Find Peak Element from given Java Array with Explanation | O(log n) Time Complexity

Java code for finding peak element from Array | Linear and Binary Search | LeetCode

Problem Description :

A peak element is an element that is strictly greater than its neighbors. 

Given a 0-indexed integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.

Example 1 :

Input : [1, 2, 3, 1]
Output : 2
Explanation : 3 is peak element in given array so it should return index 2

Example 2 : 

Input : [1, 2, 1, 3, 5, 6, 4]
Output : 5
Explanation : Here in given array two peak elements are present, we can return any of them. 6 is peak element so return index 5

Example 3 :

Input = [5, 6, 8, 1, 3, 9, 10, 8, 7, 5]
Output : 6
Explanation : 10 is peak elements so return index 6

In given array, there can be multiple peak elements so we can return  any of them.

We can solve this problem using Linear search and Binary Search. Binary Search is optimal solution because its time complexity is O(log n).

Solution 1 : Find Peak Element from Array in Java using Linear approach

   
    import java.util.Scanner;
    
    public class FindPeakElement {
    
        public static void main(String[] args) {
           
            Scanner sc = new Scanner(System.in);
           
            System.out.println("Enter array size : ");
            int size = sc.nextInt();
    
            int[] array = new int[size];
           
            // Taking user inputs
            System.out.println("Enter array elements :");
            for (int i = 0; i < array.length; i++) {
                array[i] = sc.nextInt();
            }
           
            System.out.println(findPeakElementLinear(array));
    
        }
       
        public static int findPeakElementLinear(int[] nums) {
           
            int peak = 0;
            for (int i = 0; i < nums.length; i++) {
                if (i+1 < nums.length) {
                    if (nums[i] > nums[i+1]) {
                        peak = i;
                        break;
                    } else if (nums[i] < nums[i+1]) {
                        peak = i+1;
                    }
                }
            }
            return peak;
        }
    }
 

This approach gives O(n) time complexity. We can optimize this solution from linear way to binary search. 

Lets see optimal solution of this.

Solution 2 : Find Peak Element from Array in Java using Binary Search approach

    
    import java.util.Scanner;
    
    public class FindPeakElement {
    
        public static void main(String[] args) {
           
            Scanner sc = new Scanner(System.in);
           
            System.out.println("Enter array size : ");
            int size = sc.nextInt();
    
            int[] array = new int[size];
           
            // Taking user inputs
            System.out.println("Enter array elements :");
            for (int i = 0; i < array.length; i++) {
                array[i] = sc.nextInt();
            }
           
            System.out.println(findPeakElement(array));
    
        }
    
        public static int findPeakElement(int[] nums) {
           
            int start = 0;
            int end = nums.length-1;
           
            while (start < end) {
                int mid = (start + end) / 2;
               
                if (nums[mid] > nums[mid+1]) {
                    end = mid;
                } else {
                    start = mid+1;
                }
            }
            return start;
        }
    }

Solution Explanation :

• First declare start and end, start value to 0 and end to length of given nums array.
• Loop until start is less than end
• Assign middle index of array to mid variable.
• If mid element is greater than next of mid, then store mid element to end
• else assign mid+1 to start.
• return start

Output Explanation :

nums = [1 2 1 3 5 6 4]

• start = 0, end = 6
• 0 < 6 becomes true
• mid = 3
• nums[3] > nums[4] | 3 > 5 becomes false
• start = 3 + 1 = 4
• start = 4, end = 6, mid = 3
  
• 4 < 6
• mid = 5
• 6 > 4 becomes true
• end = 5
• start = 4, end = 5, mid = 5
• 4 < 5 
• mid = 4
• 5 > 6 becomes false
• start = 5
• start = 5, end = 5, mid = 4
  
• 5 < 5 becomes false
• return 5

This approach gives O(log n) time complexity. 

 

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