# Find subarray of Length K with Maximum Peak

• Difficulty Level : Medium
• Last Updated : 26 May, 2021

Given an array arr[] of length n and a positive integer K, we have to find a subarray of length K which has maximum peak inside in it.
Peaks of the segment [l, r] are those indexes such that l < i < r, a[i-1] < a[i] and a[i+1] < a[i]
Note: The boundary indexes l and r for the segment are not peaks. If there are many subarrays with maximum peak, then print that subarray whose has minimum left index.
Examples:

Input :
arr = {3, 1, 4, 1, 5, 9, 2, 6}, k = 7
Output:
Left = 1
Right = 7
Peak = 2
Explanation:
There are two subarray with length 7 i.e [1, 7] and [2, 8]. Both subarray has 2 peak inside it i.e 3 and 6 index are the peak in both the subarray. We have to return the subarray with minimum l and maximum peak i.e l = 1 and peak = 2.
Input:
arr = {3, 2, 3, 2, 1}, k = 3
Output :
Left = 2
Right = 4
Peak = 1
Explanation:
Only one subarray whose length is 3 and number of peak inside it is 1 i.e. l =2 and peak is i = 3.

Approach:
The approach to solve this problem is to use a sliding window, where we slide across window size of K, and find the total count of peaks in every window, whichever windows gives the maximum number of peaks will be the answer. While moving the right index, we check if an index added is a peak, we increase the count, and while moving the left index, we check if the removed index is a peak, if it is, then decrease the count. We have a window of size K always.
Below is the implementation of the above approach:

## C++

 `// C++ implementation to Find subarray``// of Length K with Maximum Peak` `#include ``using` `namespace` `std;` `// Function to find the subarray``void` `findSubArray(``int``* a, ``int` `n, ``int` `k)``{``    ``// Make prefix array to store``    ``// the prefix sum of peak count``    ``int` `pref[n];` `    ``pref[0] = 0;` `    ``for` `(``int` `i = 1; i < n - 1; ++i) {` `        ``// Count peak for previous index``        ``pref[i] = pref[i - 1];` `        ``// Check if this element is a peak``        ``if` `(a[i] > a[i - 1] && a[i] > a[i + 1])` `            ``// Increment the count``            ``pref[i]++;``    ``}` `    ``int` `peak = 0, left = 0;` `    ``for` `(``int` `i = 0; i + k - 1 < n; ++i)` `        ``// Check if number of peak in the sub array``        ``// whose l = i is greater or not``        ``if` `(pref[i + k - 2] - pref[i] > peak) {``            ``peak = pref[i + k - 2] - pref[i];``            ``left = i;``        ``}` `    ``// Print the result``    ``cout << ``"Left = "` `<< left + 1 << endl;``    ``cout << ``"Right = "` `<< left + k << endl;``    ``cout << ``"Peak = "` `<< peak << endl;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 3, 2, 3, 2, 1 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``int` `k = 3;` `    ``findSubArray(arr, n, k);` `    ``return` `0;``}`

## Java

 `// Java implementation to Find subarray``// of Length K with Maximum Peak` `class` `GFG{`` ` `// Function to find the subarray``static` `void` `findSubArray(``int` `[]a, ``int` `n, ``int` `k)``{``    ``// Make prefix array to store``    ``// the prefix sum of peak count``    ``int` `[]pref = ``new` `int``[n];`` ` `    ``pref[``0``] = ``0``;`` ` `    ``for` `(``int` `i = ``1``; i < n - ``1``; ++i) {`` ` `        ``// Count peak for previous index``        ``pref[i] = pref[i - ``1``];`` ` `        ``// Check if this element is a peak``        ``if` `(a[i] > a[i - ``1``] && a[i] > a[i + ``1``])`` ` `            ``// Increment the count``            ``pref[i]++;``    ``}`` ` `    ``int` `peak = ``0``, left = ``0``;`` ` `    ``for` `(``int` `i = ``0``; i + k - ``1` `< n; ++i)`` ` `        ``// Check if number of peak in the sub array``        ``// whose l = i is greater or not``        ``if` `(pref[i + k - ``2``] - pref[i] > peak) {``            ``peak = pref[i + k - ``2``] - pref[i];``            ``left = i;``        ``}`` ` `    ``// Print the result``    ``System.out.print(``"Left = "` `+  (left + ``1``) +``"\n"``);``    ``System.out.print(``"Right = "` `+  (left + k) +``"\n"``);``    ``System.out.print(``"Peak = "` `+  peak +``"\n"``);``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``3``, ``2``, ``3``, ``2``, ``1` `};`` ` `    ``int` `n = arr.length;`` ` `    ``int` `k = ``3``;`` ` `    ``findSubArray(arr, n, k);`` ` `}``}` `// This code contributed by Princi Singh`

