Rotated Matrix Pivot Validator 6 — Problem Statement & Solution Guide

HeapMediumReorganize String Frequency

TimeO(N*M)

SpaceO(K)

Problem Description

You are given an N × M matrix of integers. A *pivot* is defined as the element that stays in the same cell after rotating the matrix 90 degrees clockwise. Such a cell can exist only when the matrix is square (N = M) and its size is odd; in that case the pivot is the unique central cell at position (⌊N/2⌋, ⌊N/2⌋). A pivot is considered *valid* if its value is the most frequent number in the entire matrix. If several numbers share the highest frequency, the larger number is regarded as the most frequent. Return the value of the valid pivot; if no valid pivot exists, return -1.

Input format:

- The first line contains two integers N and M.

- The next N lines each contain M space‑separated integers representing the matrix rows.

Output format:

- A single integer: the value of the valid pivot or -1 if the conditions are not satisfied.

DSA Pattern Breakdown

"Rotated Matrix Pivot Validator 6"

medium

WHY DOES IT MATTER?

This problem tests the ability to combine geometric reasoning with data structure selection. It highlights the importance of pre-processing conditions (checking for square/odd dimensions) before diving into complex algorithms. It also reinforces the concept that the 'most frequent' element problem is fundamentally a frequency counting problem, best solved with Hash Maps, not Heaps, unless k-selection is involved.

OPTIMIZATION CHALLENGE

The key optimization is recognizing that you don't need to find the global most frequent element if you can prove the pivot is not the most frequent early. However, in the worst case, you must count all frequencies. The real optimization is avoiding unnecessary work by first checking the geometric conditions. If the matrix is not square or even-sized, return false immediately in O(1) time.

REAL-WORLD CONNECTION

In image processing, identifying invariant features under rotation is crucial for object recognition. Similarly, in database indexing, understanding which keys remain stable under schema transformations (like partitioning) helps in designing efficient query plans. The frequency counting aspect mirrors log analysis, where identifying the most common error code or user action is a standard task.

In interviews, always start by clarifying the constraints. If the matrix is guaranteed to be square and odd, you can skip the geometric check. If not, include it. Also, be prepared to discuss the trade-offs between Hash Map and Heap-based solutions, even if the Hash Map is superior here. This shows depth of understanding.

COMPLEXITY AT A GLANCE

⏱ Time:O(N*M)

💾 Space:O(K)

Core Theory — Why This Approach?

The problem combines geometric invariance with frequency analysis. First, we must identify the pivot cell. A 90-degree clockwise rotation maps cell (i, j) in an N x M matrix to (j, N-1-i) in the M x N matrix. For a cell to remain in the same position, we require i = j and j = N-1-i, which implies i = N-1-i, or 2i = N-1. This equation has an integer solution only if N-1 is even, meaning N is odd. Furthermore, for the matrix to be square (N=M), the dimensions must match. Thus, the pivot exists only if N=M and N is odd, located at (N//2, N//2). If these conditions are not met, no pivot exists, and the validation fails immediately.

Once the pivot value is identified, the problem reduces to determining if this value is the most frequent in the entire matrix. A naive approach would involve counting frequencies of all elements and comparing them, which is O(N*M) time and space. However, the problem context suggests an interest in Heap-based solutions or efficient frequency tracking. While a Hash Map is the standard O(N*M) solution for frequency counting, understanding why a Heap might be considered (e.g., for finding the k-th largest frequency) is crucial for interview depth. In this specific case, since we only need to check if one specific value is the maximum frequency, a single-pass frequency count with a Hash Map is optimal. The 'Heap' tag in the topic likely refers to the broader category of selection problems or is a distractor, but the core algorithmic challenge is efficient frequency verification.

Interview Questions on This Problem

Q1How would you modify this solution if the matrix was not square, but you still needed to find the element that remains in the same relative position after a 90-degree rotation?

For a non-square matrix, no element remains in the exact same cell index (i,j) after rotation because the dimensions swap. However, if the question implies 'relative position' in a normalized sense, it's ill-defined. More likely, the interviewer is testing if you recognize that the pivot condition (i = j and j = N-1-i) strictly requires a square matrix. You should explain that for non-square matrices, the set of fixed points is empty, so the validation returns false immediately without frequency counting.

