# Sliding Window Maximum : Set 2

Given an array arr of size N and an integer K, the task is to 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
All ccontigious subarrays of size k are
{1, 2, 3} => 3
{2, 3, 1} => 3
{3, 1, 4} => 4
{1, 4, 5} => 5
{4, 5, 2} => 5
{5, 2, 3} => 5
{2, 3, 6} => 6

Input: arr[] = {8, 5, 10, 7, 9, 4, 15, 12, 90, 13}, K = 4
Output: 10 10 10 15 15 90 90

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: To solve this in lesser space complexity we can use two pointer technique.

• First variable pointer iterates through the subarray and finds maximum element from given size K
• Second variable pointer marks the ending index of the first variable pointer i.e., (i + K – 1)th index.
• When the first variable pointer reaches the index of second variable pointer, maximum of that subarray has been computed and will be printed.
• The process is repeated until second variable pointer reach last array index (i.e array_size – 1).

Below is the implementation of the above approach:

## C++

 `// C++ program to find the maximum for each ` `// and every contiguous subarray of size K ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the maximum for each ` `// and every contiguous subarray of size k ` `void` `printKMax(``int` `a[], ``int` `n, ``int` `k) ` `{ ` `    ``// If k = 1, print all elements ` `    ``if` `(k == 1) { ` `        ``for` `(``int` `i = 0; i < n; i += 1) ` `            ``cout << a[i] << ``" "``; ` `        ``return``; ` `    ``} ` ` `  `    ``// Using p and q as variable pointers ` `    ``// where p iterates through the subarray ` `    ``// and q marks end of the subarray. ` `    ``int` `p = 0, ` `        ``q = k - 1, ` `        ``t = p, ` `        ``max = a[k - 1]; ` ` `  `    ``// Iterating through subarray. ` `    ``while` `(q <= n - 1) { ` ` `  `        ``// Finding max ` `        ``// from the subarray. ` `        ``if` `(a[p] > max) ` `            ``max = a[p]; ` ` `  `        ``p += 1; ` ` `  `        ``// Printing max of subarray ` `        ``// and shifting pointers ` `        ``// to next index. ` `        ``if` `(q == p && p != n) { ` `            ``cout << max << ``" "``; ` `            ``q++; ` `            ``p = ++t; ` ` `  `            ``if` `(q < n) ` `                ``max = a[q]; ` `        ``} ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `a[] = { 1, 2, 3, 4, 5, ` `                ``6, 7, 8, 9, 10 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` `    ``int` `K = 3; ` ` `  `    ``printKMax(a, n, K); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the maximum for each ` `// and every contiguous subarray of size K ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function to find the maximum for each ` `// and every contiguous subarray of size k ` `static` `void` `printKMax(``int` `a[], ``int` `n, ``int` `k) ` `{ ` `    ``// If k = 1, print all elements ` `    ``if` `(k == ``1``) { ` `        ``for` `(``int` `i = ``0``; i < n; i += ``1``) ` `            ``System.out.print(a[i]+ ``" "``); ` `        ``return``; ` `    ``} ` `  `  `    ``// Using p and q as variable pointers ` `    ``// where p iterates through the subarray ` `    ``// and q marks end of the subarray. ` `    ``int` `p = ``0``, ` `        ``q = k - ``1``, ` `        ``t = p, ` `        ``max = a[k - ``1``]; ` `  `  `    ``// Iterating through subarray. ` `    ``while` `(q <= n - ``1``) { ` `  `  `        ``// Finding max ` `        ``// from the subarray. ` `        ``if` `(a[p] > max) ` `            ``max = a[p]; ` `  `  `        ``p += ``1``; ` `  `  `        ``// Printing max of subarray ` `        ``// and shifting pointers ` `        ``// to next index. ` `        ``if` `(q == p && p != n) { ` `            ``System.out.print(max+ ``" "``); ` `            ``q++; ` `            ``p = ++t; ` `  `  `            ``if` `(q < n) ` `                ``max = a[q]; ` `        ``} ` `    ``} ` `} ` `  `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `a[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ` `                ``6``, ``7``, ``8``, ``9``, ``10` `}; ` `    ``int` `n = a.length; ` `    ``int` `K = ``3``; ` `  `  `    ``printKMax(a, n, K); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 program to find the maximum for each ` `# and every contiguous subarray of size K ` ` `  `# Function to find the maximum for each ` `# and every contiguous subarray of size k ` `def` `printKMax(a, n, k): ` `     `  `    ``# If k = 1, prall elements ` `    ``if` `(k ``=``=` `1``): ` `        ``for` `i ``in` `range``(n): ` `            ``print``(a[i], end``=``" "``); ` `        ``return``; ` `         `  `    ``# Using p and q as variable pointers ` `    ``# where p iterates through the subarray ` `    ``# and q marks end of the subarray. ` `    ``p ``=` `0``; ` `    ``q ``=` `k ``-` `1``; ` `    ``t ``=` `p; ` `    ``max` `=` `a[k ``-` `1``]; ` ` `  `    ``# Iterating through subarray. ` `    ``while` `(q <``=` `n ``-` `1``): ` ` `  `        ``# Finding max ` `        ``# from the subarray. ` `        ``if` `(a[p] > ``max``): ` `            ``max` `=` `a[p]; ` `        ``p ``+``=` `1``; ` ` `  `        ``# Printing max of subarray ` `        ``# and shifting pointers ` `        ``# to next index. ` `        ``if` `(q ``=``=` `p ``and` `p !``=` `n): ` `            ``print``(``max``, end``=``" "``); ` `            ``q ``+``=` `1``; ` `            ``p ``=` `t ``+` `1``; ` ` `  `            ``if` `(q < n): ` `                ``max` `=` `a[q]; ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``a ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``8``, ``9``, ``10``]; ` `    ``n ``=` `len``(a); ` `    ``K ``=` `3``; ` ` `  `    ``printKMax(a, n, K); ` ` `  `# This code is contributed by Princi Singh `

