Find subarray of Length K with Maximum Peak

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 <bits/stdc++.h>
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

<script>
// Javascript implementation to Find subarray
// of Length K with Maximum Peak

// Function to find the subarray
function findSubArray(a, n, k)
{

    // Make prefix array to store
    // the prefix sum of peak count
    let pref = new Array(n);

    pref[0] = 0;

    for (let 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]++;
    }

    let peak = 0, left = 0;

    for (let 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
    document.write("Left = " + (left + 1) + "<br>");
    document.write("Right = " + (left + k) + "<br>");
    document.write("Peak = " + peak + "<br>");
}

// Driver code
let arr = [3, 2, 3, 2, 1];
let n = arr.length;
let k = 3;

findSubArray(arr, n, k);

// This code is contributed by gfgking
</script>

Output:

Left = 2
Right = 4
Peak = 1

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