    def query(self, x):
+        s = 0
+        while x > 0:
+            s += self.c[x]
+            x -= BinaryIndexedTree.lowbit(x)
+        return s
+
+class Solution:
+    def kEmptySlots(self, bulbs: List[int], k: int) -> int:
+        n = len(bulbs)
+        tree = BinaryIndexedTree(n)
+        for i, x in enumerate(bulbs, 1):
+            tree.update(x, 1)
+            case1 = x - k - 1 > 0 and tree.query(x - k - 1) - tree.query(x - k - 2) == 1 and tree.query(x - 1) - tree.query(x - k - 1) == 0
+            case2 = x + k + 1 <= n and tree.query(x + k + 1) - tree.query(x + k) == 1 and tree.query(x + k) - tree.query(x) == 0
+            if case1 or case2:
+                return i
+        return -1
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int kEmptySlots(int[] bulbs, int k) {
+        int n = bulbs.length;
+        BinaryIndexedTree tree = new BinaryIndexedTree(n);
+        for (int i = 0; i < n; ++i) {
+            int x = bulbs[i];
+            tree.update(x, 1);
+            boolean case1 = x - k - 1 > 0 && tree.query(x - k - 1) - tree.query(x - k - 2) == 1 && tree.query(x - 1) - tree.query(x - k - 1) == 0;
+            boolean case2 = x + k + 1 <= n && tree.query(x + k + 1) - tree.query(x + k) == 1 && tree.query(x + k) - tree.query(x) == 0;
+            if (case1 || case2) {
+                return i + 1;
+            }
+        }
+        return -1;
+    }
+}
+
+class BinaryIndexedTree {
+    private int n;
+    private int[] c;
+
+    public BinaryIndexedTree(int n) {
+        this.n = n;
+        c = new int[n + 1];
+    }
+
+    public void update(int x, int delta) {
+        while (x <= n) {
+            c[x] += delta;
+            x += lowbit(x);
+        }
+    }
+
+    public int query(int x) {
+        int s = 0;
+        while (x > 0) {
+            s += c[x];
+            x -= lowbit(x);
+        }
+        return s;
+    }
+
+    public static int lowbit(int x) {
+        return x & -x;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class BinaryIndexedTree {
+public:
+    int n;
+    vector<int> c;
+
+    BinaryIndexedTree(int _n): n(_n), c(_n + 1){}
+
+    void update(int x, int delta) {
+        while (x <= n)
+        {
+            c[x] += delta;
+            x += lowbit(x);
+        }
+    }
+
+    int query(int x) {
+        int s = 0;
+        while (x > 0)
+        {
+            s += c[x];
+            x -= lowbit(x);
+        }
+        return s;
+    }
+
+    int lowbit(int x) {
+        return x & -x;
+    }
+};
+
+class Solution {
+public:
+    int kEmptySlots(vector<int>& bulbs, int k) {
+        int n = bulbs.size();
+        BinaryIndexedTree* tree = new BinaryIndexedTree(n);
+        for (int i = 0; i < n; ++i)
+        {
+            int x = bulbs[i];
+            tree->update(x, 1);
+            bool case1 = x - k - 1 > 0 && tree->query(x - k - 1) - tree->query(x - k - 2) == 1 && tree->query(x - 1) - tree->query(x - k - 1) == 0;
+            bool case2 = x + k + 1 <= n && tree->query(x + k + 1) - tree->query(x + k) == 1 && tree->query(x + k) - tree->query(x) == 0;
+            if (case1 || case2) return i + 1;
+        }
+        return -1;
+    }
+};
+```
+
+### **Go**
+
+```go
+type BinaryIndexedTree struct {
+	n int
+	c []int
+}
+
+func newBinaryIndexedTree(n int) *BinaryIndexedTree {
+	c := make([]int, n+1)
+	return &BinaryIndexedTree{n, c}
+}
+
+func (this *BinaryIndexedTree) lowbit(x int) int {
+	return x & -x
+}
+
+func (this *BinaryIndexedTree) update(x, delta int) {
+	for x <= this.n {
+		this.c[x] += delta
+		x += this.lowbit(x)
+	}
+}
+
+func (this *BinaryIndexedTree) query(x int) int {
+	s := 0
+	for x > 0 {
+		s += this.c[x]
+		x -= this.lowbit(x)
+	}
+	return s
+}
 
+func kEmptySlots(bulbs []int, k int) int {
+	n := len(bulbs)
+	tree := newBinaryIndexedTree(n)
+	for i, x := range bulbs {
+		tree.update(x, 1)
+		case1 := x-k-1 > 0 && tree.query(x-k-1)-tree.query(x-k-2) == 1 && tree.query(x-1)-tree.query(x-k-1) == 0
+		case2 := x+k+1 <= n && tree.query(x+k+1)-tree.query(x+k) == 1 && tree.query(x+k)-tree.query(x) == 0
+		if case1 || case2 {
+			return i + 1
+		}
+	}
+	return -1
+}
 ```
 
 ### **...**