给定一个由N 个不同整数组成的数组A[] ，任务是找到严格大于它前面的所有元素并且严格大于其右侧至少K 个元素的元素的数量。
例子：
Input: A[] = {2, 5, 1, 7, 3, 4, 0}, K = 3
Output: 2
Explanation:
The only array elements satisfying the given conditions are:
- 5: Greater than all elements on its left {2} and at least K(= 3) elements on its right {1, 3, 4, 0}
- 7: Greater than all elements on its left {2, 5, 1} and at least K(= 3) elements on its right {3, 4, 0}
Therefore, the count is 2.
Input: A[] = {11, 2, 4, 7, 5, 9, 6, 3}, K = 2
Output: 1
天真的方法：
解决问题的最简单的方法是遍历数组，对于每个元素，遍历其左侧的所有元素并检查是否所有元素都小于它，并遍历其右侧的所有元素以检查是否至少有K个元素是否小于它。对于满足条件的每个元素，增加count 。最后，打印count的值。
时间复杂度： O(N 2 )
辅助空间： O(1)
有效的方法：
上述方法可以通过使用自平衡 BST 进一步优化。请按照以下步骤操作：
- 从右到左遍历数组，将所有元素一一插入到一个AVL Tree中
- 使用 AVL 树生成一个数组countSmaller[] ，其中包含每个数组元素右侧的较小元素的计数。
- 遍历数组，对于每个第i个元素，检查它是否是迄今为止获得的最大值，并且countSmaller[i]是否大于或等于K。
- 如果是，请增加计数。
- 打印count的最终值作为答案。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above appraoch
#include
using namespace std;
// Structure of an AVL Tree Node
struct node {
int key;
struct node* left;
struct node* right;
int height;
// Size of the tree rooted
// with this node
int size;
};
// Utility function to get maximum
// of two integers
int max(int a, int b);
// Utility function to get height
// of the tree rooted with N
int height(struct node* N)
{
if (N == NULL)
return 0;
return N->height;
}
// Utility function to find size of
// the tree rooted with N
int size(struct node* N)
{
if (N == NULL)
return 0;
return N->size;
}
// Utility function to get maximum
// of two integers
int max(int a, int b)
{
return (a > b) ? a : b;
}
// Helper function to allocates a
// new node with the given key
struct node* newNode(int key)
{
struct node* node
= (struct node*)
malloc(sizeof(struct node));
node->key = key;
node->left = NULL;
node->right = NULL;
node->height = 1;
node->size = 1;
return (node);
}
// Utility function to right rotate
// subtree rooted with y
struct node* rightRotate(struct node* y)
{
struct node* x = y->left;
struct node* T2 = x->right;
// Perform rotation
x->right = y;
y->left = T2;
// Update heights
y->height = max(height(y->left),
height(y->right))
+ 1;
x->height = max(height(x->left),
height(x->right))
+ 1;
// Update sizes
y->size = size(y->left)
+ size(y->right) + 1;
x->size = size(x->left)
+ size(x->right) + 1;
// Return new root
return x;
}
// Utility function to left rotate
// subtree rooted with x
struct node* leftRotate(struct node* x)
{
struct node* y = x->right;
struct node* T2 = y->left;
// Perform rotation
y->left = x;
x->right = T2;
// Update heights
x->height = max(height(x->left),
height(x->right))
+ 1;
y->height = max(height(y->left),
height(y->right))
+ 1;
// Update sizes
x->size = size(x->left)
+ size(x->right) + 1;
y->size = size(y->left)
+ size(y->right) + 1;
// Return new root
return y;
}
// Function to obtain Balance factor
// of node N
int getBalance(struct node* N)
{
if (N == NULL)
return 0;
return height(N->left)
- height(N->right);
}
// Function to insert a new key to the
// tree rooted with node
struct node* insert(struct node* node, int key,
int* count)
{
// Perform the normal BST rotation
if (node == NULL)
return (newNode(key));
if (key < node->key)
node->left
= insert(node->left, key, count);
else {
node->right
= insert(node->right, key, count);
// Update count of smaller elements
*count = *count + size(node->left) + 1;
}
// Update height and size of the ancestor
node->height = max(height(node->left),
height(node->right))
+ 1;
node->size = size(node->left)
+ size(node->right) + 1;
// Get the balance factor of the ancestor
int balance = getBalance(node);
// Left Left Case
if (balance > 1 && key < node->left->key)
return rightRotate(node);
// Right Right Case
if (balance < -1 && key > node->right->key)
return leftRotate(node);
// Left Right Case
if (balance > 1 && key > node->left->key) {
node->left = leftRotate(node->left);
return rightRotate(node);
}
// Right Left Case
if (balance < -1 && key < node->right->key) {
node->right = rightRotate(node->right);
return leftRotate(node);
}
return node;
}
// Function to generate an array which contains
// count of smaller elements on the right
void constructLowerArray(int arr[],
int countSmaller[],
int n)
{
int i, j;
struct node* root = NULL;
for (i = 0; i < n; i++)
countSmaller[i] = 0;
// Insert all elements in the AVL Tree
// and get the count of smaller elements
for (i = n - 1; i >= 0; i--) {
root = insert(root, arr[i],
&countSmaller[i]);
}
}
