Rotated Matrix Pivot Validator 8 — Problem Statement & Solution Guide

GraphsEasyUnion-Find Disjoint Set

TimeO(N * α(N))

SpaceO(N)

Problem Description

Given a complex dataset of length N representing system constraints and values, calculate the rotated matrix pivot using the Union-Find Disjoint Set methodology.

DSA Pattern Breakdown

"Rotated Matrix Pivot Validator 8"

easy

WHY DOES IT MATTER?

The connectivity‑via‑DSU pattern abstracts many seemingly geometric or matrix‑based problems into simple equivalence classes, allowing O(N α(N)) solutions where brute force would be quadratic. Recognizing this pattern lets engineers solve large‑scale constraint validation tasks with minimal code.

OPTIMIZATION CHALLENGE

The key insight is to avoid explicit rotation simulation; instead, treat each rotation rule as an edge and union the endpoints. Path compression then guarantees that the final find for the pivot is essentially O(1).

REAL-WORLD CONNECTION

In distributed systems, DSU mirrors the way cluster membership protocols merge partitions after network partitions heal – each node belongs to a partition (set) and merges happen when connectivity is restored, just like rotating matrix indices merge into a single component.

During an interview, first sketch the implicit graph, then immediately propose DSU. Implement union‑by‑size and path compression; a one‑liner find after all unions often suffices to validate the pivot.

COMPLEXITY AT A GLANCE

⏱ Time:O(N * α(N))

💾 Space:O(N)

Core Theory — Why This Approach?

The rotated matrix pivot problem can be modeled as a connectivity question on an implicit graph where each index of the dataset represents a node and a valid rotation creates an edge between two nodes. By interpreting the constraints as equivalence relations, the Union‑Find (Disjoint Set Union, DSU) structure efficiently merges groups of indices that belong to the same rotation cycle. Naïve solutions that simulate each rotation step-by-step incur O(N²) time because each element may be visited repeatedly for every possible pivot. DSU, however, compresses paths and unions sets in near‑constant amortized time, yielding an almost linear solution even for massive N. The optimal paradigm therefore consists of a single pass to union related indices followed by a find operation to verify whether the proposed pivot belongs to the same component as the expected rotated position.

Interview Questions on This Problem

Q1How does Union‑Find achieve near‑constant time per operation, and why is it preferable to a DFS/BFS approach for this problem?

Union‑Find uses path compression during find and union by rank/size, which flattens the tree structure so subsequent finds are O(α(N)). Unlike DFS/BFS, which requires O(N) traversal for each query, DSU maintains connectivity information incrementally, making it ideal when many pivot validations are required.

Q2If the dataset contains duplicate values that map to the same rotated index, how does the DSU handle these collisions?

Duplicate values simply result in multiple union operations on the same pair of nodes. DSU’s union operation is idempotent – if the nodes are already in the same set, the operation is a no‑op, preserving correctness without extra overhead.

Q3Explain how you would extend the solution to support dynamic updates (inserting or deleting constraints) while still answering pivot queries efficiently.

Dynamic updates can be handled by maintaining a DSU that supports roll‑backs or using a link‑cut tree for fully dynamic connectivity. For insertions, we perform a new union; for deletions, we need a reversible DSU or rebuild the structure periodically, balancing update cost against query frequency.

Examples

Example 1

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Output

Explanation: Step-by-step: Given the array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we first create a Union-Find Disjoint Set data structure. Then, we iterate over the array, and for each element, we find its parent in the Union-Find Disjoint Set. If the parent is not the element itself, we update the parent to be the parent of the parent. Finally, we return the sum of the array elements, which is the sum of the elements in the Union-Find Disjoint Set.

Example 2

Input

[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]

Output

Explanation: Step-by-step: Given the array [10, 9, 8, 7, 6, 5, 4, 3, 2, 1], we first create a Union-Find Disjoint Set data structure. Then, we iterate over the array, and for each element, we find its parent in the Union-Find Disjoint Set. If the parent is not the element itself, we update the parent to be the parent of the parent. Finally, we return the sum of the array elements, which is the sum of the elements in the Union-Find Disjoint Set.

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

Build a DSU, union each index with its rotated partner in one pass, then verify the pivot with a single find operation, achieving O(N α(N)) time.

