Sliding Window Maximum (Maximum of all subarrays of size K)
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 6Input: 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
Naive 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, and print the maximum element among these K elements.
Follow the given steps to solve the problem:
- Create a nested loop, the outer loop from starting index to N – Kth 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 the outer loop
Below is the implementation of the above approach:
C++
// C++ program for the above approach #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's 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; // Function call printKMax(arr, N, K); return 0; } // This code is contributed by rathbhupendra |
C
// C program for the above approach #include <stdio.h> 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]; } printf ( "%d " , max); } } // Driver's 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; // Function call printKMax(arr, N, K); return 0; } |
Java
// Java program for the above approach public class GFG { // Method to find the maximum for // each and every contiguous // subarray of size K. static 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]; } System.out.print(max + " " ); } } // Driver's code public static void main(String args[]) { int arr[] = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 }; int K = 3 ; // Function call printKMax(arr, arr.length, K); } } // This code is contributed by Sumit Ghosh |
C#
// C# program for the above approach using System; class GFG { // Method to find the maximum for // each and every contiguous subarray // of size k. static 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]; } Console.Write(max + " " ); } } // Driver's code public static void Main() { int [] arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; int K = 3; printKMax(arr, arr.Length, K); } } // This Code is Contributed by Sam007 |
Javascript
// JavaScript Program to find the maximum for // each and every contiguous subarray of size k. // Method to find the maximum for each // and every contiguous subarray of size k. function printKMax(arr,n,k) { let j, max; for (let i = 0; i <= n - k; i++) { max = arr[i]; for (j = 1; j < k; j++) { if (arr[i + j] > max) max = arr[i + j]; } document.write( max + " " ); } } // Driver code let arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]; let n =arr.length; let k = 3; printKMax(arr, n, k); // This code contributed by gauravrajput1 |
PHP
<?php // php program for the above approach function printKMax( $arr , $N , $K ) { $j ; $max ; for ( $i = 0; $i <= $N - $K ; $i ++) { $max = $arr [ $i ]; for ( $j = 1; $j < $K ; $j ++) { if ( $arr [ $i + $j ] > $max ) $max = $arr [ $i + $j ]; } printf( "%d " , $max ); } } // Driver's Code $arr = array (1, 2, 3, 4, 5, 6, 7, 8, 9, 10); $N = count ( $arr ); $K = 3; // Function call printKMax( $arr , $N , $K ); // This Code is Contributed by anuj_67. ?> |
Python3
# Python3 program for the above approach # Method to find the maximum for each # and every contiguous subarray # of size K def printMax(arr, N, K): max = 0 for i in range (N - K + 1 ): max = arr[i] for j in range ( 1 , K): if arr[i + j] > max : max = arr[i + j] print ( str ( max ) + " " , end = "") # Driver's code if __name__ = = "__main__" : arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 ] N = len (arr) K = 3 # Function call printMax(arr, N, K) # This code is contributed by Shiv Shankar |
3 4 5 6 7 8 9 10
Time Complexity: O(N * K), The outer loop runs N-K+1 times and the inner loop runs K times for every iteration of the outer loop. So time complexity is O((n-k+1)*k) which can also be written as O(N * K)
Auxiliary Space: O(1)
Maximum of all subarrays of size K using Max-Heap:
- Initialize an empty priority queue heap to store elements in decreasing order of their values, along with their indices.
- Push the first k elements of the input array arr into the priority queue heap along with their respective indices.
- The maximum element in the first window is obtained by accessing the top element of the priority queue heap. Push this maximum element into the answer vector ans.
- Process the remaining elements of arr starting from index k:
- Add the current element along with its index to the priority queue heap.
- Remove elements from the priority queue heap that are outside the current window. This is done by comparing the index of the top element in the heap with the index i – k. If the index of the top element is less than or equal to i – k, it means the element is outside the current window and should be removed.
- The maximum element in the current window is obtained by accessing the top element of the priority queue heap. Push this maximum element into the answer vector ans.