## Python3

 `# Python3 implementation to Find subarray``# of Length K with Maximum Peak` `# Function to find the subarray``def` `findSubArray(a, n, k):` `    ``# Make prefix array to store``    ``# the prefix sum of peak count``    ``pref ``=` `[``0` `for` `i ``in` `range``(n)]` `    ``pref[``0``] ``=` `0` `    ``for` `i ``in` `range``(``1``, n ``-` `1``, ``1``):``        ``# Count peak for previous index``        ``pref[i] ``=` `pref[i ``-` `1``]` `        ``# Check if this element is a peak``        ``if` `(a[i] > a[i ``-` `1``] ``and` `a[i] > a[i ``+` `1``]):``            ``# Increment the count``            ``pref[i] ``+``=` `1` `    ``peak ``=` `0``    ``left ``=` `0` `    ``for` `i ``in` `range``(``0``, n ``-` `k ``+` `1``, ``1``):` `        ``# Check if number of peak in the sub array``        ``# whose l = i is greater or not``        ``if` `(pref[i ``+` `k ``-` `2``] ``-` `pref[i] > peak):``            ``peak ``=` `pref[i ``+` `k ``-` `2``] ``-` `pref[i]``            ``left ``=` `i` `    ``# Print the result``    ``print``(``"Left ="``,left ``+` `1``)``    ``print``(``"Right ="``,left ``+` `k)``    ``print``(``"Peak ="``,peak)` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``3``, ``2``, ``3``, ``2``, ``1``]` `    ``n ``=` `len``(arr)``    ``k ``=` `3``    ``findSubArray(arr, n, k)` `# This code is contributed by Surendra_Gangwar`

## C#

 `// C# implementation to Find subarray``// of Length K with Maximum Peak``using` `System;` `class` `GFG{` `// Function to find the subarray``static` `void` `findSubArray(``int` `[]a, ``int` `n, ``int` `k)``{``    ` `    ``// Make prefix array to store``    ``// the prefix sum of peak count``    ``int` `[]pref = ``new` `int``[n];``    ` `    ``pref[0] = 0;` `    ``for``(``int` `i = 1; i < n - 1; ++i)``    ``{``        ` `       ``// Count peak for previous index``       ``pref[i] = pref[i - 1];``       ` `       ``// Check if this element is a peak``       ``if` `(a[i] > a[i - 1] && a[i] > a[i + 1])``       ``{``           ``// Increment the count``           ``pref[i]++;``       ``}``    ``}` `    ``int` `peak = 0;``    ``int` `left = 0;``    ` `    ``for``(``int` `i = 0; i + k - 1 < n; ++i)``    ``{``        ` `       ``// Check if number of peak in the sub array``       ``// whose l = i is greater or not``       ``if` `(pref[i + k - 2] - pref[i] > peak)``       ``{``           ``peak = pref[i + k - 2] - pref[i];``           ``left = i;``       ``}``    ``}``       ` `    ``// Print the result``    ``Console.Write(``"Left = "` `+ (left + 1) + ``"\n"``);``    ``Console.Write(``"Right = "` `+ (left + k) + ``"\n"``);``    ``Console.Write(``"Peak = "` `+ peak + ``"\n"``);``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 3, 2, 3, 2, 1 };``    ``int` `n = arr.Length;``    ``int` `k = 3;` `    ``findSubArray(arr, n, k);``}``}` `// This code is contributed by Rohit_ranjan`

## Javascript

 ``

Output:

```Left = 2
Right = 4
Peak = 1```

Time Complexity: O(N)

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