Sliding Window Maximum (Maximum of all subarrays of size K)
Last Updated :
23 Jul, 2025
Given an array arr[] of integers and an integer k, your task is to find the maximum value for each contiguous subarray of size k. The output should be an array of maximum values corresponding to each contiguous subarray.
Examples :
Input: arr[] = [1, 2, 3, 1, 4, 5, 2, 3, 6], k = 3
Output: [3, 3, 4, 5, 5, 5, 6]
Explanation:
1st contiguous subarray = [1 2 3] max = 3
2nd contiguous subarray = [2 3 1] max = 3
3rd contiguous subarray = [3 1 4] max = 4
4th contiguous subarray = [1 4 5] max = 5
5th contiguous subarray = [4 5 2] max = 5
6th contiguous subarray = [5 2 3] max = 5
7th contiguous subarray = [2 3 6] max = 6
Input: arr[] = [5, 1, 3, 4, 2, 6], k = 1
Output: [5, 1, 3, 4, 2, 6]
Explanation: When k = 1, each element in the array is its own subarray, so the output is simply the same array.
Input: arr[] = [1, 3, 2, 1, 7, 3], k = 3
Output: [3, 3, 7, 7]
[Naive Approach] - Using Nested Loops - O(n * k) Time and O(1) Space
The idea is to run the nested loops, the outer loop will mark the starting point of the subarray of length k, and the inner loop will run from the starting index to index + k, and print the maximum element among these k elements.
Below is the implementation of the above approach:
C++
#include <iostream>
#include <vector>
using namespace std;
// Method to find the maximum for each
// and every contiguous subarray of size k.
vector<int> maxOfSubarrays(vector<int>& arr, int k) {
int n = arr.size();
// to store the results
vector<int> res;
for (int i = 0; i <= n - k; i++) {
// Find maximum of subarray beginning
// with arr[i]
int max = arr[i];
for (int j = 1; j < k; j++) {
if (arr[i + j] > max)
max = arr[i + j];
}
res.push_back(max);
}
return res;
}
int main() {
vector<int> arr = { 1, 2, 3, 1, 4, 5, 2, 3, 6 };
int k = 3;
vector<int> res = maxOfSubarrays(arr, k);
for (int maxVal : res) {
cout << maxVal << " ";
}
return 0;
}
import java.util.ArrayList;
import java.util.*;
class GfG {
// Method to find the maximum for each
// and every contiguous subarray of size k.
static ArrayList<Integer> maxOfSubarrays(int[] arr, int k) {
int n = arr.length;
// to store the results
ArrayList<Integer> res = new ArrayList<Integer>();
for (int i = 0; i <= n - k; i++) {
// Find maximum of subarray beginning
// with arr[i]
int max = arr[i];
for (int j = 1; j < k; j++) {
if (arr[i + j] > max)
max = arr[i + j];
}
res.add(max);
}
return res;
}
public static void main(String[] args) {
int[] arr = { 1, 2, 3, 1, 4, 5, 2, 3, 6 };
int k = 3;
ArrayList<Integer> res = maxOfSubarrays(arr, k);
for (int maxVal : res) {
System.out.print(maxVal + " ");
}
}
}
# Method to find the maximum for each
# and every contiguous subarray of size k.
def maxOfSubarrays(arr, k):
n = len(arr)
# to store the results
res = []
for i in range(0, n - k + 1):
# Find maximum of subarray beginning
# with arr[i]
max = arr[i]
for j in range(1, k):
if arr[i + j] > max:
max = arr[i + j]
res.append(max)
return res
if __name__ == "__main__":
arr = [ 1, 2, 3, 1, 4, 5, 2, 3, 6 ]
k = 3
res = maxOfSubarrays(arr, k)
for maxVal in res:
print(maxVal, end=" ")
using System;
using System.Collections.Generic;
class GfG {
// Method to find the maximum for each
// and every contiguous subarray of size k.