Brute Force Approach

Simulate every possible rotation by moving elements one step at a time and checking the pivot after each move, leading to O(N²) time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N * α(N))

function solution(nums) {
   const unionFind = new UnionFind(nums.length);
   let sum = 0;
   for (let num of nums) {
       unionFind.union(num, num);
       sum += num;
   }
   return sum;
}

class UnionFind {
   constructor(size) {
       this.parent = new Array(size).fill(0).map((_, i) => i);
       this.rank = new Array(size).fill(0);
   }

   find(x) {
       if (this.parent[x] !== x) {
           this.parent[x] = this.find(this.parent[x]);
       }
       return this.parent[x];
   }

   union(x, y) {
       const rootX = this.find(x);
       const rootY = this.find(y);
       if (rootX !== rootY) {
           if (this.rank[rootX] > this.rank[rootY]) {
               this.parent[rootY] = rootX;
           } else if (this.rank[rootX] < this.rank[rootY]) {
               this.parent[rootX] = rootY;
           } else {
               this.parent[rootY] = rootX;
               this.rank[rootX]++;
           }
       }
   }
}

class Solution {
public:
   int solution(vector<int>& nums) {
       UnionFind unionFind(nums.size());
       int sum = 0;
       for (int num : nums) {
           unionFind.union(num, num);
           sum += num;
       }
       return sum;
   }
};

class UnionFind {
public:
   UnionFind(int size) {
       parent = new int[size];
       rank = new int[size];
       for (int i = 0; i < size; i++) {
           parent[i] = i;
       }
   }

   ~UnionFind() {
       delete[] parent;
       delete[] rank;
   }

   int find(int x) {
       if (parent[x] != x) {
           parent[x] = find(parent[x]);
       }
       return parent[x];
   }

   void union(int x, int y) {
       int rootX = find(x);
       int rootY = find(y);
       if (rootX != rootY) {
           if (rank[rootX] > rank[rootY]) {
               parent[rootY] = rootX;
           } else if (rank[rootX] < rank[rootY]) {
               parent[rootX] = rootY;
           } else {
               parent[rootY] = rootX;
               rank[rootX]++;
           }
       }
   }

private:
   int* parent;
   int* rank;
}

class Solution {
   public int solution(int[] nums) {
       UnionFind unionFind = new UnionFind(nums.length);
       int sum = 0;
       for (int num : nums) {
           unionFind.union(num, num);
           sum += num;
       }
       return sum;
   }
}

class UnionFind {
   private int[] parent;
   private int[] rank;

   public UnionFind(int size) {
       parent = new int[size];
       rank = new int[size];
       for (int i = 0; i < size; i++) {
           parent[i] = i;
       }
   }

   public int find(int x) {
       if (parent[x] != x) {
           parent[x] = find(parent[x]);
       }
       return parent[x];
   }

   public void union(int x, int y) {
       int rootX = find(x);
       int rootY = find(y);
       if (rootX != rootY) {
           if (rank[rootX] > rank[rootY]) {
               parent[rootY] = rootX;
           } else if (rank[rootX] < rank[rootY]) {
               parent[rootX] = rootY;
           } else {
               parent[rootY] = rootX;
               rank[rootX]++;
           }
       }
   }
}

def solution(nums):
   union_find = UnionFind(len(nums))
   sum_ = 0
   for num in nums:
       union_find.union(num, num)
       sum_ += num
   return sum_

class UnionFind:
   def __init__(self, size):
       self.parent = [i for i in range(size)]
       self.rank = [0] * size

   def find(self, x):
       if self.parent[x] != x:
           self.parent[x] = self.find(self.parent[x])
       return self.parent[x]

   def union(self, x, y):
       root_x = self.find(x)
       root_y = self.find(y)
       if root_x != root_y:
           if self.rank[root_x] > self.rank[root_y]:
               self.parent[root_y] = root_x
           elif self.rank[root_x] < self.rank[root_y]:
               self.parent[root_x] = root_y
           else:
               self.parent[root_y] = root_x
               self.rank[root_x] += 1

function solution(nums) {
   const unionFind = new UnionFind(nums.length);
   let sum = 0;
   for (let num of nums) {
       unionFind.union(num, num);
       sum += num;
   }
   return sum;
}

class UnionFind {
   constructor(size) {
       this.parent = new Array(size).fill(0).map((_, i) => i);
       this.rank = new Array(size).fill(0);
   }

   find(x) {
       if (this.parent[x] !== x) {
           this.parent[x] = this.find(this.parent[x]);
       }
       return this.parent[x];
   }

   union(x, y) {
       const rootX = this.find(x);
       const rootY = this.find(y);
       if (rootX !== rootY) {
           if (this.rank[rootX] > this.rank[rootY]) {
               this.parent[rootY] = rootX;
           } else if (this.rank[rootX] < this.rank[rootY]) {
               this.parent[rootX] = rootY;
           } else {
               this.parent[rootY] = rootX;
               this.rank[rootX]++;
           }
       }
   }
}

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Rotated Matrix Pivot Validator 8 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Graphs

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Rotated Matrix Pivot Validator 8 — 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