Subarray Rotation Checker — Problem Statement & Solution Guide

ArraysMediumFundamentals

TimeO(n)

SpaceO(1)

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Arrays and solve the Subarray Rotation Checker problem optimally.

TopicArrays

PatternFundamentals

TimeO(n)

SpaceO(1)

Problem Description

Given an array of integers and two pointers, left and right, representing a subarray, determine if there exists a rotation of the subarray such that the sum of the absolute differences between consecutive elements is minimized.

DSA Pattern Breakdown

"Subarray Rotation Checker"

medium

WHY DOES IT MATTER?

Recognizing the rotation as a cut on a cycle allows the problem to be solved in linear time, turning an O(n^2) brute force into O(n). This pattern is common in problems involving circular buffers, ring topologies, and cyclic permutations, where the invariant total cost can be leveraged to optimize linearized computations.

OPTIMIZATION CHALLENGE

The key insight is that the linear sum after rotation equals the total circular sum minus the edge that is removed. Thus, the optimization reduces to finding the maximum adjacent difference, not recomputing sums for each rotation.

REAL-WORLD CONNECTION

In distributed systems, a token ring network processes data in a circular fashion. Determining the optimal point to break the ring (e.g., for load balancing or maintenance) mirrors this problem: you want to minimize the cost of traversing the ring by cutting the most expensive link.

When explaining this to a hiring manager, emphasize that the solution is O(n) with constant extra space, making it scalable to subarrays of millions of elements—critical for real‑time analytics pipelines.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to a classic circular array observation: the sum of absolute differences between consecutive elements in a rotated subarray equals the total sum of absolute differences around the cycle minus the edge that is cut by the rotation. Since a rotation merely chooses a starting point, the set of edges remains the same; only one edge (between the last and first element in the linearized array) is omitted. Therefore, to minimize the linear sum we should maximize the value of the omitted edge, i.e., find the maximum absolute difference between any two adjacent elements in the circular view. Naïve solutions that explicitly generate every rotation and recompute the sum run in O(n^2) time, which is infeasible for large subarrays. The optimal paradigm computes the total circular sum and the maximum adjacent difference in a single pass, achieving O(n) time and O(1) extra space.

Interview Questions on This Problem

Q1How would you explain the relationship between a circular array and its linear rotations to a candidate during a technical interview?

I would describe the subarray as a cycle where each element has a neighbor on both sides. Rotating the subarray is equivalent to choosing a cut point on this cycle; the linear array then consists of all edges except the one that was cut. This perspective shows that the total sum of absolute differences around the cycle is invariant, and the linear sum depends only on which edge is omitted.

Q2A fintech company asks: "Given a subarray, how can you determine if rotating it can reduce the total absolute difference sum?" What key insight would you highlight?

I would highlight that the only way to reduce the sum is to cut the largest edge in the circular representation. Thus, compute the maximum |a[i] - a[(i+1)%n]|; if this value is positive, rotating to cut that edge yields the minimal sum. If all edges are zero, any rotation yields the same sum.

Q3During a startup interview, you’re asked to implement this efficiently. What data structure or algorithmic pattern would you use and why?

I would use a single-pass linear scan (a simple for-loop) to compute both the total circular sum and the maximum adjacent difference. This is a classic example of the "prefix-sum with a running maximum" pattern, which is both time‑efficient and space‑light.

Examples

Example 1

Input

[1, 2, 3, 4, 5], 1, 3

Output

true

Explanation: Step-by-step: Given the subarray [1, 2, 3, 4, 5] and the rotation [2, 3, 4, 5, 1], we calculate the sum of absolute differences between consecutive elements: |2-3| + |3-4| + |4-5| + |5-1| + |1-2| = 10. This is the minimum possible sum, so the function returns true.

Example 2

Input

[1, 2, 3, 4, 5], 2, 4

Output

false

Explanation: Step-by-step: Given the subarray [1, 2, 3, 4, 5] and the rotation [3, 4, 5, 1, 2], we calculate the sum of absolute differences between consecutive elements: |3-4| + |4-5| + |5-1| + |1-2| + |2-3| = 12. This is not the minimum possible sum, so the function returns false.

Constraints

1 <= array length <= 10^5
0 <= left < right < array length
All elements in the array are integers between -10^5 and 10^5
The input array is not empty

Optimal Approach & Strategy

Compute the total sum of absolute differences around the circular subarray and the maximum adjacent difference in a single linear scan. The minimal linear sum equals total minus that maximum, achieving O(n) time and O(1) space.

Brute Force Approach

Generate every rotation of the subarray, compute the sum of absolute differences for each, and pick the minimum. This requires O(n^2) time and O(n) space for storing rotations.

Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums, left, right) {
      const n = right - left + 1;
      let minSum = Infinity;
      for (let i = 0; i < n; i++) {
         let sum = 0;
         for (let j = 0; j < n; j++) {
            const index = (left + i + j) % n;
            sum += Math.abs(nums[index] - nums[(left + i + j + 1) % n]);
         }
         minSum = Math.min(minSum, sum);
      }
      return minSum === 0;
   }

class Solution {
   public:
      bool solution(vector<int>& nums, int left, int right) {
         int n = right - left + 1;
         int minSum = INT_MAX;
         for (int i = 0; i < n; i++) {
            int sum = 0;
            for (int j = 0; j < n; j++) {
               int index = (left + i + j) % n;
               sum += abs(nums[index] - nums[(left + i + j + 1) % n]);
            }
            minSum = min(minSum, sum);
         }
         return minSum == 0;
      }
   };

class Solution {
      public boolean solution(int[] nums, int left, int right) {
         int n = right - left + 1;
         int minSum = Integer.MAX_VALUE;
         for (int i = 0; i < n; i++) {
            int sum = 0;
            for (int j = 0; j < n; j++) {
               int index = (left + i + j) % n;
               sum += Math.abs(nums[index] - nums[(left + i + j + 1) % n]);
            }
            minSum = Math.min(minSum, sum);
         }
         return minSum == 0;
      }
   }

def solution(nums, left, right):
      n = right - left + 1
      min_sum = float('inf')
      for i in range(n):
         sum_ = 0
         for j in range(n):
            index = (left + i + j) % n
            sum_ += abs(nums[index] - nums[(left + i + j + 1) % n])
         min_sum = min(min_sum, sum_)
      return min_sum == 0

function solution(nums, left, right) {
      const n = right - left + 1;
      let minSum = Infinity;
      for (let i = 0; i < n; i++) {
         let sum = 0;
         for (let j = 0; j < n; j++) {
            const index = (left + i + j) % n;
            sum += Math.abs(nums[index] - nums[(left + i + j + 1) % n]);
         }
         minSum = Math.min(minSum, sum);
      }
      return minSum === 0;
   }

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Microsoft

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