Frequency Window Constraint Resolver 6 — 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 frequency window constraint using the Union-Find Disjoint Set methodology.

DSA Pattern Breakdown

"Frequency Window Constraint Resolver 6"

easy

WHY DOES IT MATTER?

Transforming range‑based constraints into connectivity problems lets us leverage DSU’s near‑constant operations, turning quadratic sliding‑window checks into linear passes. This pattern appears in many graph‑like grouping tasks such as island counting, friend circles, and clustering based on proximity.

OPTIMIZATION CHALLENGE

The key insight is that the window condition is *local* – it only depends on adjacent elements. By uniting only neighboring indices that satisfy the condition, we propagate the property transitively, eliminating the need to examine every possible pair.

REAL-WORLD CONNECTION

Think of a distributed cache where nodes with similar load (within a tolerance) are grouped into the same shard for load‑balancing. Union‑Find acts like the controller that merges shards when their load difference falls below the threshold, ensuring balanced clusters without scanning every node pair.

During an interview, implement DSU with path compression and union‑by‑size in a single class, and keep the size array up‑to‑date during unions. This lets you answer the “size of each window” query instantly without a second pass.

COMPLEXITY AT A GLANCE

⏱ Time:O(N α(N))

💾 Space:O(N)

Core Theory — Why This Approach?

Union‑Find (Disjoint Set Union, DSU) is a data‑structure that maintains a partition of a set into disjoint subsets and supports two operations in near‑constant amortized time: **find** – retrieve the representative (root) of the subset containing an element, and **union** – merge two subsets. The classic implementation uses path compression during find and union by rank/size, yielding an inverse‑Ackermann α(N) bound, which for all practical N behaves like O(1). In the Frequency Window Constraint problem we treat each position in the dataset as a node in a graph; edges are implicitly defined by the constraint that two positions belong to the same frequency window if their values differ by at most a given threshold. By iterating over the array and uniting adjacent indices that satisfy the window condition, we can collapse the entire dataset into a collection of maximal windows, each represented by a DSU root. The size of each component directly gives the frequency count for that window.

A naïve solution would scan every possible sub‑array, checking the max‑min difference for each window – an O(N²) or worse approach that quickly explodes for N up to 10⁵ or 10⁶. Moreover, recomputing the window condition for overlapping intervals leads to redundant work. The DSU‑based paradigm eliminates this redundancy: each element is processed once, and unions are performed only when the local constraint holds, guaranteeing linear traversal. Because the union operation merges whole components, the algorithm automatically propagates the window property across the entire connected region without revisiting individual pairs.

The optimal paradigm therefore combines a single linear pass with DSU’s almost‑constant operations. After the pass, a final pass over the roots aggregates the frequencies, delivering the answer in O(N α(N)) time and O(N) auxiliary space. This pattern is a textbook example of converting a sliding‑window or range‑constraint problem into a connectivity problem, where DSU shines.

Interview Questions on This Problem

Q1How would you use Union‑Find to compute the size of all maximal frequency windows in a one‑dimensional array where adjacent elements belong to the same window if their absolute difference ≤ K?

Iterate the array once; for each index i, if |arr[i]‑arr[i‑1]| ≤ K, call union(i, i‑1). After the loop, iterate over all indices, find their root, and maintain a hashmap root→size (or use DSU size array). The hashmap values are the sizes of maximal windows.

Q2Why does a naïve O(N²) sliding‑window check fail for N = 10⁵, and how does DSU improve the asymptotic complexity?

A naïve check examines every possible sub‑array, leading to ~N² comparisons, which is infeasible for N = 10⁵ (≈10¹⁰ operations). DSU reduces work by merging only adjacent pairs that satisfy the constraint, performing at most N‑1 unions; each union/find is O(α(N)), so total time becomes O(N α(N)), effectively linear.

Q3In a distributed system, how could the DSU approach be adapted to compute frequency windows across data shards?

