Number of local extrema in an array

14 Jun, 2022
You are given an array on n-elements. An extrema is an elements which is either greater than its both of neighbors or less than its both neighbors. You have to calculate the number of local extrema in given array. Note : 1st and last elements are not extrema.Examples :  
Input : a[] = {1, 5, 2, 5}
Output : 2

Input : a[] = {1, 2, 3}
Output : 0
 
Approach :For calculating number of extrema we have to check whether an element is maxima or minima i.e. whether it is greater than both of its neighbors or less than both neighbors. For this simply iterate over the array and for each elements check its possibility of being an extrema.Note: a[0] and a[n-1] has exactly one neighbour each, they are neither minima nor maxima. 
C++
Java
Python3
C#
PHP
Javascript
// CPP to find number// of extrema#include <bits/stdc++.h>using namespace std; // function to find// local extremumint extrema(int a[], int n){    int count = 0;     // start loop from position 1    // till n-1    for (int i = 1; i < n - 1; i++)    {         // only one condition        // will be true at a         // time either a[i]        // will be greater than        // neighbours or less        // than neighbours         // check if a[i] is greater        // than both its neighbours        // then add 1 to x        count += (a[i] > a[i - 1] && a[i] > a[i + 1]);         // check if a[i] is        // less than both its        // neighbours, then        // add 1 to x        count += (a[i] < a[i - 1] && a[i] < a[i + 1]);    }     return count;} // driver programint main(){    int a[] = { 1, 0, 2, 1 };    int n = sizeof(a) / sizeof(a[0]);    cout << extrema(a, n);    return 0;}
// Java to find// number of extremaimport java.io.*; class GFG {         // function to find    // local extremum    static int extrema(int a[], int n)    {        int count = 0;             // start loop from        // position 1 till n-1        for (int i = 1; i < n - 1; i++)        {                 // only one condition            // will be true at a             // time either a[i]            // will be greater than            // neighbours or less            // than neighbours                 // check if a[i] is greater            // than both its neighbours            // then add 1 to x            if(a[i] > a[i - 1] && a[i] > a[i + 1])                count += 1;                         // check if a[i] is            // less than both its            // neighbours, then            // add 1 to x            if(a[i] < a[i - 1] && a[i] < a[i + 1])                count += 1;        }             return count;    }         // driver program    public static void main(String args[])                            throws IOException    {        int a[] = { 1, 0, 2, 1 };        int n = a.length;        System.out.println(extrema(a, n));    }}  /* This code is contributed by Nikita Tiwari.*/
# Python 3 to find# number of extrema # function to find# local extremumdef extrema(a, n):     count = 0    # start loop from    # position 1 till n-1    for i in range(1, n - 1) :        # only one condition        # will be true        # at a time either        # a[i] will be greater        # than neighbours or        # less than neighbours         # check if a[i] if        # greater than both its        # neighbours, then add        # 1 to x        count += (a[i] > a[i - 1] and a[i] > a[i + 1]);         # check if a[i] if        # less than both its        # neighbours, then        # add 1 to x        count += (a[i] < a[i - 1] and a[i] < a[i + 1]);     return count # driver programa = [1, 0, 2, 1 ]n = len(a)print(extrema(a, n)) # This code is contributed by Smitha Dinesh Semwal
// C# to find// number of extremausing System; class GFG {         // function to find    // local extremum    static int extrema(int []a, int n)    {        int count = 0;             // start loop from        // position 1 till n-1        for (int i = 1; i < n - 1; i++)        {                 // only one condition            // will be true at a            // time either a[i]            // will be greater than            // neighbours or less            // than neighbours                 // check if a[i] is greater            // than both its neighbours            // then add 1 to x            if(a[i] > a[i - 1] && a[i] > a[i + 1])                count += 1;                         // check if a[i] is            // less than both its            // neighbours, then            // add 1 to x            if(a[i] < a[i - 1] && a[i] < a[i + 1])                count += 1;        }             return count;    }         // Driver program    public static void Main()                                 {        int []a = { 1, 0, 2, 1 };        int n = a.Length;    Console.WriteLine(extrema(a, n));    }}  /* This code is contributed by vt_m.*/
<?php// PHP to find number// of extrema  // function to find// local extremumfunction extrema($a, $n){    $count = 0;     // start loop from position 1    // till n-1    for ($i = 1; $i < $n - 1; $i++)    {         // only one condition        // will be true at a        // time either a[i]        // will be greater than        // neighbours or less        // than neighbours         // check if a[i] is greater        // than both its neighbours        // then add 1 to x        $count += ($a[$i] > $a[$i - 1] and                   $a[$i] > $a[$i + 1]);         // check if a[i] is        // less than both its        // neighbours, then        // add 1 to x        $count += ($a[$i] < $a[$i - 1] and                   $a[$i] < $a[$i + 1]);    }     return $count;} // Driver Code$a = array( 1, 0, 2, 1 );$n = count($a);echo extrema($a, $n); // This code is contributed by anuj_67.?>
<script> // JavaScript program to find// number of extrema     // function to find    // local extremum   function extrema(a,  n)    {        let count = 0;               // start loop from        // position 1 till n-1        for (let i = 1; i < n - 1; i++)        {                   // only one condition            // will be true at a             // time either a[i]            // will be greater than            // neighbours or less            // than neighbours                   // check if a[i] is greater            // than both its neighbours            // then add 1 to x            if(a[i] > a[i - 1] && a[i] > a[i + 1])                count += 1;                           // check if a[i] is            // less than both its            // neighbours, then            // add 1 to x            if(a[i] < a[i - 1] && a[i] < a[i + 1])                count += 1;        }               return count;    } // Driver code                 let a = [ 1, 0, 2, 1 ];        let n = a.length;        document.write(extrema(a, n));         </script>
Output : 
2
Time Complexity: O(n)
Auxiliary Space: O(1)
vt_m
souravghosh0416
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