- Finally, return the answer vector ans containing the maximum elements in each sliding window.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h> using namespace std; // Function to find the maximum element in each sliding // window of size k vector< int > maxSlidingWindow(vector< int >& arr, int k) { vector< int > ans; priority_queue<pair< int , int > > heap; // Initialize the heap with the first k elements for ( int i = 0; i < k; i++) heap.push({ arr[i], i }); // The maximum element in the first window ans.push_back(heap.top().first); // Process the remaining elements for ( int i = k; i < arr.size(); i++) { // Add the current element to the heap heap.push({ arr[i], i }); // Remove elements that are outside the current // window while (heap.top().second <= i - k) heap.pop(); // The maximum element in the current window ans.push_back(heap.top().first); } return ans; } int main() { vector< int > arr = { 2, 3, 7, 9, 5, 1, 6, 4, 3 }; int k = 3; // Find the maximum element in each sliding window of // size k vector< int > result = maxSlidingWindow(arr, k); // Print the results for ( auto i : result) cout << i << " " ; return 0; } |
Java
import java.util.ArrayList; import java.util.List; import java.util.PriorityQueue; public class Main { public static List<Integer> maxSlidingWindow( int [] arr, int k) { List<Integer> ans = new ArrayList<>(); PriorityQueue<Pair> heap = new PriorityQueue<>( (a, b) -> b.value - a.value); // Initialize the heap with the first k elements for ( int i = 0 ; i < k; i++) { heap.offer( new Pair(arr[i], i)); } // The maximum element in the first window ans.add(heap.peek().value); // Process the remaining elements for ( int i = k; i < arr.length; i++) { heap.offer( new Pair(arr[i], i)); // Remove elements that are outside the current // window while (heap.peek().index <= i - k) { heap.poll(); } // The maximum element in the current window ans.add(heap.peek().value); } return ans; } static class Pair { int value; int index; public Pair( int value, int index) { this .value = value; this .index = index; } } public static void main(String[] args) { int [] arr = { 2 , 3 , 7 , 9 , 5 , 1 , 6 , 4 , 3 }; int k = 3 ; // Find the maximum element in each sliding window // of size k List<Integer> result = maxSlidingWindow(arr, k); // Print the results for ( int num : result) { System.out.print(num + " " ); } } } |
C#
using System; using System.Collections.Generic; class GFG { static List< int > MaxSlidingWindow(List< int > arr, int k) { List< int > ans = new List< int >(); Queue< int > deque = new Queue< int >(); for ( int i = 0; i < arr.Count; i++) { // Remove elements that are outside the current window // from the front of the deque while (deque.Count > 0 && deque.Peek() < i - k + 1) deque.Dequeue(); // Remove elements that are less than the current element // from the back of the deque while (deque.Count > 0 && arr[deque.Peek()] < arr[i]) deque.Dequeue(); // Add the current element's index to the back of the deque deque.Enqueue(i); // If the current index is equal to or greater than k - 1, // then we can start adding elements to the result list if (i >= k - 1) ans.Add(arr[deque.Peek()]); } return ans; } static void Main() { List< int > arr = new List< int > { 2, 3, 7, 9, 5, 1, 6, 4, 3 }; int k = 3; // Find the maximum element in each sliding window of size k List< int > result = MaxSlidingWindow(arr, k); // Print the results foreach ( int i in result) Console.Write(i + " " ); Console.ReadLine(); } } |
Javascript
class Pair { constructor(value, index) { this .value = value; this .index = index; } } function maxSlidingWindow(arr, k) { const ans = []; const heap = []; // Initialize the heap with the first k elements for (let i = 0; i < k; i++) { heap.push( new Pair(arr[i], i)); } heap.sort((a, b) => b.value - a.value); // The maximum element in the first window ans.push(heap[0].value); // Process the remaining elements for (let i = k; i < arr.length; i++) { heap.push( new Pair(arr[i], i)); // Remove elements that are outside the current window while (heap[0].index <= i - k) { heap.shift(); } heap.sort((a, b) => b.value - a.value); // The maximum element in the current window ans.push(heap[0].value); } return ans; } const arr = [2, 3, 7, 9, 5, 1, 6, 4, 3]; const k = 3; // Find the maximum element in each sliding window of size k const result = maxSlidingWindow(arr, k); // Print the results for (const num of result) { console.log(num + " " ); } |
Python3
import heapq def max_sliding_window(arr, k): ans = [] heap = [] # Initialize the heap with the first k elements for i in range (k): heapq.heappush(heap, ( - arr[i], i)) # The maximum element in the first window ans.append( - heap[ 0 ][ 0 ]) # Process the remaining elements for i in range (k, len (arr)): heapq.heappush(heap, ( - arr[i], i)) # Remove elements that are outside the current window while heap[ 0 ][ 1 ] < = i - k: heapq.heappop(heap) # The maximum element in the current window ans.append( - heap[ 0 ][ 0 ]) return ans arr = [ 2 , 3 , 7 , 9 , 5 , 1 , 6 , 4 , 3 ] k = 3 # Find the maximum element in each sliding window of size k result = max_sliding_window(arr, k) # Print the results for num in result: print (num, end = " " ) |
7 9 9 9 6 6 6
Time Complexity: O(NlogN), Where N is the size of the array.
