34. Search for a Range

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Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,

return [3, 4].

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
         int left=0,right=nums.size()-1;

        int  ant;

        vector<int> res{-1,-1};
        if(nums.empty()) return res;
        while(left<=right){
            int mid=left+(right-left)/2;
            if(nums[mid]>=target){
                right=mid-1;  
            }else
                left=mid+1;
        }
        if(left>=nums.size()||nums[left]!=target) return res;
        res[0]=left;
               right=nums.size()-1;
        while(left<=right){
            int mid=left+(right-left)/2;
            if(nums[mid]>target){
                right=mid-1;  
            }else
                left=mid+1;
        }
        res[1]=right;
        return res;
    }
};

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