/*

 if( nums[mid] == target && (mid == 0 or nums[mid-1]!=target))
 in this piece of code ,if mid turns out to 0 then nums[mid-1] is not evaluated and this will not lead to run time error .
 this is called short circuit evaluation .
 Similary 
 if(nums[mid] == target && (mid +1== nums.size() or nums[mid+1]!=target))
 the code nums[mid+1] is not evaluated if mid+1 is already equal to nums.size()
 



*/



class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        
        
        
        if(nums.size() == 0 ){
            return {-1,-1};
        }
        int left =0 ;
        int right = nums.size()-1;
        
        int start = -1 ;
        while(left <=right){
            int mid = (left+right)/2;
            if( nums[mid] == target && (mid == 0 or nums[mid-1]!=target)){
                start= mid;
                break;
            }
            else if(nums[mid]>=target){
                right = mid-1;
            }
            else{
                left = mid+1;
            }
        }
        
        int last = -1 ;
        left =0;
        right = nums.size() -1;
        while(left <= right){
            int mid =(left+right)/2 ;
            if(nums[mid] == target && (mid +1== nums.size() or nums[mid+1]!=target)){
                last = mid ;
                break;
            }
            else if(nums[mid] <= target){
                left = mid+1;
            }
            else {
                right = mid -1 ;
            }
        }
        return {start,last};
        
    }
};