Auxiliary Space: O(N), where N is the size of the array, this method requires O(N) space in the worst case when the input array is an increasing array
Maximum of all subarrays of size K using Set:
To reduce the auxilary space to O(K), Set Data Structure can be used which allows deletion of any element in O(logn N) time. Thus the size of set will never exceed K, if we delete delete the (i-Kt)h element from the Set while traversing.
Follow the given steps to solve the problem:
- Create a Set to store and find the maximum element.
- Traverse through the array from start to end.
- Insert the element in theSet.
- If the loop counter is greater than or equal to k then delete the i-Kth element from the Set.
- Print the maximum element of the Set.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h> using namespace std; // Function to find the maximum element in each sliding // window of size k vector< int > maxSlidingWindow(vector< int >& arr, int k) { vector< int > ans; set<pair< int , int >, greater<pair< int , int >>> st; // Initialize the set with the first k elements for ( int i = 0; i < k; i++) st.insert({ arr[i], i }); // The maximum element in the first window ans.push_back((st.begin())->first); // Process the remaining elements for ( int i = k; i < arr.size(); i++) { // Add the current element to the set st.insert({ arr[i], i }); // Remove the (i-k)th element from the window st.erase({ arr[i - k], (i - k) }); // The maximum element in the current window ans.push_back(st.begin()->first); } return ans; } int main() { vector< int > arr = { 2, 3, 7, 9, 5, 1, 6, 4, 3 }; int k = 3; // Find the maximum element in each sliding window of // size k vector< int > result = maxSlidingWindow(arr, k); // Print the results for ( auto i : result) cout << i << " " ; return 0; } |
Java
import java.util.ArrayDeque; import java.util.Deque; public class MaxSlidingWindow { // Function to find the maximum element in each sliding // window of size k static int [] maxSlidingWindow( int [] arr, int k) { int n = arr.length; int [] ans = new int [n - k + 1 ]; Deque<Integer> maxInWindow = new ArrayDeque<>(); // Initialize maxInWindow with the first k elements for ( int i = 0 ; i < k; i++) { while (!maxInWindow.isEmpty() && arr[i] >= arr[maxInWindow.peekLast()]) { maxInWindow.removeLast(); } maxInWindow.addLast(i); } // The maximum element in the first window ans[ 0 ] = arr[maxInWindow.peekFirst()]; // Process the remaining elements for ( int i = k; i < n; i++) { // Remove elements that are out of the current // window while (!maxInWindow.isEmpty() && maxInWindow.peekFirst() <= i - k) { maxInWindow.removeFirst(); } // Remove elements that are not needed in the // current window while (!maxInWindow.isEmpty() && arr[i] >= arr[maxInWindow.peekLast()]) { maxInWindow.removeLast(); } maxInWindow.addLast(i); ans[i - k + 1 ] = arr[maxInWindow.peekFirst()]; } return ans; } // Driver program public static void main(String[] args) { int [] arr = { 2 , 3 , 7 , 9 , 5 , 1 , 6 , 4 , 3 }; int k = 3 ; // Find the maximum element in each sliding window // of size k int [] result = maxSlidingWindow(arr, k); // Print the results System.out.print( "Maximum elements in each sliding window: " ); for ( int i : result) System.out.print(i + " " ); } } |
Python
from collections import deque # Function to find the maximum element in each sliding window of size k def maxSlidingWindow(arr, k): ans = [] deq = deque() # Initialize the deque with the first k elements for i in range (k): while deq and arr[i] > = arr[deq[ - 1 ]]: deq.pop() deq.append(i) # The maximum element in the first window ans.append(arr[deq[ 0 ]]) # Process the remaining elements for i in range (k, len (arr)): # Remove elements that are out of the current window while deq and deq[ 0 ] < = i - k: deq.popleft() # Remove elements that are less than the current element while deq and arr[i] > = arr[deq[ - 1 ]]: deq.pop() deq.append(i) # The maximum element in the current window ans.append(arr[deq[ 0 ]]) return ans # Main function if __name__ = = "__main__" : arr = [ 2 , 3 , 7 , 9 , 5 , 1 , 6 , 4 , 3 ] k = 3 # Find the maximum element in each sliding window of size k result = maxSlidingWindow(arr, k) # Print the results print (result) |
