Leetcode 33 Search in Rotated Sorted Array Java Solution
// SOLVING THIS WITH AN AI ASSISTANT (2026)
If you are working through this problem with an AI coding assistant — Claude, ChatGPT, Cursor chat, Gemini, GitHub Copilot, Aider, or any agent — the goal isn’t to ask for the answer. It is to use the tool to understand the pattern. The prompt sequence I’d run:
- Spec it back to me first. “In your own words, what is this problem actually testing? What’s the smallest example that fails the naive approach?”
- Brute-force first, optimize after. “Write the simplest correct solution, even if it’s O(n²). Don’t optimize. Just make it correct, with comments explaining each step.”
- Ask for the upgrade. “Now show me the optimal solution. What insight makes it possible? What pattern is this an instance of?”
- Stress-test it. “Generate 10 edge cases — empty input, single element, duplicates, max size, sorted, reverse-sorted. Run my solution against each.”
The pattern matters more than the answer. If the agent just hands you optimized code, you’ve trained yourself to lose interviews.
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 algorithm’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 1:
public class Solution {
public int search(int[] nums, int target) {
return binarySearch(nums,target,0,nums.length-1);
}
public int binarySearch(int[] nums, int target, int start, int end){
if(start>end)return -1;
int mid = start+(end-start)/2;
if(nums[mid] == target) return mid;
if(nums[start]<=nums[mid]){
if(nums[start]<=target && target<nums[mid]){
return binarySearch(nums,target,start,mid-1);
}else{
return binarySearch(nums,target,mid+1,end);
}
}else{
if(nums[mid]<target && target<=nums[end]){
return binarySearch(nums,target,mid+1,end);
}else{
return binarySearch(nums,target,start,mid-1);
}
}
}
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