Accelerated Node Cluster — Problem Statement & Solution Guide

GraphsMediumDepth-First Search

TimeO(n)

SpaceO(1)

Problem Description

You are given an array of integers nums with length N. Two indices i and j (0‑based) are considered connected if the following two conditions hold:

1. The absolute difference of the indices does not exceed a given integer K: |i - j| <= K.

2. The absolute difference of the values at those indices does not exceed a given integer D: |nums[i] - nums[j]| <= D.

Treating each index as a node and each pair that satisfies both conditions as an undirected edge, the graph may split into several connected components. The *accelerated node cluster* is defined as the size (number of nodes) of the largest connected component in this graph.

Your task is to compute and output the size of this largest component.

**Input**

The first line contains three space‑separated integers: N, K, and D.

The second line contains N space‑separated integers representing the array nums.

**Output**

Print a single integer: the size of the largest connected component.

**Example**

If N = 5, K = 1, D = 2 and nums = [1, 3, 5, 2, 4], the largest component contains the nodes at indices 0, 1, and 2, so the output is 3.

Core Theory — Why This Approach?

The 'Accelerated Node Cluster' problem models a functional graph (or successor graph) where each array index represents a node and its value represents its sole outgoing edge. To identify the size and membership of reachable clusters (cycles) efficiently, we must perform a Depth-First Search (DFS). However, a standard DFS requires O(N) auxiliary space for either a recursion stack or a visited set, which violates the strict O(1) auxiliary space constraint. To resolve this, we must perform an iterative, in-place DFS by repurposing the input array as its own state-tracking mechanism.

By mutating the values in the array during traversal—such as negating indices or offsetting values to represent 'visiting' and 'visited' states—we can track our active search path and detect cycle boundaries without external memory. Once a cycle (the core of the cluster) is detected, we can backtrack or run a secondary pointer to compute the cluster's size and propagate this value to all lead-in nodes. This approach elegantly balances the O(N) linear time complexity with O(1) auxiliary space by leveraging the mathematical property that every node in a functional graph eventually terminates in a single cycle.

Interview Questions on This Problem

Q1What is the core trick to tracking visited nodes and detecting cycles during the DFS without using an external set or recursion stack?

The core trick is in-place state mutation of the input array. As we traverse from node to node, we can temporarily modify the array values (for example, by negating them, adding a large sentinel offset, or swapping them) to mark them as 'currently visiting'. If we encounter a node already marked with the current traversal's identifier, we have found a cycle. Once the cycle length is computed, we do a second pass to write the final cluster size into these elements, restoring or converting their values to mark them as 'fully processed'.

Q2How do you prove that this iterative DFS achieves O(N) time complexity and O(1) auxiliary space?

The auxiliary space is O(1) because we only maintain a few scalar pointers (e.g., current, next, and cycle-start) and modify the input array in-place, avoiding recursion stacks. The time complexity is O(N) because each node is visited at most three times: once during the initial discovery traversal, once to resolve and count the cycle size when a cycle is detected, and once to propagate the cluster sizes to the precursor nodes that lead into the cycle.

Q3How does your O(1) space algorithm handle the edge cases of self-loops (A[i] = i) and long linear paths that do not contain a cycle until the very end?

A self-loop is detected immediately on the first step of the traversal because the next state matches the current index. The algorithm records a cluster size of 1 and marks it processed. For long linear chains leading to a cycle (a 'rho' shape), our in-place DFS tracks the path. Once the cycle at the end of the chain is detected and sized, we backtrack or re-traverse the linear portion, assigning each precursor node the cycle's size (plus its distance to the cycle) and marking them as processed, ensuring we never re-traverse them.

Q4If the input array were strictly read-only, how would you modify the algorithm to find the largest node cluster while keeping O(1) auxiliary space, and what would be the trade-off?

If the array is read-only, we cannot perform in-place mutations to mark visited states. To maintain O(1) auxiliary space, we would have to use Floyd's Cycle Detection algorithm (Tortoise and Hare) starting from each node. However, without a global visited array to skip previously processed nodes, we would end up re-traversing overlapping paths multiple times. This would degrade our worst-case time complexity from O(N) to O(N^2).

Examples

Example 1

Input

5 1 2
1 3 5 2 4

Output

Explanation: Indices 0 and 1 are connected because |0-1|=1≤1 and |1-3|=2≤2. Indices 1 and 2 are connected because |1-2|=1≤1 and |3-5|=2≤2. Indices 2 and 3 are not connected because |5-2|=3>2. Indices 3 and 4 are connected because |3-4|=1≤1 and |2-4|=2≤2. Thus we have two components: {0,1,2} of size 3 and {3,4} of size 2. The largest size is 3.

Example 2

Input

6 2 1
10 9 8 7 6 5

Output

Explanation: With K=2 and D=1, every consecutive pair of indices differs by at most 2 and the values differ by exactly 1, so each adjacent pair is connected. This forms a single chain that includes all 6 nodes. Therefore the largest component size is 6.

Example 3

Input

4 3 0
5 5 5 5

Output

Explanation: All values are equal, so |nums[i]-nums[j]|=0 for any pair. Since K=3, any two indices with distance up to 3 are connected. The graph is fully connected, giving a single component of size 4.

Constraints

1 <= N <= 100000
1 <= K <= N
0 <= D <= 1000000000
-1000000000 <= nums[i] <= 1000000000

Optimal Approach & Strategy

Use Depth-First Search to maintain a running state in O(N) time and O(1) auxiliary space.

Brute Force Approach

Iterate over all pairs/subarrays using nested loops and calculate the metric in O(N^2) time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums) {
    if (nums.length < 2) {
      return 0;
    }
    nums.sort((a, b) => b - a);
    return nums[0] + nums[1];
}

class Solution {
public:
    int solution(vector<int>& nums) {
      if (nums.size() < 2) {
        return 0;
      }
      sort(nums.rbegin(), nums.rend());
      return nums[0] + nums[1];
    }
  };

class Solution {
    public int solution(int[] nums) {
      if (nums.length < 2) {
        return 0;
      }
      Arrays.sort(nums);
      for (int i = nums.length - 1; i >= 0; i--) {
        if (i == nums.length - 1 || nums[i] != nums[i - 1]) {
          return nums[i] + nums[i - 1];
        }
      }
      return 0;
    }
  }

def solution(nums):
    if len(nums) < 2:
      return 0
    nums.sort(reverse=True)
    return nums[0] + nums[1]

function solution(nums) {
    if (nums.length < 2) {
      return 0;
    }
    nums.sort((a, b) => b - a);
    return nums[0] + nums[1];
}

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Accelerated Node Cluster Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Graphs

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Accelerated Node Cluster — Problem Statement & Solution Guide

Problem Description

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