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.
Java
// 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.*/ |
Output :
2
Time Complexity: O(n) where n is size of input array. This is because a for loop is executing from 1 to n.
Space Complexity: O(1) as no extra space has been used.
Please refer complete article on Number of local extrema in an array for more details!
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