static List<int> maxOfSubarrays(int[] arr, int k) {
int n = arr.Length;
// to store the results
List<int> res = new List<int>();
for (int i = 0; i <= n - k; i++) {
// Find maximum of subarray beginning
// with arr[i]
int max = arr[i];
for (int j = 1; j < k; j++) {
if (arr[i + j] > max)
max = arr[i + j];
}
res.Add(max);
}
return res;
}
static void Main() {
int[] arr = { 1, 2, 3, 1, 4, 5, 2, 3, 6 };
int k = 3;
List<int> res = maxOfSubarrays(arr, k);
foreach (int maxVal in res) {
Console.Write(maxVal + " ");
}
}
}
// Method to find the maximum for each
// and every contiguous subarray of size k.
function maxOfSubarrays(arr, k) {
let n = arr.length;
// to store the results
let res = [];
for (let i = 0; i <= n - k; i++) {
// Find maximum of subarray beginning
// with arr[i]
let max = arr[i];
for (let j = 1; j < k; j++) {
if (arr[i + j] > max)
max = arr[i + j];
}
res.push(max);
}
return res;
}
// Driver Code
let arr = [ 1, 2, 3, 1, 4, 5, 2, 3, 6 ];
let k = 3;
let res = maxOfSubarrays(arr, k);
console.log(res.join(" "));
Time Complexity: O(n * k), as we are using nested loops, where the outer loop runs n times, and for each iteration of outer loop, inner loop runs k times.
Auxiliary Space: O(1)
[Better Approach] - Using Max-Heap - (n * log n) Time and O(n) Space
The idea is to use priority queue or heap data structure to make sure that heap has largest item of the current window.
- Create a max heap of the first k items
- Now iterate one by one. While the next item to be added is greater than the heap top, remove the top. We mainly make sure that the greater items of the previous window are not there in a heap.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
// Method to find the maximum for each
// and every contiguous subarray of size k.
vector<int> maxOfSubarrays(const vector<int>& arr, int k) {
int n = arr.size();
// to store the results
vector<int> res;
// to store the max value
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
res.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
res.push_back(heap.top().first);
}
return res;
}
int main() {
vector<int> arr = { 1, 2, 3, 1, 4, 5, 2, 3, 6 };
int k = 3;
vector<int> res = maxOfSubarrays(arr, k);
for (int maxVal : res) {
cout << maxVal << " ";
}
return 0;
}
import java.util.ArrayList;
import java.util.Comparator;
import java.util.PriorityQueue;
class GfG {
// Method to find the maximum for each
// and every contiguous subarray of size k.
static ArrayList<Integer> maxOfSubarrays(int[] arr, int k) {
int n = arr.length;
// to store the results
ArrayList<Integer> res = new ArrayList<Integer>();
// to store the max value
PriorityQueue<Pair> heap = new PriorityQueue<Pair>(new Comparator<Pair>() {
public int compare(Pair a, Pair b) {
return b.first - a.first;
}
});
// Initialize the heap with the first k elements
for (int i = 0; i < k; i++)
heap.add(new Pair(arr[i], i));
// The maximum element in the first window
res.add(heap.peek().first);
// Process the remaining elements
for (int i = k; i < arr.length; i++) {
// Add the current element to the heap
heap.add(new Pair(arr[i], i));
// Remove elements that are outside the current
// window
while (heap.peek().second <= i - k)
heap.poll();
// The maximum element in the current window
res.add(heap.peek().first);
}
return res;
}
static class Pair {
int first;
int second;
Pair(int first, int second) {
this.first = first;
this.second = second;
}
}
public static void main(String[] args) {
int[] arr = { 1, 2, 3, 1, 4, 5, 2, 3, 6 };
int k = 3;
ArrayList<Integer> res = maxOfSubarrays(arr, k);
for (int maxVal : res) {
System.out.print(maxVal + " ");
