Dynamic Interval Alignment Resolver 6 — Problem Statement & Solution Guide

Binary SearchMediumRotated Array Pivot Search

TimeO(log n)

SpaceO(1)

Problem Description

You are provided with a sequence of distinct integers that was originally sorted in strictly ascending order. This sequence has been rotated at an unknown pivot index, meaning the first k elements have been moved to the end of the array. Your task is to determine the index of the minimum element in this rotated sequence.

The rotation can occur in two primary scenarios: either the array remains in its original sorted state (pivot at index 0), or it has been rotated such that the smallest element is no longer at the beginning. Note that the problem guarantees all elements are unique, ensuring a single, unambiguous minimum value.

Design an algorithm that locates the index of the minimum value in O(log n) time complexity. You must not simply scan the entire array, as the input size can be substantial. The solution should efficiently narrow down the search space by comparing the midpoint value against the boundaries of the current search interval.

DSA Pattern Breakdown

"Dynamic Interval Alignment Resolver 6"

medium

WHY DOES IT MATTER?

Binary search on a rotated array demonstrates how to maintain logarithmic time even when the input violates global order, a pattern that appears in many interview questions involving partially sorted data.

OPTIMIZATION CHALLENGE

The insight is that the pivot is the only place where the sorted order breaks. By comparing mid to the rightmost element, we can discard half the array each iteration, reducing time from O(n) to O(log n).

REAL-WORLD CONNECTION

Consider a circular buffer in a high-frequency trading system where the oldest price tick is needed. The buffer is logically rotated; finding the oldest tick efficiently is analogous to locating the minimum in a rotated array.

Always guard against infinite loops by updating low and high correctly; a common pitfall is using mid = (low + high) / 2 without handling overflow or misplacing the update of low/high after comparison.

COMPLEXITY AT A GLANCE

⏱ Time:O(log n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem of finding the minimum element in a rotated sorted array is a classic illustration of how binary search can be adapted to non-trivial search spaces. In a strictly ascending array, the minimum is trivially at index 0. When the array is rotated, the order is preserved in two contiguous segments: a suffix that is still sorted and a prefix that has been moved to the end. The key observation is that the pivot point—the boundary between these two segments—must be the location of the smallest element. Naive approaches that scan the entire array in O(n) time or that attempt to sort the array again in O(n log n) time are unnecessary and wasteful, especially for large inputs where n can be millions.

Binary search exploits the fact that, although the array is not globally sorted, it is still partially sorted. By comparing the middle element with the rightmost element, we can determine which half contains the pivot. If the middle element is greater than the rightmost, the pivot lies to the right; otherwise, it lies to the left (or could be the middle itself). Repeating this halving process reduces the search space logarithmically, yielding an O(log n) solution with constant extra space. This optimal paradigm is essential for interview settings where time and space efficiency are scrutinized.

Interview Questions on This Problem

Q1How would you modify a standard binary search to find the minimum in a rotated sorted array, and why does this modification work?

You compare the middle element with the rightmost element. If mid > right, the minimum is in the right half; otherwise, it is in the left half (including mid). This works because the right half is sorted and the pivot must be in the unsorted part.

Q2What edge case must you handle when the array is not rotated at all, and how does your algorithm detect it?

If the first element is less than the last, the array is already sorted and the minimum is at index 0. The algorithm can detect this by checking if nums[0] < nums[-1] before starting the binary search.

Q3Can you explain a real-world scenario where finding the rotation point of a sorted list is useful, and how the binary search approach scales in that context?

In distributed systems, a log of events may be rotated when a new server takes over. Quickly locating the earliest event (minimum timestamp) is critical for consistency checks. Binary search scales to millions of log entries, ensuring low latency in production.

Examples

Example 1

Input

nums = [15, 18, 22, 3, 6, 9, 12]

Output

Explanation: The array is rotated. The minimum value is 3. The index of 3 is 3. Therefore, the output is 3.

Example 2

Input

nums = [4, 5, 6, 7, 8, 9, 10]

Output

Explanation: The array is already sorted in ascending order. The minimum value is 4, which is at index 0. Therefore, the output is 0.

Example 3

Input

nums = [25, 28, 31, 34, 37, 40, 43, 1, 4, 7, 10, 13, 16, 19, 22]

Output

Explanation: The array is rotated with the smallest element 1. The index of 1 is 7. Therefore, the output is 7.

Example 4

Input

nums = [100]

Output

Explanation: The array contains a single element. The minimum value is 100, which is at index 0. Therefore, the output is 0.

Constraints

1 <= nums.length <= 10^5
All elements in nums are distinct.
-10^9 <= nums[i] <= 10^9
The array nums is a rotated version of a sorted array in ascending order.

Optimal Approach & Strategy

Use binary search: compare mid with the rightmost element to decide which half contains the pivot, halving the search space each step for O(log n) time and O(1) space.

Brute Force Approach

Scan the entire array to find the minimum element, which takes O(n) time and O(1) space.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(log n)

function solution(nums) {
   let pivot = nums[0];
   for (let i = 1; i < nums.length; i++) {
       if (nums[i] > nums[nums.length - 1]) {
           pivot = nums[i];
           break;
       }
   }
   let leftSum = 0;
   let rightSum = 0;
   for (let i = 0; i < nums.length; i++) {
       if (nums[i] <= pivot) {
           leftSum += nums[i];
       } else {
           rightSum += nums[i];
       }
   }
   return leftSum + rightSum;
}

class Solution {
   public:
   int solution(vector<int>& nums) {
       int pivot = nums[0];
       for (int i = 1; i < nums.size(); i++) {
           if (nums[i] > nums[nums.size() - 1]) {
               pivot = nums[i];
               break;
           }
       }
       int leftSum = 0;
       int rightSum = 0;
       for (int i = 0; i < nums.size(); i++) {
           if (nums[i] <= pivot) {
               leftSum += nums[i];
           } else {
               rightSum += nums[i];
           }
       }
       return leftSum + rightSum;
   }
};

class Solution {
   public int solution(int[] nums) {
       int pivot = nums[0];
       for (int i = 1; i < nums.length; i++) {
           if (nums[i] > nums[nums.length - 1]) {
               pivot = nums[i];
               break;
           }
       }
       int leftSum = 0;
       int rightSum = 0;
       for (int i = 0; i < nums.length; i++) {
           if (nums[i] <= pivot) {
               leftSum += nums[i];
           } else {
               rightSum += nums[i];
           }
       }
       return leftSum + rightSum;
   }
}

def solution(nums):
   pivot = nums[0]
   for i in range(1, len(nums)):
       if nums[i] > nums[-1]:
           pivot = nums[i]
           break
   leftSum = 0
   rightSum = 0
   for i in range(len(nums)):
       if nums[i] <= pivot:
           leftSum += nums[i]
       else:
           rightSum += nums[i]
   return leftSum + rightSum

function solution(nums) {
   let pivot = nums[0];
   for (let i = 1; i < nums.length; i++) {
       if (nums[i] > nums[nums.length - 1]) {
           pivot = nums[i];
           break;
       }
   }
   let leftSum = 0;
   let rightSum = 0;
   for (let i = 0; i < nums.length; i++) {
       if (nums[i] <= pivot) {
           leftSum += nums[i];
       } else {
           rightSum += nums[i];
       }
   }
   return leftSum + rightSum;
}

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Dynamic Interval Alignment Resolver 6 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Binary Search

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Dynamic Interval Alignment Resolver 6 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor