Check If a Number Is Majority Element in a Sorted Array

Given an array nums sorted in non-decreasing order, and a number target, return True if and only if target is a majority element.

A majority element is an element that appears more than N/2 times in an array of length N.

 

Example 1:

Input: nums = [2,4,5,5,5,5,5,6,6], target = 5
Output: true
Explanation: 
The value 5 appears 5 times and the length of the array is 9.
Thus, 5 is a majority element because 5 > 9/2 is true.
Example 2:

Input: nums = [10,100,101,101], target = 101
Output: false
Explanation: 
The value 101 appears 2 times and the length of the array is 4.
Thus, 101 is not a majority element because 2 > 4/2 is false.
 

Constraints:


Solution:

class Solution {
    public boolean isMajorityElement(int[] nums, int target) {
        int l = left(nums, target);
        if (l == nums.length || nums[l] != target) return false;
        int r = right(nums, target);
        return (r - l + 1) > (nums.length / 2);
    }
    
    private int left(int[] nums, int target) {
        int l = 0, r = nums.length - 1;
        while (l <= r) {
            int mid = (l + r) / 2;
            if (nums[mid] > target) {
                r = mid - 1;
            } else if (nums[mid] == target) {
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
    
    private int right(int[] nums, int target) {
        int l = 0, r = nums.length - 1;
        while (l <= r) {
            int mid = (l + r) / 2;
            if (nums[mid] > target) {
                r = mid - 1;
            } else if (nums[mid] == target) {
                l = mid + 1;
            } else {
                l = mid + 1;
            }
        }
        return r;
    }
}