package com.fishercoder.solutions;

/**
 * 1150. 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.
 *
 * Note:
 * 1 <= nums.length <= 1000
 * 1 <= nums[i] <= 10^9
 * 1 <= target <= 10^9
 **/
public class _1150 {
    public static class Solution1 {
        public boolean isMajorityElement(int[] nums, int target) {
            int left = 0;
            int right = nums.length - 1;
            int mid = left + (right - left) / 2;
            return nums[mid] == target;
        }
    }
}

@@ -1011,139 +1011,41 @@
     
-int left = 0;%0A            int right = nums.length - 1;%0A            int mid = left + (right - left) / 2;%0A            return nums%5Bmid
+return nums%5B(nums.length - 1) / 2
 %5D ==

refactor 1150