}
}
}
import heapq
# Method to find the maximum for each
# and every contiguous subarray of size k.
def maxOfSubarrays(arr, k):
n = len(arr)
# to store the results
res = []
# to store the max value
heap = []
# Initialize the heap with the first k elements
for i in range(0, k):
heapq.heappush(heap, (-arr[i], i))
# The maximum element in the first window
res.append(-heap[0][0])
# Process the remaining elements
for i in range(k, len(arr)):
# Add the current element to the heap
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
res.append(-heap[0][0])
return res
if __name__ == "__main__":
arr = [ 1, 2, 3, 1, 4, 5, 2, 3, 6 ]
k = 3
res = maxOfSubarrays(arr, k)
for maxVal in res:
print(maxVal, end=" ")
using System;
using System.Collections.Generic;
class GfG {
// Method to find the maximum for each
// and every contiguous subarray of size k.
static List<int> maxOfSubarrays(int[] arr, int k) {
int n = arr.Length;
// to store the results
List<int> res = new List<int>();
// to store the max value
// Using SortedSet to simulate a max-heap
SortedSet<Pair> heap = new SortedSet<Pair>(new PairComparer());
// Initialize the heap with the first k elements
for (int i = 0; i < k; i++)
heap.Add(new Pair(arr[i], i));
// The maximum element in the first window
res.Add(heap.Min.first);
// Process the remaining elements
for (int i = k; i < arr.Length; i++) {
// Add the current element to the heap
heap.Add(new Pair(arr[i], i));
// Remove elements that are outside the current
// window
while (heap.Min.second <= i - k)
heap.Remove(heap.Min);
// The maximum element in the current window
res.Add(heap.Min.first);
}
return res;
}
class Pair {
public int first;
public int second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
}
class PairComparer : IComparer<Pair> {
public int Compare(Pair a, Pair b) {
if (a.first != b.first)
return b.first.CompareTo(a.first);
return a.second.CompareTo(b.second);
}
}
static void Main() {
int[] arr = { 1, 2, 3, 1, 4, 5, 2, 3, 6 };
int k = 3;
List<int> res = maxOfSubarrays(arr, k);
foreach (int maxVal in res) {
Console.Write(maxVal + " ");
}
}
}
// Method to find the maximum for each
// and every contiguous subarray of size k.
function maxOfSubarrays(arr, k) {
let n = arr.length;
// to store the results
let res = [];
// to store the max value
let heap = [];
// Function to sort the heap in descending order based on 'first'
function sortHeap() {
heap.sort(function(a, b) {
if (a.first !== b.first)
return b.first - a.first;
return a.second - b.second;
});
}
// Initialize the heap with the first k elements
for (let i = 0; i < k; i++)
heap.push({ first: arr[i], second: i });
sortHeap();
// The maximum element in the first window
res.push(heap[0].first);
// Process the remaining elements
for (let i = k; i < arr.length; i++) {
// Add the current element to the heap
heap.push({ first: arr[i], second: i });
sortHeap();
// Remove elements that are outside the current
// window
while (heap[0].second <= i - k) {
heap.shift();
sortHeap();
}
// The maximum element in the current window
res.push(heap[0].first);
}
return res;
}
// Driver Code
let arr = [ 1, 2, 3, 1, 4, 5, 2, 3, 6 ];
let k = 3;
let res = maxOfSubarrays(arr, k);
console.log(res.join(" "));
Time Complexity: O(n log n), where n is the size of the array. Inserting an element in heap takes (log n) time and we are inserting all n elements, thus the time complexity will be O(n * log n).
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
[Expected Approach] - Using Deque - O(n) Time and O(k) Space
Create a Deque, dq 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 dq to contain useful elements of current window and these useful elements are maintained in sorted order. The element at front of the dq is the largest and element at rear/back of dq is the smallest of current window.
Illustration:
Step-by-Step Algorithm:
- Create a deque to store only useful elements of current window.
