540. Single Element in a Sorted Array 21 FEB

class Solution {
public:
    int singleNonDuplicate(vector<int>& nums) {
        int start=0;
        int end=nums.size()-1;
        int mid;
        if(end==0){
            return nums[0];   
        }else if( nums[0]!=nums[1]){  /*checking if first or last pairs may be our answer or not .
        if pair having different value then this is our target.*/
            return nums[0];
        }else if(nums[end]!=nums[end-1]){
            return nums[end];
        }

        // binary search starts from here 
        while(start<=end){
         mid=start+(end-start)/2;
         //base condition when it is unique .see first iamge above 
         if(nums[mid]!=nums[mid-1] && nums[mid]!=nums[mid+1]){
             return nums[mid];
         }
         //checking for step 5 .see step 5 from above 
        else if(((mid%2)==0 && nums[mid]==nums[mid+1]) || ((mid%2)==1 && nums[mid]==nums[mid-1])){
             start=mid+1;
         }
        else{        //step 6
             end=mid-1;
        }
        }
        return -1;
    }
};