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Java

/*This is a Java Program to find peak element of an array. A peak element of an array is that element which is not smaller than its neighbors. Consider only one neighbour for corner elements. The time complexity of the following program is O (n).
Brute Force Algorithm is as follows :
for i in range (n) :
if A[i - 1] <= A[i] >= A[i + 1] :
print A[i]
end if
end for*/
/*
* Java Program to Find the peak element of an array O(n) time (Naive Method)
*/
import java.util.Scanner;
public class PeakElement1
{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
System.out.println("Enter size of array");
int N = scan.nextInt();
int[] arr = new int[N + 2];
/* set corner values to -infinity */
arr[0] = Integer.MIN_VALUE;
arr[N + 1] = Integer.MIN_VALUE;
/* Accept N elements */
System.out.println("Enter "+ N +" elements");
for (int i = 1; i <= N; i++)
arr[i] = scan.nextInt();
/* Find All Peak Elements */
System.out.println("\nAll Peak Elements : ");
for (int i = 1; i <= N; i++)
if (arr[i - 1] <= arr[i] && arr[i] >= arr[i + 1])
System.out.println(arr[i] +" at position "+ i);
System.out.println();
}
}
/*
Enter size of array
6
Enter 6 elements
1 2 5 5 4 1
All Peak Elements :
5 at position 3
5 at position 4
Enter size of array
7
Enter 7 elements
6 24 15 2 23 99 67
All Peak Elements :
24 at position 2
99 at position 6
Enter size of array
10
Enter 10 elements
10 9 8 24 8 7 97 28 17 63
All Peak Elements :
10 at position 1
24 at position 4
97 at position 7
63 at position 10