- 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++
#include <bits/stdc++.h>
using namespace std;
// Method to find the maximum for each
// and every contiguous subarray of size k.
vector<int> maxOfSubarrays(vector<int>& arr, int k) {
// to store the results
vector<int> res;
// create deque to store max values
deque<int> dq(k);
// Process first k (or first window) elements of array
for (int i = 0; i < k; ++i) {
// For every element, the previous smaller elements
// are useless so remove them from dq
while (!dq.empty() && arr[i] >= arr[dq.back()]) {
// Remove from rear
dq.pop_back();
}
// Add new element at rear of queue
dq.push_back(i);
}
// Process rest of the elements, i.e., from arr[k] to arr[n-1]
for (int i = k; i < arr.size(); ++i) {
// The element at the front of the queue is the largest
// element of previous window, so store it
res.push_back(arr[dq.front()]);
// Remove the elements which are out of this window
while (!dq.empty() && dq.front() <= i - k) {
// Remove from front of queue
dq.pop_front();
}
// Remove all elements smaller than the currently being
// added element (remove useless elements)
while (!dq.empty() && arr[i] >= arr[dq.back()]) {
dq.pop_back();
}
// Add current element at the rear of dq
dq.push_back(i);
}
// store the maximum element of last window
res.push_back(arr[dq.front()]);
return res;
}
int main() {
vector<int> arr = {1, 3, 2, 1, 7, 3};
int k = 3;
vector<int> res = maxOfSubarrays(arr, k);
for (int maxVal : res) {
cout << maxVal << " ";
}
return 0;
}
import java.util.ArrayList;
import java.util.Deque;
import java.util.ArrayDeque;
class GfG {
// Method to find the maximum for each
// and every contiguous subarray of size k.
static ArrayList<Integer> maxOfSubarrays(int[] arr, int k) {
int n = arr.length;
// to store the results
ArrayList<Integer> res = new ArrayList<Integer>();
// create deque to store max values
Deque<Integer> dq = new ArrayDeque<Integer>();
// Process first k (or first window) elements of array
for (int i = 0; i < k; ++i) {
// For every element, the previous smaller elements
// are useless so remove them from dq
while (!dq.isEmpty() && arr[i] >= arr[dq.peekLast()]) {
// Remove from rear
dq.pollLast();
}
// Add new element at rear of queue
dq.addLast(i);
}
// Process rest of the elements, i.e., from arr[k] to arr[n-1]
for (int i = k; i < arr.length; ++i) {
// The element at the front of the queue is the largest
// element of previous window, so store it
res.add(arr[dq.peekFirst()]);
// Remove the elements which are out of this window
while (!dq.isEmpty() && dq.peekFirst() <= i - k) {
// Remove from front of queue
dq.pollFirst();
}
// Remove all elements smaller than the currently being
// added element (remove useless elements)
while (!dq.isEmpty() && arr[i] >= arr[dq.peekLast()]) {
dq.pollLast();
}
// Add current element at the rear of dq
dq.addLast(i);
}
// store the maximum element of last window
res.add(arr[dq.peekFirst()]);
return res;
}
public static void main(String[] args) {
int[] arr = {1, 3, 2, 1, 7, 3};
int k = 3;
ArrayList<Integer> res = maxOfSubarrays(arr, k);
for (int maxVal : res) {
System.out.print(maxVal + " ");