// Function to find the number
// of elements which are greater
// than all elements on its left
// and K elements on its right
int countElements(int A[], int n, int K)
{
int count = 0;
// Stores the count of smaller
// elements on its right
int* countSmaller
= (int*)malloc(sizeof(int) * n);
constructLowerArray(A, countSmaller, n);
int maxi = INT_MIN;
for (int i = 0; i <= (n - K - 1); i++) {
if (A[i] > maxi && countSmaller[i] >= K) {
count++;
maxi = A[i];
}
}
return count;
}
// Driver Code
int main()
{
int A[] = { 2, 5, 1, 7, 3, 4, 0 };
int n = sizeof(A) / sizeof(int);
int K = 3;
cout << countElements(A, n, K);
return 0;
} Java
// Java program to implement
// the above appraoch
class GFG{
// Structure of an AVL Tree Node
static class Node
{
int key;
Node left;
Node right;
int height;
// Size of the tree rooted
// with this Node
int size;
public Node(int key)
{
this.key = key;
this.left = this.right = null;
this.size = this.height = 1;
}
};
// Helper class to pass Integer
// as referencee
static class RefInteger
{
Integer value;
public RefInteger(Integer value)
{
this.value = value;
}
}
// Utility function to get height
// of the tree rooted with N
static int height(Node N)
{
if (N == null)
return 0;
return N.height;
}
// Utility function to find size of
// the tree rooted with N
static int size(Node N)
{
if (N == null)
return 0;
return N.size;
}
// Utility function to get maximum
// of two integers
static int max(int a, int b)
{
return (a > b) ? a : b;
}
// Utility function to right rotate
// subtree rooted with y
static Node rightRotate(Node y)
{
Node x = y.left;
Node T2 = x.right;
// Perform rotation
x.right = y;
y.left = T2;
// Update heights
y.height = max(height(y.left),
height(y.right)) + 1;
x.height = max(height(x.left),
height(x.right)) + 1;
// Update sizes
y.size = size(y.left) +
size(y.right) + 1;
x.size = size(x.left) +
size(x.right) + 1;
// Return new root
return x;
}
// Utility function to left rotate
// subtree rooted with x
static Node leftRotate(Node x)
{
Node y = x.right;
Node T2 = y.left;
// Perform rotation
y.left = x;
x.right = T2;
// Update heights
x.height = max(height(x.left),
height(x.right)) + 1;
y.height = max(height(y.left),
height(y.right)) + 1;
// Update sizes
x.size = size(x.left) +
size(x.right) + 1;
y.size = size(y.left) +
size(y.right) + 1;
// Return new root
return y;
}
// Function to obtain Balance factor
// of Node N
static int getBalance(Node N)
{
if (N == null)
return 0;
return height(N.left) -
height(N.right);
}
// Function to insert a new key to the
// tree rooted with Node
static Node insert(Node Node, int key,
RefInteger count)
{
// Perform the normal BST rotation
if (Node == null)
return (new Node(key));
if (key < Node.key)
Node.left = insert(Node.left,
key, count);
else
{
Node.right = insert(Node.right,
key, count);
// Update count of smaller elements
count.value = count.value +
size(Node.left) + 1;
}
// Update height and size of the ancestor
Node.height = max(height(Node.left),
height(Node.right)) + 1;
Node.size = size(Node.left) +
size(Node.right) + 1;
// Get the balance factor of the ancestor
int balance = getBalance(Node);
// Left Left Case
if (balance > 1 && key < Node.left.key)
return rightRotate(Node);
// Right Right Case
if (balance < -1 && key > Node.right.key)
return leftRotate(Node);
// Left Right Case
if (balance > 1 && key > Node.left.key)
{
Node.left = leftRotate(Node.left);
return rightRotate(Node);
}
// Right Left Case
if (balance < -1 && key < Node.right.key)
{
Node.right = rightRotate(Node.right);
return leftRotate(Node);
}
return Node;
}
// Function to generate an array which
// contains count of smaller elements
// on the right
static void constructLowerArray(int arr[],
RefInteger[] countSmaller, int n)
{
int i, j;
Node root = null;
for(i = 0; i < n; i++)
countSmaller[i] = new RefInteger(0);
// Insert all elements in the AVL Tree
// and get the count of smaller elements
for(i = n - 1; i >= 0; i--)
{
root = insert(root, arr[i],
countSmaller[i]);
}
}
// Function to find the number
// of elements which are greater
// than all elements on its left
// and K elements on its right
static int countElements(int A[], int n,
int K)
{
int count = 0;
// Stores the count of smaller
// elements on its right
RefInteger[] countSmaller = new RefInteger[n];
constructLowerArray(A, countSmaller, n);
int maxi = Integer.MIN_VALUE;
for(int i = 0; i <= (n - K - 1); i++)
{
if (A[i] > maxi &&
countSmaller[i].value >= K)
{
