leetcode Question 95: Search in Rotated Sorted Array II

Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

Analysis:
Similar to the previous question (see here).
Just when A[mid]==A[st], we cannot say the previous part is ordered, then we just go to the next element and check again.


Code:

class Solution {
public:

    bool se(int st, int ed, int target, int A[]){
        if (st>ed) {return false;}
        else{
            int mid = st+(ed-st)/2;
            if (A[mid]==target){return true;}
            
            if (A[mid]>A[st]){
                if (target<=A[mid] && target>=A[st]){
                    return se(st,mid-1,target,A);
                }else{
                    return se(mid+1,ed,target,A);
                }
            }
            
            if (A[mid]<A[st]){
                
                if (target<=A[mid] || target >= A[st]){
                    return se(st,mid-1,target,A);
                }else{
                    return se(mid+1,ed,target,A);
                }
                
            }
            
            if (A[mid]==A[st]){return se(st+1,ed,target,A);}
            
            return false;
        }
        
    }

    bool search(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (n==0){return false;}
        return  se(0,n-1,target,A);
    }
};