Rotated Matrix Pivot Analyzer 6 — Problem Statement & Solution Guide

Binary SearchMediumSearch Peak Element

TimeO(log n)

SpaceO(1)

Problem Description

You are provided with an array of integers representing a sequence of sensor readings that has been rotated at an unknown pivot point. The original sequence was strictly increasing, but after rotation, it consists of two contiguous increasing segments. Your task is to identify the index of the minimum element in this rotated array, which corresponds to the pivot point where the rotation occurred. If the array is not rotated (i.e., it is already sorted in ascending order), the pivot is at index 0.

The input is a single array nums of length N. The output should be the index of the smallest element in the array. You must solve this in O(log N) time complexity using binary search, as the array size can be very large. The array contains distinct integers, ensuring a unique minimum value.

DSA Pattern Breakdown

"Rotated Matrix Pivot Analyzer 6"

medium

WHY DOES IT MATTER?

Binary search on rotated arrays demonstrates how to preserve logarithmic efficiency even when the input violates global order. It teaches candidates to identify hidden structure and to adapt classic algorithms to non‑standard conditions, a skill highly valued in production systems that must handle partially corrupted or out‑of‑order data.

OPTIMIZATION CHALLENGE

The pivotal insight is that the pivot is the only place where the ascending order breaks. By comparing the middle element to the high‑end element, we can decide which side to discard, reducing the search space by half each iteration and achieving O(log n) time.

REAL-WORLD CONNECTION

Consider a distributed log replication system where logs are rotated daily. Each replica holds a rotated sequence of entries; finding the earliest entry quickly is essential for consistency checks and roll‑backs. The same binary‑search logic applies to locate the rotation point in the log without scanning the entire dataset.

When explaining this in an interview, start by visualizing the array as two mountains: a high plateau and a low valley. Point out that the valley’s bottom is the pivot, and that the mountain’s slope tells you which side to explore. This mental model helps you articulate the logic quickly and confidently.

COMPLEXITY AT A GLANCE

⏱ Time:O(log n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem of finding the pivot (minimum element) in a rotated strictly increasing array is a classic application of binary search on a partially sorted structure. In a normal sorted array, the minimum is at index 0, but rotation splits the array into two ascending segments: a suffix that contains the largest elements and a prefix that contains the smallest. The key observation is that the array is still “mostly sorted” – every element is greater than its predecessor except at the pivot where the order breaks. By comparing the middle element with the high‑end element, we can determine which half contains the pivot: if mid > high, the pivot lies to the right; otherwise it lies to the left or at mid. Repeating this halving process reduces the search space logarithmically.

Naive approaches, such as scanning linearly to find the point where a[i] < a[i-1], run in O(n) time. While acceptable for small inputs, this linear scan becomes a bottleneck for large datasets (e.g., millions of sensor readings) and violates the performance expectations of high‑throughput systems. The binary search strategy leverages the array’s structure to achieve O(log n) time while using only constant extra space, making it the optimal solution for both time‑critical and memory‑constrained environments.

The algorithmic paradigm here is “divide and conquer” applied to a rotated sorted array. By exploiting the invariant that one half of the array remains sorted, we can discard half of the candidates in each iteration. This pattern generalizes to many problems—searching in rotated arrays, finding peaks, or locating a target in a sorted matrix—where a global property is preserved despite local disorder.

Interview Questions on This Problem

Q1How would you modify the algorithm if the array could contain duplicate values?

With duplicates, the simple comparison mid > high no longer guarantees that the pivot is on the right, because equal values can appear on both sides. The safe approach is to shrink the search window by one when nums[mid] == nums[high], reducing to O(n) worst‑case but still O(log n) average. Alternatively, you can use a modified binary search that checks both halves when equality occurs.

Q2A fintech platform needs to detect the earliest transaction timestamp in a rotated log file. What considerations would you discuss with the engineering team?

