Another good example of modifiying classical algorithm in order to reach the goal. In this example we have twisted task where binary-search algorithm should be modified with additional edge case checks. leetcode

# Given

Suppose an array sorted in ascending order 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.

Your algorithmâ€™s runtime complexity must be in the order of O(log n).

Example 1:

``````Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
``````

Example 2:

``````Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
``````

# Solution

``````class Solution {
public int search(int[] nums, int target) {
if (nums.length == 0)
return -1;
return searchArr(nums, target, 0, nums.length-1);
}

public int searchArr(int[] nums, int target, int startIndex, int endIndex) {

if (startIndex == endIndex) {
if (target == nums[startIndex])
return startIndex;

return -1;
}

int halfIndex = (endIndex + startIndex) / 2;

if (nums[halfIndex] == target)
return halfIndex;

if (target < nums[halfIndex]) { // left-side
if (target < nums[startIndex] && nums[startIndex] <= nums[halfIndex]) { // go-for right side
return searchArr(nums, target, halfIndex+1, endIndex);
} else { // go for left
return searchArr(nums, target, startIndex, halfIndex);
}
} else { // right-side
if (target > nums[endIndex] && nums[endIndex] >= nums[halfIndex]) { // go-for left side
return searchArr(nums, target, startIndex, halfIndex);
} else { // go-for right
return searchArr(nums, target, halfIndex+1, endIndex);
}
}
}
}
``````