<p>A peak element is an element that is strictly greater than its neighbors.</p>

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

<p>You may imagine that <code>nums[-1] = nums[n] = -&infin;</code>. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.</p>

<p>You must write an algorithm that runs in <code>O(log n)</code> time.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 3 is a peak element and your function should return the index number 2.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,1,3,5,6,4]
<strong>Output:</strong> 5
<strong>Explanation:</strong> Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>nums[i] != nums[i + 1]</code> for all valid <code>i</code>.</li>
</ul>

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int sz = nums.size();
        // if there is only one element then that's the peak.
        if(sz == 1)
            return 0;
        // check if first element can be peak
        if(nums[0]>nums[1])
            return 0;
        // check if last element can be peak
        if(nums[sz-1] > nums[sz-2])
            return sz-1;
        // binary seach for peak excluding first and last element
        int left=1, right=sz-2;
        while(left<=right){
            int mid = left + (right-left)/2;
            if(nums[mid]>nums[mid-1] && nums[mid]>nums[mid+1])
                return mid;
            // check if mid lies in increasing side(expension)
            else if(nums[mid]>nums[mid-1])
                left=mid+1;
            // if the mid lies in decreasing side
            // when mid is at negative peak(trough) peak can be both sides so
            // left/right any of them can be updated(here we combine it with
            // else case)
            else
                right=mid-1;
        }
        return -1;
    }
};
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