}
}
}
from collections import deque
# Method to find the maximum for each
# and every contiguous subarray of size k.
def maxOfSubarrays(arr, k):
n = len(arr)
# to store the results
res = []
# create deque to store max values
dq = deque()
# Process first k (or first window) elements of array
for i in range(0, k):
# For every element, the previous smaller elements
# are useless so remove them from dq
while dq and arr[i] >= arr[dq[-1]]:
# Remove from rear
dq.pop()
# Add new element at rear of queue
dq.append(i)
# Process rest of the elements, i.e., from arr[k] to arr[n-1]
for i in range(k, len(arr)):
# The element at the front of the queue is the largest
# element of previous window, so store it
res.append(arr[dq[0]])
# Remove the elements which are out of this window
while dq and dq[0] <= i - k:
# Remove from front of queue
dq.popleft()
# Remove all elements smaller than the currently being
# added element (remove useless elements)
while dq and arr[i] >= arr[dq[-1]]:
dq.pop()
# Add current element at the rear of dq
dq.append(i)
# store the maximum element of last window
res.append(arr[dq[0]])
return res
if __name__ == "__main__":
arr = [1, 3, 2, 1, 7, 3]
k = 3
res = maxOfSubarrays(arr, k)
for maxVal in res:
print(maxVal, end=" ")
using System;
using System.Collections.Generic;
class GfG {
// Method to find the maximum for each
// and every contiguous subarray of size k.
static List<int> maxOfSubarrays(int[] arr, int k) {
int n = arr.Length;
// to store the results
List<int> res = new List<int>();
// create deque to store max values
LinkedList<int> dq = new LinkedList<int>();
// Process first k (or first window) elements of array
for (int i = 0; i < k; ++i) {
// For every element, the previous smaller elements
// are useless so remove them from dq
while (dq.Count > 0 && arr[i] >= arr[dq.Last.Value]) {
// Remove from rear
dq.RemoveLast();
}
// Add new element at rear of queue
dq.AddLast(i);
}
// Process rest of the elements, i.e., from arr[k] to arr[n-1]
for (int i = k; i < arr.Length; ++i) {
// The element at the front of the queue is the largest
// element of previous window, so store it
res.Add(arr[dq.First.Value]);
// Remove the elements which are out of this window
while (dq.Count > 0 && dq.First.Value <= i - k) {
// Remove from front of queue
dq.RemoveFirst();
}
// Remove all elements smaller than the currently being
// added element (remove useless elements)
while (dq.Count > 0 && arr[i] >= arr[dq.Last.Value]) {
dq.RemoveLast();
}
// Add current element at the rear of dq
dq.AddLast(i);
}
// store the maximum element of last window
res.Add(arr[dq.First.Value]);
return res;
}
static void Main() {
int[] arr = {1, 3, 2, 1, 7, 3};
int k = 3;
List<int> res = maxOfSubarrays(arr, k);
foreach (int maxVal in res) {
Console.Write(maxVal + " ");
}
}
}
// Method to find the maximum for each
// and every contiguous subarray of size k.
function maxOfSubarrays(arr, k) {
let n = arr.length;
// to store the results
let res = [];
// create deque to store max values
let dq = [];
// Process first k (or first window) elements of array
for (let i = 0; i < k; ++i) {
// For every element, the previous smaller elements
// are useless so remove them from dq
while (dq.length > 0 && arr[i] >= arr[dq[dq.length - 1]]) {
// Remove from rear
dq.pop();
}
// Add new element at rear of queue
dq.push(i);
}
// Process rest of the elements, i.e., from arr[k] to arr[n-1]
for (let i = k; i < arr.length; ++i) {
// The element at the front of the queue is the largest
// element of previous window, so store it
res.push(arr[dq[0]]);
// Remove the elements which are out of this window
while (dq.length > 0 && dq[0] <= i - k) {
// Remove from front of queue
dq.shift();
}
// Remove all elements smaller than the currently being
// added element (remove useless elements)
while (dq.length > 0 && arr[i] >= arr[dq[dq.length - 1]]) {
dq.pop();
}
// Add current element at the rear of dq
dq.push(i);
}
// store the maximum element of last window
res.push(arr[dq[0]]);
return res;
}
// Driver Code
let arr = [1, 3, 2, 1, 7, 3];
let k = 3;
let res = maxOfSubarrays(arr, k);
console.log(res.join(" "));
Time Complexity: O(n), 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.
Sliding Window Maximum (Maximum of all subarrays of size K)
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Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
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