feat: add solutions to lc problem: No.1409

No.1409.Queries on a Permutation With Key

Author: yanglbme

solution/1400-1499/1409.Queries on a Permutation With Key/README.md

@@ -55,7 +55,24 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-题目数据规模不大，直接模拟即可。
+**方法一：模拟**
+
+题目数据规模不大，可以直接模拟。
+
+**方法一：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)`。
+
+树状数组最基本的功能就是求比某点 x 小的点的个数（这里的比较是抽象的概念，可以是数的大小、坐标的大小、质量的大小等等）。
+
+比如给定数组 `a[5] = {2, 5, 3, 4, 1}`，求 `b[i] = 位置 i 左边小于等于 a[i] 的数的个数`。对于此例，`b[5] = {0, 1, 1, 2, 0}`。
+
+解决方案是直接遍历数组，每个位置先求出 `query(a[i])`，然后再修改树状数组 `update(a[i], 1)` 即可。当数的范围比较大时，需要进行离散化，即先进行去重并排序，然后对每个数字进行编号。
 
 <!-- tabs:start -->
 
@@ -66,13 +83,55 @@
 ```python
 class Solution:
     def processQueries(self, queries: List[int], m: int) -> List[int]:
-        nums = list(range(1, m + 1))
-        res = []
-        for num in queries:
-            res.append(nums.index(num))
-            nums.remove(num)
-            nums.insert(0, num)
-        return res
+        p = list(range(1, m + 1))
+        ans = []
+        for v in queries:
+            j = p.index(v)
+            ans.append(j)
+            p.pop(j)
+            p.insert(0, v)
+        return ans
+```
+
+```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 processQueries(self, queries: List[int], m: int) -> List[int]:
+        n = len(queries)
+        pos = [0] * (m + 1)
+        tree = BinaryIndexedTree(m + n)
+        for i in range(1, m + 1):
+            pos[i] = n + i
+            tree.update(n + i, 1)
+
+        ans = []
+        for i, v in enumerate(queries):
+            j = pos[v]
+            tree.update(j, -1)
+            ans.append(tree.query(j))
+            pos[v] = n - i
+            tree.update(n - i, 1)
+        return ans
 ```
 
 ### **Java**
@@ -82,18 +141,74 @@ class Solution:
 ```java
 class Solution {
     public int[] processQueries(int[] queries, int m) {
-        List<Integer> nums = new LinkedList<>();
-        for (int i = 0; i < m; ++i) {
-            nums.add(i + 1);
+        List<Integer> p = new LinkedList<>();
+        for (int i = 1; i <= m; ++i) {
+            p.add(i);
         }
-        int[] res = new int[queries.length];
+        int[] ans = new int[queries.length];
         int i = 0;
-        for (int num : queries) {
-            res[i++] = nums.indexOf(num);
-            nums.remove(Integer.valueOf(num));
-            nums.add(0, num);
+        for (int v : queries) {
+            int j = p.indexOf(v);
+            ans[i++] = j;
+            p.remove(j);
+            p.add(0, v);
+        }
+        return ans;
+    }
+}
+```
+
+```java
+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;
+    }
+}
+
+class Solution {
+    public int[] processQueries(int[] queries, int m) {
+        int n = queries.length;
+        BinaryIndexedTree tree = new BinaryIndexedTree(m + n);
+        int[] pos = new int[m + 1];
+        for (int i = 1; i <= m; ++i) {
+            pos[i] = n + i;
+            tree.update(n + i, 1);
+        }
+        int[] ans = new int[n];
+        int k = 0;
+        for (int i = 0; i < n; ++i) {
+            int v = queries[i];
+            int j = pos[v];
+            tree.update(j, -1);
+            ans[k++] = tree.query(j);
+            pos[v] = n - i;
+            tree.update(n - i, 1);
         }
-        return res;
+        return ans;
     }
 }
 ```
@@ -104,24 +219,82 @@ class Solution {
 class Solution {
 public:
     vector<int> processQueries(vector<int>& queries, int m) {
-        vector<int> nums(m);
-        iota(nums.begin(), nums.end(), 1);
-        vector<int> res;
-        for (int num : queries)
+        vector<int> p(m);
+        iota(p.begin(), p.end(), 1);
+        vector<int> ans;
+        for (int v : queries)
         {
-            int idx = -1;
+            int j = 0;
             for (int i = 0; i < m; ++i)
             {
-                if (nums[i] == num) {
-                    idx = i;
+                if (p[i] == v)
+                {
+                    j = i;
                     break;
                 }
             }
-            res.push_back(idx);
-            nums.erase(nums.begin() + idx);
-            nums.insert(nums.begin(), num);
+            ans.push_back(j);
+            p.erase(p.begin() + j);
+            p.insert(p.begin(), v);
+        }
+        return ans;
+    }
+};
+```
+
+```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:
+    vector<int> processQueries(vector<int>& queries, int m) {
+        int n = queries.size();
+        vector<int> pos(m + 1);
+        BinaryIndexedTree* tree = new BinaryIndexedTree(m + n);
+        for (int i = 1; i <= m; ++i)
+        {
+            pos[i] = n + i;
+            tree->update(n + i, 1);
+        }
+        vector<int> ans;
+        for (int i = 0; i < n; ++i)
+        {
+            int v = queries[i];
+            int j = pos[v];
+            tree->update(j, -1);
+            ans.push_back(tree->query(j));
+            pos[v] = n - i;
+            tree->update(n - i, 1);
         }
-        return res;
+        return ans;
     }
 };
 ```
@@ -130,24 +303,75 @@ public:
 
 ```go
 func processQueries(queries []int, m int) []int {
-	nums := make([]int, m)
-	for i := 0; i < m; i++ {
-		nums[i] = i + 1
+	p := make([]int, m)
+	for i := range p {
+		p[i] = i + 1
 	}
-	var res []int
-	for _, num := range queries {
-		idx := -1
-		for i := 0; i < m; i++ {
-			if nums[i] == num {
-				idx = i
+	ans := []int{}
+	for _, v := range queries {
+		j := 0
+		for i := range p {
+			if p[i] == v {
+				j = i
 				break
 			}
 		}
-		res = append(res, idx)
-		nums = append(nums[:idx], nums[idx+1:]...)
-		nums = append([]int{num}, nums...)
+		ans = append(ans, j)
+		p = append(p[:j], p[j+1:]...)
+		p = append([]int{v}, p...)
+	}
+	return ans
+}
+```
+
+```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 processQueries(queries []int, m int) []int {
+	n := len(queries)
+	pos := make([]int, m+1)
+	tree := newBinaryIndexedTree(m + n)
+	for i := 1; i <= m; i++ {
+		pos[i] = n + i
+		tree.update(n+i, 1)
+	}
+	ans := []int{}
+	for i, v := range queries {
+		j := pos[v]
+		tree.update(j, -1)
+		ans = append(ans, tree.query(j))
+		pos[v] = n - i
+		tree.update(n-i, 1)
 	}
-	return res
+	return ans
 }
 ```