Rotated Matrix Pivot Validator 3 — Problem Statement & Solution Guide

HeapMediumReorganize String Frequency

TimeO(N log K)

SpaceO(K)

Problem Description

You are tasked with validating a sequence of integer weights that represent pivot candidates in a rotated matrix configuration. The validation process relies on a frequency-based reorganization strategy to determine if a stable pivot can be established. Given an array nums of length N, you must determine if it is possible to rearrange the elements such that no two adjacent elements are identical. If such a rearrangement exists, return the maximum frequency of any single element in the original array, which serves as the 'pivot weight'. If no valid rearrangement is possible (i.e., the most frequent element appears more than half the time, rounded up), return -1.

The core logic involves counting the frequency of each distinct value. Let maxFreq be the highest frequency among all values and N be the total length of the array. A valid non-adjacent rearrangement is possible if and only if maxFreq <= (N + 1) / 2. This condition ensures that the most frequent element can be distributed across the array without forcing two identical elements to be neighbors.

Your function should take the array nums as input and return an integer. If the pivot condition is satisfied, return maxFreq. Otherwise, return -1. This problem tests your ability to apply frequency analysis and greedy placement logic to validate structural constraints in data sequences.

DSA Pattern Breakdown

"Rotated Matrix Pivot Validator 3"

medium

WHY DOES IT MATTER?

The pattern exemplifies frequency‑driven greedy reordering, a staple for problems where adjacency constraints exist (e.g., task scheduling, reorganizing strings). Mastery of this pattern enables engineers to design solutions that avoid exponential blow‑up by focusing on the most constrained element first.

OPTIMIZATION CHALLENGE

The key insight is that only the relative ordering of the most frequent elements matters; by using a max‑heap we reduce the selection of the next candidate from O(K) to O(log K), turning a potentially quadratic scan into a linearithmic process.

REAL-WORLD CONNECTION

Think of a distributed job scheduler that must assign high‑frequency tasks to different time slots to avoid resource contention. The scheduler repeatedly picks the two busiest queues and interleaves their jobs, mirroring the heap‑based reorganization of matrix pivots.

During an interview, first compute the frequency map and immediately check the (N+1)/2 bound. If it passes, push all (count, value) pairs into a max‑heap and pop two at a time—this pattern is quick to code and hard to mess up.

COMPLEXITY AT A GLANCE

⏱ Time:O(N log K)

💾 Space:O(K)

Core Theory — Why This Approach?

The core of this problem is a frequency‑based greedy strategy that can be efficiently implemented with a max‑heap (priority queue). By always selecting the two most frequent remaining elements and placing them consecutively, we guarantee that the most “dangerous” element (the one with the highest remaining count) is spaced out as much as possible, preventing identical adjacency. This approach works because the feasibility condition reduces to checking whether the maximum frequency exceeds (N+1)/2; if it does, no arrangement can avoid adjacent duplicates. Naïve brute‑force permutations explode factorially (O(N!)) and are infeasible for N beyond ~10, while the heap‑based method runs in O(N log K) where K is the number of distinct values, typically far smaller than N. The optimal paradigm thus combines counting (O(N)) with a priority queue to enforce the greedy choice, yielding an overall linearithmic solution that scales to large inputs.

Interview Questions on This Problem

Q1How can you determine in O(N) time whether a valid rearrangement exists without actually constructing it?

Count the frequency of each number and find the maximum count maxFreq. If maxFreq <= (N + 1) / 2, a valid arrangement exists; otherwise it is impossible.

Q2Why does the greedy choice of always picking the two most frequent remaining elements guarantee a correct solution?

Choosing the two highest frequencies maximally separates the most frequent element from itself. If a solution exists, this greedy step never blocks future placements because any alternative would place a less frequent element earlier, leaving the high‑frequency element with even fewer safe slots.

Q3Can this problem be solved using a bucket sort instead of a heap? If so, outline the approach.

Yes. After counting frequencies, place each number into buckets indexed by its count. Iterate from the highest bucket down, repeatedly pulling the top two non‑empty buckets and appending their elements to the result. This mimics the heap’s behavior in O(N) time when the range of frequencies is bounded by N.

Examples

Example 1

Input

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

Output

Explanation: The frequencies are: 1 appears 3 times, 2 appears 2 times, 3 appears 1 time. Max frequency is 3. Total length N is 6. The threshold is (6 + 1) / 2 = 3.5, which floors to 3. Since 3 <= 3, a valid rearrangement exists (e.g., 1, 2, 1, 3, 1, 2). Return the max frequency, which is 3.

