Date: May 11, 2020\nLocation: Seattle, USA\n\nGiven a sorted array of size `n`, find the majority element. The majority element is the element that appears more than n/4 times. You may assume that the array is non-empty and the majority element always exist in the array.\n\n**Example 1:**\n```\nInput: [1, 1, 2, 2, 3, 3, 3, 4, 5, 6, 7]\nOutput: 3\n```\n\n**Example 2:**\n```\nInput: [1, 1, 2, 2, 3, 3, 4, 5, 7, 7, 7]\nOutput: 7\n```\n\n**Example 3:**\n```\nInput: [1, 1, 2, 3]\nOutput: 1\n```\n\nInitialy, I\'ve offer 2 solution\n1. Use Map with key and count pair and go through the Map and return the true, if any of the count >= 1/4*length;\n2. Use for loop with two pointer(first, prev) , increase the count, if values are same and return true when you see the count >=1/4* length;\n\nBoth the solutions are O(N) time complexity. But, she was expecting the O(log n) solution and she given below HINT: \nint windowSize = arr.length*(1/4)\na[0] == a[0+windowSize-1]\na[0] != a[0+windowSize-1]\n\nI understood it was a binary search apporach, but couldn\'t make it happen and time is crossed.\n\n---\nRelated problems:\n* [Majority Element](https://leetcode.com/discuss/interview-question/344289/)

Date: May 11, 2020\nLocation: Seattle, USA\n\nGiven a sorted array of size n, find the majority element. The majority element is the element that appears more t