15 November

q1 Search in Rotated sorted array

//IN THIS QUESTION, WE APPLIED THE SIMPLE LOGIC OF BINARY SEARCH IN AN ARRAY. DIRECT APPLICATION.

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int st = 0, end= nums.size() -1;

        while(st <= end){
            int mid= st + (end-st)/2;

            if(nums[mid] == target){
                return mid;
            }

            if(nums[st] <= nums[mid]) {

                if(nums[st]<= target && target <= nums[mid]){
                    end = mid -1;
                } else {
                    st = mid +1;
                }

            } else{ 
                if(nums[mid] <= target && target <= nums[end]){
                    st = mid+1;
                } else {
                    end = mid -1;
                }
                
            }

        }

        return -1 ;


    }
};



Q2