Example 2

Input

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

Output

-1

Explanation: The frequencies are: 1 appears 3 times, 2 appears 1 time, 3 appears 1 time. Max frequency is 3. Total length N is 5. The threshold is (5 + 1) / 2 = 3. Since 3 <= 3, a valid rearrangement exists (e.g., 1, 2, 1, 3, 1). Wait, let's re-evaluate. If maxFreq is 3 and N is 5, (5+1)/2 = 3. 3 <= 3 is true. So it should return 3. Let's pick a failing case. Let's use nums = [1, 1, 1, 1, 2]. Frequencies: 1 appears 4 times, 2 appears 1 time. Max frequency is 4. N is 5. Threshold is (5+1)/2 = 3. Since 4 > 3, no valid rearrangement exists. Return -1.

Example 3

Input

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

Output

-1

Explanation: The frequencies are: 5 appears 4 times, 2 appears 1 time, 3 appears 1 time. Max frequency is 4. Total length N is 6. The threshold is (6 + 1) / 2 = 3.5, which floors to 3. Since 4 > 3, it is impossible to arrange the elements such that no two 5s are adjacent. Return -1.

Example 4

Input

nums = [7, 8, 7, 8, 7, 8, 7]

Output

Explanation: The frequencies are: 7 appears 4 times, 8 appears 3 times. Max frequency is 4. Total length N is 7. The threshold is (7 + 1) / 2 = 4. Since 4 <= 4, a valid rearrangement exists (e.g., 7, 8, 7, 8, 7, 8, 7). Return the max frequency, which is 4.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
The array will contain at least one element.
Time complexity should be O(N) where N is the length of the array.
Space complexity should be O(N) to store the frequency map.

Optimal Approach & Strategy

Count frequencies, verify the max‑frequency condition, then use a max‑heap to greedily place the two most frequent remaining numbers at each step, achieving O(N log K) time.

Brute Force Approach

Generate every permutation of the array and check each one for adjacent equal elements; this runs in O(N!) time and is only viable for tiny N.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N log K)

/**
 * @param {number[]} nums
 * @return {number}
 */
var validatePivot = function(nums) {
    const freq = new Map();
    for (let num of nums) {
        freq.set(num, (freq.get(num) || 0) + 1);
    }
    
    const heap = [];
    for (let [val, f] of freq) {
        heap.push([f, val]);
    }
    
    heap.sort((a, b) => b[0] - a[0]);
    
    let count = 0;
    for (let [f, val] of heap) {
        count += f;
        if (count > nums.length / 2) {
            return val;
        }
    }
    return -1;
};

class Solution {
public:
    int validatePivot(vector<int>& nums) {
        unordered_map<int, int> freq;
        for (int num : nums) {
            freq[num]++;
        }
        
        priority_queue<pair<int, int>> maxHeap;
        for (auto& p : freq) {
            maxHeap.push({p.second, p.first});
        }
        
        int count = 0;
        while (!maxHeap.empty()) {
            auto [f, val] = maxHeap.top();
            maxHeap.pop();
            count += f;
            if (count > nums.size() / 2) {
                return val;
            }
        }
        return -1;
    }
};

class Solution {
    public int validatePivot(int[] nums) {
        Map<Integer, Integer> freq = new HashMap<>();
        for (int num : nums) {
            freq.put(num, freq.getOrDefault(num, 0) + 1);
        }
        
        PriorityQueue<int[]> maxHeap = new PriorityQueue<>((a, b) -> b[0] - a[0]);
        for (Map.Entry<Integer, Integer> entry : freq.entrySet()) {
            maxHeap.offer(new int[]{entry.getValue(), entry.getKey()});
        }
        
        int count = 0;
        while (!maxHeap.isEmpty()) {
            int[] top = maxHeap.poll();
            count += top[0];
            if (count > nums.length / 2) {
                return top[1];
            }
        }
        return -1;
    }
}

class Solution:
    def validatePivot(self, nums: List[int]) -> int:
        from collections import Counter
        import heapq
        
        freq = Counter(nums)
        heap = []
        for val, f in freq.items():
            heapq.heappush(heap, (-f, val))
        
        count = 0
        while heap:
            f, val = heapq.heappop(heap)
            count += -f
            if count > len(nums) / 2:
                return val
        return -1

/**
 * @param {number[]} nums
 * @return {number}
 */
var validatePivot = function(nums) {
    const freq = new Map();
    for (let num of nums) {
        freq.set(num, (freq.get(num) || 0) + 1);
    }
    
    const heap = [];
    for (let [val, f] of freq) {
        heap.push([f, val]);
    }
    
    heap.sort((a, b) => b[0] - a[0]);
    
    let count = 0;
    for (let [f, val] of heap) {
        count += f;
        if (count > nums.length / 2) {
            return val;
        }
    }
    return -1;
};

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Rotated Matrix Pivot Validator 3 Solution

Problem Statement

Input

Output

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 3 — 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