Rotated Sequence Search — Problem Statement & Solution Guide
Problem Description
You are provided with a sequence of distinct integers that was originally sorted in strictly ascending order. This sequence has undergone an unknown number of cyclic rotations, resulting in a partitioned structure where the first segment is greater than the second segment. Your task is to determine the index of a specific target value within this rotated sequence. If the target value is not present in the sequence, return -1.
The input consists of the rotated array and the target integer. You must design an algorithm that efficiently locates the target by leveraging the sorted properties of the sub-arrays created by the rotation. The solution should ideally operate in logarithmic time complexity relative to the length of the sequence.
Return the zero-based index of the target if found. If the target does not exist in the sequence, return -1.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Rotated Sequence Search"
WHY DOES IT MATTER?
The rotated‑array search pattern exemplifies how to exploit hidden order within seemingly chaotic data, a skill crucial for designing efficient algorithms in constrained environments.
OPTIMIZATION CHALLENGE
The key insight is that any sub‑range still contains a fully sorted half; by discarding the unsorted half each recursion, we cut the search space by half, achieving logarithmic complexity.
REAL-WORLD CONNECTION
Think of a circular buffer in a high‑throughput logging system where the newest entries wrap around; locating a specific timestamp mirrors searching a rotated sequence.
During an interview, first state the invariant (one half is sorted), then walk through the decision tree with concrete index examples before writing code; this demonstrates clarity of thought.
COMPLEXITY AT A GLANCE
O(log n)O(1)Core Theory — Why This Approach?
When a strictly ascending array is rotated an unknown number of times, it becomes a piecewise monotonic sequence where one sub‑array remains sorted and the other is also sorted but all its elements are smaller than those in the first sub‑array. A naive linear scan would examine every element, leading to O(n) time, which is prohibitive for large n (e.g., n ≈ 10⁷) especially under tight interview time constraints. The optimal paradigm leverages the fact that at least one half of any sub‑range is still sorted; by comparing the target with the boundary values of the current range we can decide which half to discard, achieving logarithmic search. While binary search on a fully sorted array is straightforward, the rotated variant requires an extra conditional check to identify the sorted half before recursing, preserving O(log n) time while using only O(1) extra space.
Interview Questions on This Problem
Q1How would you modify classic binary search to work on a rotated sorted array without using extra memory?
Identify the sorted half by comparing the low and mid elements; if the target lies within that half, recurse into it, otherwise recurse into the other half. This retains O(log n) time and O(1) space.
Q2What is the worst‑case time complexity if the array contains duplicate values, and how does it affect the algorithm?
With duplicates, the guarantee that one half is strictly sorted breaks, potentially degrading to O(n) because we may need to linearly skip equal elements to decide the sorted side.
Q3Explain how you could find the rotation pivot (minimum element) in O(log n) time and why that might be useful before searching for a target.
Perform a modified binary search where you compare mid with high; if mid > high, the pivot lies to the right, else to the left. Knowing the pivot lets you convert the rotated index to a virtual sorted index, simplifying the target lookup.
Examples
Input
nums = [4, 5, 6, 7, 0, 1, 2], target = 0
Output
4
Explanation: The array is rotated such that the pivot is at index 4 (value 0). The left half [4, 5, 6, 7] is sorted and the right half [0, 1, 2] is sorted. Since target 0 is less than nums[0] (4), it must be in the right half. Checking the right half, we find 0 at index 4.
Input
nums = [4, 5, 6, 7, 0, 1, 2], target = 3
Output
-1
Explanation: The target 3 is not present in the array. The search space is divided, but neither the left nor the right sorted segment contains the value 3. Thus, the function returns -1.
Input
nums = [1], target = 1
Output
0
Explanation: The array contains a single element which is the target. The index is 0.
Input
nums = [2, 3, 4, 5, 6, 7, 0, 1], target = 7
Output
5
Explanation: The array is rotated at index 6. The target 7 is in the left sorted segment [2, 3, 4, 5, 6, 7]. Binary search within this segment identifies 7 at index 5.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- All values in nums are unique.
- nums is guaranteed to be a rotated version of a sorted array.
- -10^9 <= target <= 10^9
Optimal Approach & Strategy
Use a modified binary search that first determines which half of the current range is sorted, then recursively (or iteratively) search only the half that could contain the target.
Brute Force Approach
Iterate through the array from start to finish, comparing each element with the target until you find a match or reach the end.
Verified Code Solutions
function solution(nums) {
if (!nums) return -1;
let left = 0, right = nums.length - 1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] === target) return mid;
if (nums[left] <= nums[mid]) {
if (nums[left] <= target && target < nums[mid]) right = mid - 1;
else left = mid + 1;
} else {
if (nums[mid] < target && target <= nums[right]) left = mid + 1;
else right = mid - 1;
}
}
return -1;
}class Solution {
public:
int solution(vector<int>& nums, int target) {
if (!nums.size()) return -1;
int left = 0, right = nums.size() - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) return mid;
if (nums[left] <= nums[mid]) {
if (nums[left] <= target && target < nums[mid]) right = mid - 1;
else left = mid + 1;
} else {
if (nums[mid] < target && target <= nums[right]) left = mid + 1;
else right = mid - 1;
}
}
return -1;
}
};class Solution {
public int solution(int[] nums, int target) {
if (nums == null) return -1;
int left = 0, right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) return mid;
if (nums[left] <= nums[mid]) {
if (nums[left] <= target && target < nums[mid]) right = mid - 1;
else left = mid + 1;
} else {
if (nums[mid] < target && target <= nums[right]) left = mid + 1;
else right = mid - 1;
}
}
return -1;
}
}def solution(nums, target):
if not nums: return -1
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == target: return mid
if nums[left] <= nums[mid]:
if nums[left] <= target and target < nums[mid]: right = mid - 1
else: left = mid + 1
else:
if nums[mid] < target and target <= nums[right]: left = mid + 1
else: right = mid - 1
return -1function solution(nums) {
if (!nums) return -1;
let left = 0, right = nums.length - 1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] === target) return mid;
if (nums[left] <= nums[mid]) {
if (nums[left] <= target && target < nums[mid]) right = mid - 1;
else left = mid + 1;
} else {
if (nums[mid] < target && target <= nums[right]) left = mid + 1;
else right = mid - 1;
}
}
return -1;
}Solve in Interative Editor
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