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> |
Left = 2 Right = 4 Peak = 1
Time Complexity: O(N)
Space Complexity: O(N)