## C#

 `// C# program to find the maximum for each ` `// and every contiguous subarray of size K ` `using` `System; ` ` `  `class` `GFG{ ` `   `  `// Function to find the maximum for each ` `// and every contiguous subarray of size k ` `static` `void` `printKMax(``int` `[]a, ``int` `n, ``int` `k) ` `{ ` `    ``// If k = 1, print all elements ` `    ``if` `(k == 1) { ` `        ``for` `(``int` `i = 0; i < n; i += 1) ` `            ``Console.Write(a[i]+ ``" "``); ` `        ``return``; ` `    ``} ` `   `  `    ``// Using p and q as variable pointers ` `    ``// where p iterates through the subarray ` `    ``// and q marks end of the subarray. ` `    ``int` `p = 0, ` `        ``q = k - 1, ` `        ``t = p, ` `        ``max = a[k - 1]; ` `   `  `    ``// Iterating through subarray. ` `    ``while` `(q <= n - 1) { ` `   `  `        ``// Finding max ` `        ``// from the subarray. ` `        ``if` `(a[p] > max) ` `            ``max = a[p]; ` `   `  `        ``p += 1; ` `   `  `        ``// Printing max of subarray ` `        ``// and shifting pointers ` `        ``// to next index. ` `        ``if` `(q == p && p != n) { ` `            ``Console.Write(max+ ``" "``); ` `            ``q++; ` `            ``p = ++t; ` `   `  `            ``if` `(q < n) ` `                ``max = a[q]; ` `        ``} ` `    ``} ` `} ` `   `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]a = { 1, 2, 3, 4, 5, ` `                ``6, 7, 8, 9, 10 }; ` `    ``int` `n = a.Length; ` `    ``int` `K = 3; ` `   `  `    ``printKMax(a, n, K); ` `} ` `} ` `  `  `// This code is contributed by Rajput-Ji `

Output:

```3 4 5 6 7 8 9 10
```

Time Complexity: O(N)
Auxiliary Space Complexity: O(1)

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