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// aman jain mca 1st year 2nd shift

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int ans = 0, left = 0, right = nums.size() - 1;
        while(left <= right) {
            int mid = (left + right)/2;
            if(mid - 1 >= 0 && mid + 1 <= nums.size() - 1) { // both neighbours in range
                if(nums[mid - 1] < nums[mid] && nums[mid + 1] < nums[mid]) { // peak made
                    ans = mid;
                    break;
                }
                else if(nums[mid - 1] > nums[mid]) { // possible peak to the left of mid
                    right = mid - 1;
                }
                else { // possible peak to the right of mid
                    left = mid + 1;
                }
            }
            else if(mid - 1 < 0 && mid + 1 >= nums.size()) { // both neighbours out of range
                ans = mid;
                break;
            }
            else if(mid + 1 >= nums.size()) { // right neighbour out of range
                if(nums[mid - 1] < nums[mid]) { // peak made
                    ans = mid;
                    break;
                }
                else { // possible peak to the left of mid
                    right = mid - 1;
                }
            }
            else { // left neighbour out of range
                if(nums[mid + 1] < nums[mid]) { // peak made
                    ans = mid;
                    break;
                }
                else { // possible peak to the right of mid
                    left = mid + 1;
                }
            }
        }
        return ans;
    }
};