Hello Everyone,
Given an array and an integer K, find the maximum for each and every contiguous subarray of size k.
Examples :
Input: arr[] = {1, 2, 3, 1, 4, 5, 2, 3, 6}, K = 3 Output: 3 3 4 5 5 5 6 Explanation: Maximum of 1, 2, 3 is 3 Maximum of 2, 3, 1 is 3 Maximum of 3, 1, 4 is 4 Maximum of 1, 4, 5 is 5 Maximum of 4, 5, 2 is 5 Maximum of 5, 2, 3 is 5 Maximum of 2, 3, 6 is 6 Input: arr[] = {8, 5, 10, 7, 9, 4, 15, 12, 90, 13}, K = 4 Output: 10 10 10 15 15 90 90 Explanation: Maximum of first 4 elements is 10, similarly for next 4 elements (i.e from index 1 to 4) is 10, So the sequence generated is 10 10 10 15 15 90 90
Method 1: This is a simple method to solve the above problem.
Approach:
The idea is very basic run a nested loop, the outer loop which will mark the starting point of the subarray of length k, the inner loop will run from the starting index to index+k, k elements from starting index and print the maximum element among these k elements.
Algorithm:
- Create a nested loop, the outer loop from starting index to n – k th elements. The inner loop will run for k iterations.
- Create a variable to store the maximum of k elements traversed by the inner loop.
- Find the maximum of k elements traversed by the inner loop.
- Print the maximum element in every iteration of outer loop
Implementation:
// C++ Program to find the maximum for
// each and every contiguous subarray of size k.
#include <bits/stdc++.h>
using
namespace
std;
// Method to find the maximum for each
// and every contiguous subarray of size k.
void
printKMax(
int
arr[],
int
n,
int
k)
{
int
j, max;
for
(
int
i = 0; i <= n - k; i++)
{
max = arr[i];
for
(j = 1; j < k; j++)
{
if
(arr[i + j] > max)
max = arr[i + j];
}
cout << max <<
" "
;
}
}
// Driver code
int
main()
{
int
arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int
n =
sizeof
(arr) /
sizeof
(arr[0]);
int
k = 3;
printKMax(arr, n, k);
return
0;
}
Output
3 4 5 6 7 8 9 10
Complexity Analysis:
-
Time Complexity: O(N * K).
The outer loop runs n-k+1 times and the inner loop runs k times for every iteration of outer loop. So time complexity is O((n-k+1)*k) which can also be written as O(N * K). -
Space Complexity: O(1).
No extra space is required.
Method 2: This method uses the uses the Self-Balancing BST to solve the given problem.
Approach:
To find maximum among k elements of the subarray the previous method uses a loop traversing through the elements. To reduce that time the idea is to use an AVL tree which returns the maximum element in log n time. So, traverse through the array and keep k elements in the BST and print the maximum in every iteration. AVL tree is a suitable data structure as lookup, insertion, and deletion all take O(log n) time in both the average and worst cases, where n is the number of nodes in the tree prior to the operation.
Algorithm:
- Create a Self-balancing BST (AVL tree) to store and find the maximum element.
- Traverse through the array from start to end.
- Insert the element in the AVL tree.
- If the loop counter or is greater than or equal to k then delete i-k th element from the BST
- Print the maximum element of the BST.
Implementation:
// C++ program to delete a node from AVL Tree
#include<bits/stdc++.h>
using
namespace
std;
// An AVL tree node
class
Node
{
public
:
int
key;
Node *left;
Node *right;
int
height;
};
// A utility function to get maximum
// of two integers
int
max(
int
a,
int
b);
// A utility function to get height
// of the tree
int
height(Node *N)
{
if
(N == NULL)
return
0;
return
N->height;
}
// A utility function to get maximum
// of two integers
int
max(
int
a,
int
b)
{
return
(a > b)? a : b;
}
/* Helper function that allocates a
new node with the given key and
NULL left and right pointers. */
Node* newNode(
int
key)
{
Node* node =
new
Node();
node->key = key;
node->left = NULL;
node->right = NULL;
node->height = 1;
// new node is initially
// added at leaf
return
(node);
}
// A utility function to right
// rotate subtree rooted with y
// See the diagram given above.
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;
// Return new root
return
x;
}
// A utility function to left
// rotate subtree rooted with x
// See the diagram given above.