Q2If the matrix is extremely large (e.g., 10^6 x 10^6) and stored on disk, how would you optimize the memory usage for the frequency count?

You cannot load the entire matrix into memory. You would use an external sorting algorithm or a distributed hash map (like in Hadoop/Spark) to count frequencies. Alternatively, if the value range is small, you could use a bitset or a compressed frequency table. For the specific pivot value, you could perform a single pass to count the pivot's frequency and a second pass to find the maximum frequency of any other element, but this still requires O(N*M) time. The key insight is that you don't need to store all frequencies if you can compute the max frequency in a streaming fashion, though this is complex. A more practical answer is to use a distributed system to partition the matrix and count frequencies in parallel.

Q3Why is a Heap not the optimal data structure for finding the most frequent element in this specific problem, despite the topic tag?

A Heap is useful for finding the k-th largest/smallest element or for merging sorted lists. To find the most frequent element, you first need to count frequencies, which requires a Hash Map. Once you have the frequencies, you could use a Max-Heap to find the maximum frequency in O(K log K) time where K is the number of unique elements. However, since we only need to check if a *specific* value (the pivot) is the most frequent, we can simply compare its count to the maximum count found during the Hash Map construction. This avoids the overhead of heap operations, making the Hash Map approach O(N*M) time and O(K) space, which is optimal.

Examples

Example 1

Input

Output

Explanation: The matrix is 3 × 3 (odd square), so the central cell is at (1,1) with value 5. Frequency count: 5 appears 3 times, all other numbers appear once. The most frequent value is 5, which matches the central cell, therefore the valid pivot is 5.

Example 2

Input

Output

-1

Explanation: The matrix is 5 × 5, so the central cell is (2,2) with value 9. Frequency table: 2 → 5 times, 9 → 4 times, all others fewer. The most frequent number is 2, not the central value 9, therefore no valid pivot exists and the answer is -1.

Example 3

Input

2 3
7 7 3
1 4 5

Output

-1

Explanation: The matrix is not square, so a cell cannot remain in place after a 90° rotation. Consequently, a pivot cannot exist and the result is -1.

Constraints

1 ≤ N, M ≤ 10^3
-10^9 ≤ matrix[i][j] ≤ 10^9
The total number of elements N·M does not exceed 10^6

Optimal Approach & Strategy

Check geometric conditions in O(1). If valid, traverse the matrix once to count frequencies using a Hash Map and track the max frequency. Compare the pivot's frequency to the max. This is O(N*M) time and O(K) space.

Brute Force Approach

Rotate the matrix physically and compare each cell to its original position to find the pivot, then sort the entire matrix to count frequencies. This is O(N*M log(N*M)) time and O(N*M) space, which is inefficient.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N*M)

function solution(matrix) {
   // JavaScript solution
   // Define the Reorganize String Frequency methodology
   // Define the rotated matrix pivot
   // Find the rotated matrix pivot using the Reorganize String Frequency methodology
   return 0;
}

class Solution {
   public:
      int solution(int matrix[][matrix[0].size()]) {
         // C++ solution
         // Define the Reorganize String Frequency methodology
         // Define the rotated matrix pivot
         // Find the rotated matrix pivot using the Reorganize String Frequency methodology
         return 0;
      }
   };

class Solution {
   public int solution(int[][] matrix) {
      // Java solution
      // Define the Reorganize String Frequency methodology
      // Define the rotated matrix pivot
      // Find the rotated matrix pivot using the Reorganize String Frequency methodology
      return 0;
   }
}

def solution(matrix):
   # Python solution
   # Define the Reorganize String Frequency methodology
   # Define the rotated matrix pivot
   # Find the rotated matrix pivot using the Reorganize String Frequency methodology
   return 0

function solution(matrix) {
   // JavaScript solution
   // Define the Reorganize String Frequency methodology
   // Define the rotated matrix pivot
   // Find the rotated matrix pivot using the Reorganize String Frequency methodology
   return 0;
}

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Rotated Matrix Pivot Validator 6 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Heap

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Rotated Matrix Pivot Validator 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