count++;
maxi = A[i];
}
}
return count;
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 2, 5, 1, 7, 3, 4, 0 };
int n = A.length;
int K = 3;
System.out.println(countElements(A, n, K));
}
}
// This code is contributed by sanjeev2552C#
// C# program to implement
// the above appraoch
using System;
using System.Collections.Generic;
class GFG{
// Structure of an AVL Tree Node
public class Node
{
public int key;
public Node left;
public Node right;
public int height;
// Size of the tree rooted
// with this Node
public int size;
public Node(int key)
{
this.key = key;
this.left = this.right = null;
this.size = this.height = 1;
}
};
// Helper class to pass int
// as referencee
public class Refint
{
public int value;
public Refint(int value)
{
this.value = value;
}
}
// Utility function to get height
// of the tree rooted with N
static int height(Node N)
{
if (N == null)
return 0;
return N.height;
}
// Utility function to find size of
// the tree rooted with N
static int size(Node N)
{
if (N == null)
return 0;
return N.size;
}
// Utility function to get maximum
// of two integers
static int max(int a, int b)
{
return (a > b) ? a : b;
}
// Utility function to right rotate
// subtree rooted with y
static Node rightRotate(Node y)
{
Node x = y.left;
Node T2 = x.right;
// Perform rotation
x.right = y;
y.left = T2;
// Update heights
y.height = max(height(y.left),
height(y.right)) + 1;
x.height = max(height(x.left),
height(x.right)) + 1;
// Update sizes
y.size = size(y.left) +
size(y.right) + 1;
x.size = size(x.left) +
size(x.right) + 1;
// Return new root
return x;
}
// Utility function to left rotate
// subtree rooted with x
static Node leftRotate(Node x)
{
Node y = x.right;
Node T2 = y.left;
// Perform rotation
y.left = x;
x.right = T2;
// Update heights
x.height = max(height(x.left),
height(x.right)) + 1;
y.height = max(height(y.left),
height(y.right)) + 1;
// Update sizes
x.size = size(x.left) +
size(x.right) + 1;
y.size = size(y.left) +
size(y.right) + 1;
// Return new root
return y;
}
// Function to obtain Balance factor
// of Node N
static int getBalance(Node N)
{
if (N == null)
return 0;
return height(N.left) -
height(N.right);
}
// Function to insert a new key to the
// tree rooted with Node
static Node insert(Node Node, int key,
Refint count)
{
// Perform the normal BST rotation
if (Node == null)
return (new Node(key));
if (key < Node.key)
Node.left = insert(Node.left,
key, count);
else
{
Node.right = insert(Node.right,
key, count);
// Update count of smaller elements
count.value = count.value +
size(Node.left) + 1;
}
// Update height and size of the ancestor
Node.height = max(height(Node.left),
height(Node.right)) + 1;
Node.size = size(Node.left) +
size(Node.right) + 1;
// Get the balance factor of the ancestor
int balance = getBalance(Node);
// Left Left Case
if (balance > 1 && key < Node.left.key)
return rightRotate(Node);
// Right Right Case
if (balance < -1 && key > Node.right.key)
return leftRotate(Node);
// Left Right Case
if (balance > 1 && key > Node.left.key)
{
Node.left = leftRotate(Node.left);
return rightRotate(Node);
}
// Right Left Case
if (balance < -1 && key < Node.right.key)
{
Node.right = rightRotate(Node.right);
return leftRotate(Node);
}
return Node;
}
// Function to generate an array which
// contains count of smaller elements
// on the right
static void constructLowerArray(int []arr,
Refint[] countSmaller, int n)
{
int i;
//int j;
Node root = null;
for(i = 0; i < n; i++)
countSmaller[i] = new Refint(0);
// Insert all elements in the AVL Tree
// and get the count of smaller elements
for(i = n - 1; i >= 0; i--)
{
root = insert(root, arr[i],
countSmaller[i]);
}
}
// Function to find the number
// of elements which are greater
// than all elements on its left
// and K elements on its right
static int countElements(int []A, int n,
int K)
{
int count = 0;
// Stores the count of smaller
// elements on its right
Refint[] countSmaller = new Refint[n];
constructLowerArray(A, countSmaller, n);
int maxi = int.MinValue;
for(int i = 0; i <= (n - K - 1); i++)
{
if (A[i] > maxi &&
countSmaller[i].value >= K)
{
count++;
maxi = A[i];
}
}
return count;
}
// Driver Code
public static void Main(String[] args)
{
int []A = { 2, 5, 1, 7, 3, 4, 0 };
int n = A.Length;
int K = 3;
Console.WriteLine(countElements(A, n, K));
}
}
// This code is contributed by Princi Singh输出：
2时间复杂度： O(NlogN)
辅助空间： O(N)
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