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;
// Return new root
return
y;
}
// Get Balance factor of node N
int
getBalance(Node *N)
{
if
(N == NULL)
return
0;
return
height(N->left) -
height(N->right);
}
Node* insert(Node* node,
int
key)
{
/* 1. Perform the normal BST rotation */
if
(node == NULL)
return
(newNode(key));
if
(key < node->key)
node->left = insert(node->left, key);
else
if
(key > node->key)
node->right = insert(node->right, key);
else
// Equal keys not allowed
return
node;
/* 2. Update height of this ancestor node */
node->height = 1 + max(height(node->left),
height(node->right));
/* 3. Get the balance factor of this
ancestor node to check whether
this node became unbalanced */
int
balance = getBalance(node);
// If this node becomes unbalanced,
// then there are 4 cases
// 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 the (unchanged) node pointer */
return
node;
}
/* Given a non-empty binary search tree,
return the node with minimum key value
found in that tree. Note that the entire
tree does not need to be searched. */
Node * minValueNode(Node* node)
{
Node* current = node;
/* loop down to find the leftmost leaf */
while
(current->left != NULL)
current = current->left;
return
current;
}
// Recursive function to delete a node
// with given key from subtree with
// given root. It returns root of the
// modified subtree.
Node* deleteNode(Node* root,
int
key)
{
// STEP 1: PERFORM STANDARD BST DELETE
if
(root == NULL)
return
root;
// If the key to be deleted is smaller
// than the root's key, then it lies
// in left subtree
if
( key < root->key )
root->left = deleteNode(root->left, key);
// If the key to be deleted is greater
// than the root's key, then it lies
// in right subtree
else
if
( key > root->key )
root->right = deleteNode(root->right, key);
// if key is same as root's key, then
// This is the node to be deleted
else
{
// node with only one child or no child
if
( (root->left == NULL) ||
(root->right == NULL) )
{
Node *temp = root->left ?
root->left :
root->right;
// No child case
if
(temp == NULL)
{
temp = root;
root = NULL;
}
else
// One child case
*root = *temp;
// Copy the contents of
// the non-empty child
free
(temp);
}
else
{
// node with two children: Get the inorder
// successor (smallest in the right subtree)
Node* temp = minValueNode(root->right);
// Copy the inorder successor's
// data to this node
root->key = temp->key;
// Delete the inorder successor
root->right = deleteNode(root->right,
temp->key);
}
}
// If the tree had only one node
// then return
if
(root == NULL)
return
root;
// STEP 2: UPDATE HEIGHT OF THE CURRENT NODE
root->height = 1 + max(height(root->left),
height(root->right));
// STEP 3: GET THE BALANCE FACTOR OF
// THIS NODE (to check whether this
// node became unbalanced)
int
balance = getBalance(root);
// If this node becomes unbalanced,
// then there are 4 cases
// Left Left Case
if
(balance > 1 &&
getBalance(root->left) >= 0)
return
rightRotate(root);
// Left Right Case
if
(balance > 1 &&
getBalance(root->left) < 0)
{
root->left = leftRotate(root->left);
return
rightRotate(root);
}
// Right Right Case
if
(balance < -1 &&
getBalance(root->right) <= 0)
return
leftRotate(root);
// Right Left Case
if
(balance < -1 &&
getBalance(root->right) > 0)
{
root->right = rightRotate(root->right);
return
leftRotate(root);
}
return
root;
}
// A utility function to print preorder
// traversal of the tree.
// The function also prints height
// of every node
void
preOrder(Node *root)
{
if
(root != NULL)
{
cout << root->key <<
" "
;
preOrder(root->left);
preOrder(root->right);
}
}
// Returns maximum value in a given
// Binary Tree
int
findMax(Node* root)
{
// Base case
if
(root == NULL)
return
INT_MIN;
// Return maximum of 3 values:
// 1) Root's data 2) Max in Left Subtree
// 3) Max in right subtree
int
res = root->key;
int
lres = findMax(root->left);
int
rres = findMax(root->right);
if
(lres > res)
res = lres;
if
(rres > res)
res = rres;
return
res;
}
// Method to find the maximum for each
// and every contiguous subarray of size k.
void
printKMax(
int
arr[],
int
n,
int
k)
{
int
c = 0,l=0;
Node *root = NULL;
//traverse the array ;
for
(
int
i=0; i<n; i++)
{
c++;
//insert the element in BST
root = insert(root, arr[i]);
//size of subarray greater than k
if
(c > k)
{
root = deleteNode(root, arr[l++]);
c--;
}
//size of subarray equal to k
if
(c == k)
{
cout<<findMax(root)<<
" "
;
}
}
}
// Driver code
int
main()
{
int
arr[] = {8, 5, 10, 7, 9, 4, 15, 12, 90, 13}, k = 4;
int
n =
sizeof
(arr) /
sizeof
(arr[0]);
printKMax(arr, n, k);
return
0;
}