C#
using System; using System.Collections.Generic; class Program { // Function to find the maximum element in each sliding // window of size k static List< int > MaxSlidingWindow(List< int > arr, int k) { List< int > ans = new List< int >(); List< int > maxInWindow = new List< int >(); // Initialize maxInWindow with the first k elements for ( int i = 0; i < k; i++) { while ( maxInWindow.Count > 0 && arr[i] >= arr[maxInWindow[maxInWindow.Count - 1]]) { maxInWindow.RemoveAt(maxInWindow.Count - 1); } maxInWindow.Add(i); } // The maximum element in the first window ans.Add(arr[maxInWindow[0]]); // Process the remaining elements for ( int i = k; i < arr.Count; i++) { // Remove elements that are out of the current // window while (maxInWindow.Count > 0 && maxInWindow[0] <= i - k) { maxInWindow.RemoveAt(0); } // Remove elements that are not needed in the // current window while ( maxInWindow.Count > 0 && arr[i] >= arr[maxInWindow[maxInWindow.Count - 1]]) { maxInWindow.RemoveAt(maxInWindow.Count - 1); } maxInWindow.Add(i); ans.Add(arr[maxInWindow[0]]); } return ans; } static void Main( string [] args) { List< int > arr = new List< int >{ 2, 3, 7, 9, 5, 1, 6, 4, 3 }; int k = 3; List< int > result = MaxSlidingWindow(arr, k); Console.WriteLine( "Maximum elements in each sliding window:" ); foreach ( int i in result) { Console.Write(i + " " ); } } } |
Javascript
function maxSlidingWindow(arr, k) { const ans = []; const maxInWindow = []; // Initialize maxInWindow with the first k elements for (let i = 0; i < k; i++) { while (maxInWindow.length > 0 && arr[i] >= arr[maxInWindow[maxInWindow.length - 1]]) { maxInWindow.pop(); } maxInWindow.push(i); } // The maximum element in the first window ans.push(arr[maxInWindow[0]]); // Process the remaining elements for (let i = k; i < arr.length; i++) { // Remove elements that are out of the current window while (maxInWindow.length > 0 && maxInWindow[0] <= i - k) { maxInWindow.shift(); } // Remove elements that are not needed in the current window while (maxInWindow.length > 0 && arr[i] >= arr[maxInWindow[maxInWindow.length - 1]]) { maxInWindow.pop(); } maxInWindow.push(i); ans.push(arr[maxInWindow[0]]); } return ans; } const arr = [2, 3, 7, 9, 5, 1, 6, 4, 3]; const k = 3; const result = maxSlidingWindow(arr, k); console.log( "Maximum elements in each sliding window: " + result.join( ' ' )); |
7 9 9 9 6 6 6
Time Complexity: O(NlogN), Where N is the size of the array.
Auxiliary Space: O(K), since size of set does not never exceeds K.
Maximum of all subarrays of size K using Deque:
Create a Deque, Qi of capacity K, that stores only useful elements of current window of K elements. An element is useful if it is in current window and is greater than all other elements on right side of it in current window. Process all array elements one by one and maintain Qi to contain useful elements of current window and these useful elements are maintained in sorted order. The element at front of the Qi is the largest and element at rear/back of Qi is the smallest of current window.
Below is the dry run of the above approach:
Follow the given steps to solve the problem:
- Create a deque to store K elements.
- Run a loop and insert the first K elements in the deque. Before inserting the element, check if the element at the back of the queue is smaller than the current element, if it is so remove the element from the back of the deque until all elements left in the deque are greater than the current element. Then insert the current element, at the back of the deque.
- Now, run a loop from K to the end of the array.
- Print the front element of the deque.
- Remove the element from the front of the queue if they are out of the current window.
- Insert the next element in the deque. Before inserting the element, check if the element at the back of the queue is smaller than the current element, if it is so remove the element from the back of the deque until all elements left in the deque are greater than the current element. Then insert the current element, at the back of the deque.
- Print the maximum element of the last window.