Each shard runs a local DSU to form windows within its segment. Boundary elements are exchanged; if the constraint holds across a shard boundary, a remote union is performed (e.g., via a coordinator or consensus). Finally, a reduction step merges the component sizes from all shards, yielding global window frequencies.

Examples

Example 1

Input

N = 5, constraints = [1, 2, 3, 4, 5], values = [10, 20, 30, 40, 50]

Output

Explanation: Step 1: Initialize the Union-Find Disjoint Set with N elements. Each element is its own parent. Step 2: Iterate through the constraints and values. For each constraint, find the parent of the first element and the parent of the second element. If they are different, union them. Step 3: After iterating through all constraints, find the parent of the maximum value. This will be the frequency window constraint.

Example 2

Input

N = 10, constraints = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], values = [100, 200, 300, 400, 500, 600, 700, 800, 900, 1000]

Output

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

Walk the array once, union adjacent indices that satisfy the constraint, then read component sizes – O(N α(N)) time.

Brute Force Approach

Check every possible sub‑array, compute its max‑min difference, and count it if the difference ≤ K – O(N²) time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N α(N))

function unionFindDisjointSet(N) {
   let parent = new Array(N + 1).fill(0).map((val, idx) => idx);
   let rank = new Array(N + 1).fill(0);

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

   function union(x, y) {
       let rootX = find(x);
       let 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]++;
           }
       }
   }

   function frequencyWindowConstraint(constraints, values) {
       for (let i = 0; i < constraints.length; i++) {
           union(constraints[i], values[i]);
       }

       let maxVal = Math.max(...values);
       return find(maxVal);
   }

   return frequencyWindowConstraint;
}

class Solution {
   public:
       int frequencyWindowConstraint(vector<int>& constraints, vector<int>& values) {
           int N = values.size();
           vector<int> parent(N + 1);
           vector<int> rank(N + 1);

           for (int i = 0; i <= N; i++) {
               parent[i] = i;
           }

           for (int i = 0; i < N; i++) {
               union(constraints[i], values[i]);
           }

           int maxVal = *max_element(values.begin(), values.end());
           return find(maxVal);
       }

       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]++;
               }
           }
       }
   };

class Solution {
   public int frequencyWindowConstraint(int[] constraints, int[] values) {
       int N = values.length;
       int[] parent = new int[N + 1];
       int[] rank = new int[N + 1];

       for (int i = 0; i <= N; i++) {
           parent[i] = i;
       }

       for (int i = 0; i < N; i++) {
           union(constraints[i], values[i]);
       }

       int maxVal = Arrays.stream(values).max().getAsInt();
       return find(maxVal);
   }

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

   private 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 union_find_disjoint_set(N):
   parent = [i for i in range(N + 1)]
   rank = [0] * (N + 1)

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

   def union(x, y):
       rootX = find(x)
       rootY = find(y)

       if rootX != rootY:
           if rank[rootX] > rank[rootY]:
               parent[rootY] = rootX
           elif rank[rootX] < rank[rootY]:
               parent[rootX] = rootY
           else:
               parent[rootY] = rootX
               rank[rootX] += 1

   def frequency_window_constraint(constraints, values):
       for i in range(len(constraints)):
           union(constraints[i], values[i])

       max_val = max(values)
       return find(max_val)

   return frequency_window_constraint

function unionFindDisjointSet(N) {
   let parent = new Array(N + 1).fill(0).map((val, idx) => idx);
   let rank = new Array(N + 1).fill(0);

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

   function union(x, y) {
       let rootX = find(x);
       let 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]++;
           }
       }
   }

   function frequencyWindowConstraint(constraints, values) {
       for (let i = 0; i < constraints.length; i++) {
           union(constraints[i], values[i]);
       }

       let maxVal = Math.max(...values);
       return find(maxVal);
   }

   return frequencyWindowConstraint;
}

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Frequency Window Constraint Resolver 6 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Graphs

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Frequency Window Constraint Resolver 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