Leetcode 1150

binary search left bound and right bound

class Solution {
    public boolean isMajorityElement(int[] nums, int target) {
        // search left 
        int left = 0, right = nums.length - 1;
        while(left <= right){
            int pivot = left + (right - left) / 2;
            if(nums[pivot] >= target){
                // we shrink to left 
                right = pivot - 1;
            }else{
                // it is too small
                left = pivot + 1;
            }
        }
        int p1 = left;
        left = 0; right = nums.length - 1;
        // find right boundry
        while(left <= right){
            int pivot = left + (right - left) / 2;
            if(nums[pivot] > target){
                // we shrink to left 
                right = pivot - 1;
            }else{
                // it is too small
                left = pivot + 1;
            }
        }
        int p2 = right;
        return p2 - p1 + 1 > nums.length /2;
    }
}