package binary_search;

/**

* Created by gouthamvidyapradhan on 10/07/2017.

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

Given an input array where num[i] ≠ num[i+1], find a peak element and return its index.

The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.

You may imagine that num[-1] = num[n] = -∞.

For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.

Note:

Your solution should be in logarithmic complexity.

Solution: O(log N) check if the first or the last element is the peak element, if yes then return this index.

Else binary search for the answer - check mid element if this is a peak element return this index, else if the

left element is greater than current element search left else search right.

*/

public class FindPeakElement {

public static void main(String[] args) throws Exception{

int[] nums = {3, 4, 3, 2, 1};

System.out.println(new FindPeakElement().findPeakElement(nums));

}

public int findPeakElement(int[] nums) {

if(nums.length == 1)return 0;

if(nums[0] > nums[1])

return 0;

else if(nums[nums.length - 1] > nums[nums.length - 2])

return nums.length - 1;

int l = 0, h = nums.length - 1;

int ans = 0;

while(l <= h){

int m = l + (h - l) / 2;

if(m - 1 >= 0 && m + 1 < nums.length){

if(nums[m] > nums[m - 1] && nums[m] > nums[m + 1]){

return m;

}

}

if(m - 1 >= 0 && nums[m - 1] > nums[m]){ //search left

h = m - 1;

}

else {

ans = l; //mark this as the answer and search right

l = m + 1;

}

}

return ans;

}

}