I would emphasize that the log file is essentially a rotated sorted array of timestamps. The algorithm runs in O(log n) time, which is critical for real‑time compliance checks. I’d also discuss handling edge cases like empty logs, single‑element logs, and ensuring the algorithm is stable under concurrent updates by using immutable snapshots or read‑only views.

Q3During a startup interview, you’re asked to explain why binary search works on a rotated array. What key point would you highlight?

The crux is that one half of the array remains sorted, so the pivot must lie in the unsorted half. By comparing the middle element to the high‑end element, we can determine which half contains the pivot and discard the other, guaranteeing logarithmic progress.

Examples

Example 1

Input

nums = [4, 5, 6, 7, 0, 1, 2]

Output

Explanation: The array is rotated such that the minimum element is 0. The index of 0 is 4. Binary search compares the middle element with the last element to determine which half contains the minimum. Since nums[mid] > nums[right], the minimum is in the right half. This process continues until the minimum is isolated at index 4.

Example 2

Input

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

Output

Explanation: The array is not rotated; it is already sorted in ascending order. The minimum element is 1, located at index 0. Binary search will determine that the left half is sorted and the minimum is at the start of the array.

Example 3

Input

nums = [3, 1, 2]

Output

Explanation: The array is rotated with the minimum element 1 at index 1. Binary search starts with left=0, right=2. mid=1. nums[mid]=1, nums[right]=2. Since nums[mid] < nums[right], the minimum is in the left half (including mid). right=mid=1. Now left=0, right=1. mid=0. nums[mid]=3, nums[right]=1. Since nums[mid] > nums[right], the minimum is in the right half. left=mid+1=1. Now left=1, right=1. The loop ends, and the answer is left=1.

Example 4

Input

nums = [2, 3, 4, 5, 6, 7, 1]

Output

Explanation: The minimum element is 1 at index 6. Binary search identifies that the right half contains the minimum because nums[mid] > nums[right] in the initial steps. The search narrows down to the last element, which is the smallest.

Constraints

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

Optimal Approach & Strategy

Use binary search: maintain low and high pointers; while low < high, compute mid. If nums[mid] > nums[high], set low = mid + 1; else set high = mid. When low == high, that index is the minimum. This runs in O(log n) time and O(1) space.

Brute Force Approach

Scan the array from left to right and return the first index where nums[i] < nums[i-1]. This takes O(n) time and O(1) space but is too slow for large inputs.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(log n)

function findPeakElement(nums) {
   let left = 0, right = nums.length - 1;
   while (left < right) {
       let mid = Math.floor((left + right) / 2);
       if (nums[mid] > nums[mid + 1]) {
           right = mid;
       } else {
           left = mid + 1;
       }
   }
   return left;
}

class Solution {
public:
   int findPeakElement(vector<int>& nums) {
       int left = 0, right = nums.size() - 1;
       while (left < right) {
           int mid = left + (right - left) / 2;
           if (nums[mid] > nums[mid + 1]) {
               right = mid;
           } else {
               left = mid + 1;
           }
       }
       return left;
   }
};

class Solution {
   public int findPeakElement(int[] nums) {
       int left = 0, right = nums.length - 1;
       while (left < right) {
           int mid = left + (right - left) / 2;
           if (nums[mid] > nums[mid + 1]) {
               right = mid;
           } else {
               left = mid + 1;
           }
       }
       return left;
   }
}

def find_peak_element(nums):
   left, right = 0, len(nums) - 1
   while left < right:
       mid = (left + right) // 2
       if nums[mid] > nums[mid + 1]:
           right = mid
       else:
           left = mid + 1
   return left

function findPeakElement(nums) {
   let left = 0, right = nums.length - 1;
   while (left < right) {
       let mid = Math.floor((left + right) / 2);
       if (nums[mid] > nums[mid + 1]) {
           right = mid;
       } else {
           left = mid + 1;
       }
   }
   return left;
}

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Rotated Matrix Pivot Analyzer 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

Rotated Matrix Pivot Analyzer 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