Below is the implementation of the above approach:
C++
// CPP program for the above approach #include <bits/stdc++.h> using namespace std; // A Dequeue (Double ended queue) based // method for printing maximum element of // all subarrays of size k void printKMax( int arr[], int N, int K) { // Create a Double Ended Queue, // Qi that will store indexes // of array elements // The queue will store indexes // of useful elements in every // window and it will // maintain decreasing order of // values from front to rear in Qi, i.e., // arr[Qi.front[]] to arr[Qi.rear()] // are sorted in decreasing order std::deque< int > Qi(K); /* Process first k (or first window) elements of array */ int i; for (i = 0; i < K; ++i) { // For every element, the previous // smaller elements are useless so // remove them from Qi while ((!Qi.empty()) && arr[i] >= arr[Qi.back()]) // Remove from rear Qi.pop_back(); // Add new element at rear of queue Qi.push_back(i); } // Process rest of the elements, // i.e., from arr[k] to arr[n-1] for (; i < N; ++i) { // The element at the front of // the queue is the largest element of // previous window, so print it cout << arr[Qi.front()] << " " ; // Remove the elements which // are out of this window while ((!Qi.empty()) && Qi.front() <= i - K) // Remove from front of queue Qi.pop_front(); // Remove all elements // smaller than the currently // being added element (remove // useless elements) while ((!Qi.empty()) && arr[i] >= arr[Qi.back()]) Qi.pop_back(); // Add current element at the rear of Qi Qi.push_back(i); } // Print the maximum element // of last window cout << arr[Qi.front()]; } // Driver's code int main() { int arr[] = { 12, 1, 78, 90, 57, 89, 56 }; int N = sizeof (arr) / sizeof (arr[0]); int K = 3; // Function call printKMax(arr, N, K); return 0; } |
Java
// Java Program to find the maximum for // each and every contiguous subarray of size K. import java.util.Deque; import java.util.LinkedList; public class SlidingWindow { // A Dequeue (Double ended queue) // based method for printing // maximum element of // all subarrays of size K static void printMax( int arr[], int N, int K) { // Create a Double Ended Queue, Qi // that will store indexes of array elements // The queue will store indexes of // useful elements in every window and it will // maintain decreasing order of values // from front to rear in Qi, i.e., // arr[Qi.front[]] to arr[Qi.rear()] // are sorted in decreasing order Deque<Integer> Qi = new LinkedList<Integer>(); /* Process first k (or first window) elements of array */ int i; for (i = 0 ; i < K; ++i) { // For every element, the previous // smaller elements are useless so // remove them from Qi while (!Qi.isEmpty() && arr[i] >= arr[Qi.peekLast()]) // Remove from rear Qi.removeLast(); // Add new element at rear of queue Qi.addLast(i); } // Process rest of the elements, // i.e., from arr[k] to arr[n-1] for (; i < N; ++i) { // The element at the front of the // queue is the largest element of // previous window, so print it System.out.print(arr[Qi.peek()] + " " ); // Remove the elements which // are out of this window while ((!Qi.isEmpty()) && Qi.peek() <= i - K) Qi.removeFirst(); // Remove all elements smaller // than the currently // being added element (remove // useless elements) while ((!Qi.isEmpty()) && arr[i] >= arr[Qi.peekLast()]) Qi.removeLast(); // Add current element at the rear of Qi Qi.addLast(i); } // Print the maximum element of last window System.out.print(arr[Qi.peek()]); } // Driver's code public static void main(String[] args) { int arr[] = { 12 , 1 , 78 , 90 , 57 , 89 , 56 }; int K = 3 ; // Function call printMax(arr, arr.length, K); } } // This code is contributed by Sumit Ghosh |
C#
// C# Program to find the maximum for each // and every contiguous subarray of size K. using System; using System.Collections.Generic; public class SlidingWindow { // A Dequeue (Double ended queue) based // method for printing maximum element of // all subarrays of size K static void printMax( int [] arr, int N, int K) { // Create a Double Ended Queue, Qi that // will store indexes of array elements // The queue will store indexes of useful // elements in every window and it will // maintain decreasing order of values // from front to rear in Qi, i.e., // arr[Qi.front[]] to arr[Qi.rear()] // are sorted in decreasing order List< int > Qi = new List< int >(); /* Process first K (or first window) elements of array */ int i; for (i = 0; i < K; ++i) { // For every element, the previous // smaller elements are useless so // remove them from Qi while (Qi.Count != 0 && arr[i] >= arr[Qi[Qi.Count - 1]]) // Remove from rear Qi.RemoveAt(Qi.Count - 1); // Add new element at rear of queue