Search in Rotated Sorted Array

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Example For [4, 5, 1, 2, 3] and target=1, return 2.

For [4, 5, 1, 2, 3] and target=0, return -1.

Solution: Rotated array has a half array that is sorted. So we can use binary search.

public int search(int[] A, int target) {
    if(A == null || A.length == 0) {
        return -1;
    }
    int start = 0, end = A.length - 1;
    while(start <= end) {
        int mid = start + (end - start) / 2;
        if(A[mid] == target) {
            return mid;
        }
        else if(A[start] < A[mid]) {
            if(A[start] <= target && A[mid] >= target) {
                end = mid - 1;
            }
            else {
                start = mid + 1;
            }
        }
        else {
            if(A[mid] <= target && target <= A[end]) {
                start = mid + 1;
            }
           else {
                end = mid - 1;
           }
        }
    }
    return -1;
}

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.

Example Given [1, 1, 0, 1, 1, 1] and target = 0, return true. Given [1, 1, 1, 1, 1, 1] and target = 0, return false.

Solution: Skip duplicate elements. If all elements are duplicate except target, we need search one by one and time complexity become O(N).

public boolean search(int[] A, int target) {
    if(A == null || A.length == 0) {
        return false;
    }
    int start = 0, end = A.length - 1;
    while(start <= end) {
        int mid = start + (end - start) / 2;
        if(A[mid] == target) {
            return true;
        }
        else if(A[start] < A[mid]) {
            if(A[start] <= target && A[mid] >= target) {
                end = mid - 1;
            }
            else {
                start = mid + 1;
            }
        }
        else if(A[start] > A[mid]) {
            if(A[mid] <= target && target <= A[end]) {
                start = mid + 1;
            }
            else {
                end = mid - 1;
            }
        }
        else {
            start++;
        }  
    }
    return false;
}