在二进制搜索主题下的 InterviewBit 问题中的 C++ 中的旋转排序数组中的搜索超出了时间限制

Question

The problem is to search for an element in a sorted, rotated array in C++. The approach that I have used is to find the pivot element index/ the minimum element index and then to find the key either from a[pivot......end] or a[beg,....pivot] use binary search.

The time complexity to find the minimum element index is O(log n) and then for finding the element is also O(log n) and hence the overall time complexity will be O(log n)

But I am getting Time Limit Exceeded error Here is my code:

int find_pivot(std::vector<int> a, int beg, int end)
{
    while(beg<=end)
    {
        int mid=(beg+end)/2;
        if(mid>0 && a[mid]<a[mid-1])
            return mid;
        else if(a[mid]>a[end])
            beg=mid+1;
        else
            end=mid;
    }
}

int search_element(std::vector<int> a, int key, int beg, int end)
{
    int pivot = find_pivot(a,0,a.size()-1);
    if(key>=a[pivot] && key<=a[end])
        beg=pivot;
    else
        end=pivot;
    while(beg<=end)
    {
        int mid=(beg+end)/2;
        if(key==a[mid])
            return mid;
        else if(key>a[mid])
            beg=mid+1;
        else
            end=mid-1;
    }
    return -1;
}

int Solution::search(const vector<int> &A, int B) {
    int n = A.size();
   int i = search_element(A, B, 0, n-1); 

    if (i != -1) 
    return i;
    else
    return -1;
}

Answer 1

Check your find_pivot function, Lets say array is normally sorted [1,2,3,4,5]. Now when you call your find_pivot function, there will come a time when beg=0 and end =1.

Now, since mid=(beg+end/2)=0 your function will change end=mid which means end also becomes 0.

Now your beg=0 and end=0, now when the loop starts again Mid = 0+0/2 and and this will keep on repeating.

Make some changes and I think the issue is when the array is normally sorted or when end=begin.