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feat: add solutions to lc problem: No.1409
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No.1409.Queries on a Permutation With Key
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@yanglbme
yanglbme committed Aug 5, 2022
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306 changes: 265 additions & 41 deletions solution/1400-1499/1409.Queries on a Permutation With Key/README.md
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Expand Up @@ -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 -->

Expand All @@ -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**
Expand All @@ -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;
}
}
```
Expand All @@ -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;
}
};
```
Expand All @@ -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
}
```

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