Rotated Matrix Pivot Analyzer 2 — Problem Statement & Solution Guide

Binary SearchMediumSearch Peak Element

TimeO(log N)

SpaceO(1)

Problem Description

You are given a complex dataset of length $N$ representing system constraints and values. Your task is to calculate the rotated matrix pivot using the **Search Peak Element** methodology.

DSA Pattern Breakdown

"Rotated Matrix Pivot Analyzer 2"

medium

WHY DOES IT MATTER?

Binary search on a rotated structure isolates the pivot in logarithmic steps, turning a potentially linear scan into a scalable solution. This pattern appears in many real‑world scenarios where data is cyclically shifted, such as time‑series logs, circular buffers, and versioned configuration snapshots.

OPTIMIZATION CHALLENGE

The key insight is that the array is partially ordered on both sides of the pivot; by comparing the middle element with its neighbours you can determine which half still respects the sorted order and safely discard the other half, reducing the search space by half each iteration.

REAL-WORLD CONNECTION

Consider a distributed log ring buffer where entries wrap around after reaching capacity. Finding the oldest (pivot) entry quickly is analogous to locating the rotation point, enabling fast recovery and replay without scanning the entire buffer.

During an interview, first verify the array's monotonic property on each side, then write the binary‑search loop with clear invariant comments – this demonstrates both algorithmic understanding and clean coding discipline.

COMPLEXITY AT A GLANCE

⏱ Time:O(log N)

💾 Space:O(1)

Core Theory — Why This Approach?

The rotated matrix pivot problem can be reduced to finding a peak element in a one‑dimensional representation of the matrix. A peak is an index i such that arr[i] > arr[i‑1] and arr[i] > arr[i+1] (with appropriate boundary checks). In a rotated sorted array, the pivot – the smallest element – is also the point where the monotonic increase breaks, which can be discovered by comparing middle elements with their neighbours. A naïve linear scan examines every entry, leading to O(N) time, which is prohibitive for large N (10^6+). The optimal paradigm leverages the monotonic property on either side of the pivot and applies binary search: at each step we discard half of the search space based on the relative ordering of mid and its neighbours, guaranteeing logarithmic time. This approach also uses constant extra space because we only keep index pointers.

Interview Questions on This Problem

Q1How would you modify the binary‑search pivot finder to work on a matrix that is row‑wise rotated but each row remains sorted?

Treat each row as an independent rotated sorted array and apply the same binary‑search pivot logic per row; if a global pivot across rows is required, first locate the row containing the smallest element by comparing the first element of each row (which is the rotation point) using binary search on the first column, then run the 1‑D pivot search within that row.

Q2Why does the classic "find minimum in rotated sorted array" algorithm still work when duplicate values are present, and what adjustment is needed?

Duplicates can break the strict ordering guarantee, causing arr[mid] == arr[high] and making it ambiguous which side contains the pivot. The fix is to shrink the high pointer by one (high--) when arr[mid] == arr[high], which preserves O(N) worst‑case but retains O(log N) average performance.

Q3Explain how the peak‑element binary search can be extended to a 2‑D matrix where each row and column is sorted but the matrix is rotated 90 degrees.

Map the 2‑D matrix to a virtual 1‑D array using row‑major indexing; the rotation translates to a shift in the virtual ordering. Perform binary search on the virtual indices, converting mid back to (row, col) to compare with its four neighbours, discarding quadrants that cannot contain the pivot based on monotonicity.

Examples

Example 1

Input

[100, 150, 200, 250, 300]

Output

Explanation: Step-by-step: Given the input array [100, 150, 200, 250, 300], we first find the peak element using the Search Peak Element methodology. The peak element is 300. Then, we calculate the sum of the array elements from the start to the peak element, which is 100 + 150 + 200 + 250 + 300 = 1000.

Example 2

Input

[300, 200, 250, 150, 100]

Output

Explanation: Step-by-step: Given the input array [300, 200, 250, 150, 100], we first find the peak element using the Search Peak Element methodology. The peak element is 300. Then, we calculate the sum of the array elements from the start to the peak element, which is 300 + 200 + 250 + 150 + 100 = 1000.

Constraints

1 <= N <= 2 * 10^5
-10^9 <= arr[i] <= 10^9
Time Complexity: O(N) or O(N log N)
Space Complexity: O(N) or O(1)

Optimal Approach & Strategy

Use binary search: compare the middle element with its neighbours to decide which half contains the pivot, then narrow the search interval by half each iteration.

Brute Force Approach

Scan the entire array from start to finish, tracking the smallest element or checking each index for the peak condition. This takes linear time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(log N)

function solution(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;
       }
   }
   let sum = 0;
   for (let i = 0; i <= left; i++) {
       sum += nums[i];
   }
   return sum;
}

class Solution {
   public:
   int solution(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;
           }
       }
       int sum = 0;
       for (int i = 0; i <= left; i++) {
           sum += nums[i];
       }
       return sum;
   }
};

class Solution {
   public int solution(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;
           }
       }
       int sum = 0;
       for (int i = 0; i <= left; i++) {
           sum += nums[i];
       }
       return sum;
   }
}

def solution(nums):
   left = 0
   right = len(nums) - 1
   while left < right:
       mid = (left + right) // 2
       if nums[mid] > nums[mid + 1]:
           right = mid
       else:
           left = mid + 1
   sum = 0
   for i in range(left + 1):
       sum += nums[i]
   return sum

function solution(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;
       }
   }
   let sum = 0;
   for (let i = 0; i <= left; i++) {
       sum += nums[i];
   }
   return sum;
}

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Rotated Matrix Pivot Analyzer 2 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Binary Search

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Rotated Matrix Pivot Analyzer 2 — 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