Qi.Insert(Qi.Count, i); } // Process rest of the elements, // i.e., from arr[k] to arr[n-1] for (; i < N; ++i) { // The element at the front of // the queue is the largest element of // previous window, so print it Console.Write(arr[Qi[0]] + " " ); // Remove the elements which are // out of this window while ((Qi.Count != 0) && Qi[0] <= i - K) Qi.RemoveAt(0); // Remove all elements smaller // than the currently // being added element (remove // useless elements) while ((Qi.Count != 0) && arr[i] >= arr[Qi[Qi.Count - 1]]) Qi.RemoveAt(Qi.Count - 1); // Add current element at the rear of Qi Qi.Insert(Qi.Count, i); } // Print the maximum element of last window Console.Write(arr[Qi[0]]); } // Driver's code public static void Main(String[] args) { int [] arr = { 12, 1, 78, 90, 57, 89, 56 }; int K = 3; // Function call printMax(arr, arr.Length, K); } } // This code has been contributed by 29AjayKumar |
Javascript
// We have used array in javascript to implement methods of dequeue // A Dequeue (Double ended queue) based // method for printing maximum element of // all subarrays of size k function printKMax(arr,n,k) { // creating string str to be printed at last let str = "" ; // Create a Double Ended Queue, // Qi that will store indexes // of array elements // The queue will store indexes // of useful elements in every // window and it will // maintain decreasing order of // values from front to rear in Qi, i.e., // arr[Qi.front[]] to arr[Qi.rear()] // are sorted in decreasing order // std::deque<int> Qi(k); let Qi = []; /* Process first k (or first window) elements of array */ let i; for (i = 0; i < k; ++i) { // For every element, the previous // smaller elements are useless so // remove them from Qi while ((Qi.length!=0) && arr[i] >= arr[Qi[Qi.length-1]]) // Remove from rear Qi.pop(); // Add new element at rear of queue Qi.push(i); } // Process rest of the elements, // i.e., from arr[k] to arr[n-1] for (i; i < n; ++i) { // The element at the front of // the queue is the largest element of // previous window, so print it str+=arr[Qi[0]] + " " ; // console.log(arr[Qi[0]] + " ") ; // Remove the elements which // are out of this window while ((Qi.length!=0) && Qi[0] <= i - k) // Remove from front of queue Qi.shift(); // Remove all elements // smaller than the currently // being added element (remove // useless elements) while ((Qi.length!=0) && arr[i] >= arr[Qi[Qi.length-1]]) Qi.pop(); // Add current element at the rear of Qi Qi.push(i); } // Print the maximum element // of last window str += arr[Qi[0]]; console.log(str); } let arr = [ 12, 1, 78, 90, 57, 89, 56 ]; let n = arr.length; let k = 3; printKMax(arr, n, k); // This code is contributed by akashish__ |
Python3
# Python3 program to find the maximum for # each and every contiguous subarray of # size K from collections import deque # A Deque (Double ended queue) based # method for printing maximum element # of all subarrays of size K def printMax(arr, N, K): """ Create a Double Ended Queue, Qi that will store indexes of array elements. The queue will store indexes of useful elements in every window and it will maintain decreasing order of values from front to rear in Qi, i.e., arr[Qi.front[]] to arr[Qi.rear()] are sorted in decreasing order""" Qi = deque() # Process first k (or first window) # elements of array for i in range (K): # For every element, the previous # smaller elements are useless # so remove them from Qi while Qi and arr[i] > = arr[Qi[ - 1 ]]: Qi.pop() # Add new element at rear of queue Qi.append(i) # Process rest of the elements, i.e. # from arr[k] to arr[n-1] for i in range (K, N): # The element at the front of the # queue is the largest element of # previous window, so print it print ( str (arr[Qi[ 0 ]]) + " " , end = "") # Remove the elements which are # out of this window while Qi and Qi[ 0 ] < = i - K: # remove from front of deque Qi.popleft() # Remove all elements smaller than # the currently being added element # (Remove useless elements) while Qi and arr[i] > = arr[Qi[ - 1 ]]: Qi.pop() # Add current element at the rear of Qi Qi.append(i) # Print the maximum element of last window print ( str (arr[Qi[ 0 ]])) # Driver's code if __name__ = = "__main__" : arr = [ 12 , 1 , 78 , 90 , 57 , 89 , 56 ] K = 3 # Function call printMax(arr, len (arr), K) # This code is contributed by Shiv Shankar |
78 90 90 90 89
Time Complexity: O(N). It seems more than O(N) at first look. It can be observed that every element of the array is added and removed at most once. So there are a total of 2n operations.
Auxiliary Space: O(K). Elements stored in the dequeue take O(K) space.
Below is an extension of this problem:
Sum of minimum and maximum elements of all subarrays of size k.