C ++程序计算数组元素的数量大于其左侧的所有元素且其右侧至少有K个元素
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其前面的所有元素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它们是否都小于它，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AVL树中
- 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。
- 如果是这样，请增加count 。
- 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 approach
#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;
} 输出：
2时间复杂度： O(NlogN)
辅助空间： O(N)
有关更多详细信息，请参阅有关 Array 元素的数量大于其左侧的所有元素和其右侧的至少 K 个元素的完整文章！