Monotonic Parity Sequence — Problem Statement & Solution Guide

Linked ListMediumFloyd Cycle Detection

TimeO(n)

SpaceO(1)

Problem Description

You are given a singly linked list of integers. The list is guaranteed to contain a cycle, meaning some node's next pointer eventually points back to a previous node in the list. Your task is to determine the 'Monotonic Parity Sequence' of the cycle. First, identify the starting node of the cycle using Floyd's Cycle Detection algorithm. Once the cycle is isolated, traverse it to collect the integer values in order. The 'Monotonic Parity Sequence' is defined as the length of the longest contiguous subsequence within the cycle where the parity (even or odd) of the values remains strictly monotonic (either all even or all odd) and the values themselves are non-decreasing. If no such subsequence of length greater than 1 exists, return 1. Return the length of this longest subsequence.

DSA Pattern Breakdown

"Monotonic Parity Sequence"

medium

WHY DOES IT MATTER?

Detecting cycles and extracting properties from them is a recurring pattern in interview problems because real‑world data structures (e.g., routing tables, dependency graphs) can inadvertently form loops, and algorithms must remain robust without blowing up memory or time.

OPTIMIZATION CHALLENGE

The key insight is that both cycle detection and parity extraction can be merged into a single linear traversal after the entry point is known, eliminating the need for multiple passes or auxiliary containers.

REAL-WORLD CONNECTION

Think of a network packet circulating in a ring topology; Floyd’s algorithm is analogous to two monitoring agents moving at different speeds to discover if the packet is stuck in a loop, and the parity analysis mirrors checking whether packet sizes follow a monotonic trend for congestion control.

When coding, first implement Floyd’s detection as a black‑box helper; once you have the entry node, treat the cycle as a regular array by iterating until you return to the entry, which simplifies reasoning and avoids off‑by‑one bugs.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

Floyd’s Tortoise and Hare algorithm is the cornerstone for detecting cycles in a singly linked list using O(1) extra space. Two pointers advance at different speeds (slow moves one step, fast moves two); if a cycle exists they will eventually meet inside the loop. Once a meeting point is found, resetting one pointer to the head and moving both one step at a time pinpoints the exact entry node of the cycle. After isolating the cycle, a single linear pass collects the node values. The "Monotonic Parity Sequence" can then be derived by scanning the collected values and extracting the longest contiguous subsequence where the parity (even = 0, odd = 1) never decreases (or never increases), which is a classic monotonic‑subarray problem. Naïve solutions that repeatedly traverse the list to locate the cycle entry or that recompute parity for each sub‑segment incur O(n²) time and are infeasible for large lists, whereas the combined Floyd‑plus‑single‑scan approach runs in linear time with constant auxiliary memory.

Interview Questions on This Problem

Q1How does Floyd’s cycle detection algorithm guarantee O(1) space while still finding the start of a cycle?

The algorithm uses only two pointers that move at different speeds; they meet inside the loop if one exists. After the meeting, resetting one pointer to the head and moving both one step at a time makes them converge at the cycle entry because the distance from head to entry equals the distance from meeting point to entry along the loop. No extra data structures are needed, thus O(1) space.

Q2Given a linked list with a guaranteed cycle, how would you compute the longest monotonic parity subsequence of the cycle in O(n) time?

First locate the cycle entry with Floyd’s algorithm, then traverse the cycle once, recording each node’s parity (0 for even, 1 for odd). While scanning, maintain two counters for the current non‑decreasing and non‑increasing parity runs, updating a global maximum when a run breaks. This single pass yields the longest monotonic parity subsequence in O(n) time.

Q3Why might a hash‑set based cycle detection fail in a memory‑constrained environment, and how would you adapt the solution?

A hash‑set stores every visited node, leading to O(n) extra space, which can exceed memory limits for very long lists. Replacing it with Floyd’s two‑pointer technique eliminates the need for storage, preserving correctness while meeting strict space constraints.

Examples

Example 1

Input

linked_list: 3 -> 5 -> 7 -> 9 -> 11 -> (back to 5)

Output

Explanation: 1. Detect cycle: The cycle starts at node 5. The cycle nodes are [5, 7, 9, 11]. 2. Analyze parities: All values (5, 7, 9, 11) are odd. 3. Check monotonicity: The sequence 5, 7, 9, 11 is strictly increasing (non-decreasing). 4. Longest subsequence: The entire cycle [5, 7, 9, 11] has consistent parity (odd) and is non-decreasing. Length is 4.

Example 2

Input

linked_list: 2 -> 4 -> 3 -> 6 -> 8 -> (back to 3)

Output

Explanation: 1. Detect cycle: The cycle starts at node 3. The cycle nodes are [3, 6, 8]. 2. Analyze parities: 3 (odd), 6 (even), 8 (even). 3. Check subsequences: - [3]: Length 1. - [6, 8]: Both even, and 6 <= 8. Length 2. - [3, 6]: Mixed parity. Invalid. 4. Longest valid subsequence: [6, 8] with length 2.

Example 3

Input

linked_list: 10 -> 8 -> 6 -> 4 -> 2 -> (back to 8)

Output

Explanation: 1. Detect cycle: The cycle starts at node 8. The cycle nodes are [8, 6, 4, 2]. 2. Analyze parities: All values (8, 6, 4, 2) are even. 3. Check monotonicity: The sequence is 8, 6, 4, 2. This is strictly decreasing. 4. Condition: The problem requires non-decreasing values. Since 8 > 6, the subsequence breaks at the first step. No contiguous subsequence of length > 1 is non-decreasing. 5. Result: Return 1.

Example 4

Input

linked_list: 1 -> 2 -> 3 -> 4 -> 5 -> (back to 2)

Output

Explanation: 1. Detect cycle: The cycle starts at node 2. The cycle nodes are [2, 3, 4, 5]. 2. Analyze parities: 2 (even), 3 (odd), 4 (even), 5 (odd). 3. Check subsequences: - [2]: Even. Length 1. - [3]: Odd. Length 1. - [4]: Even. Length 1. - [5]: Odd. Length 1. - [2, 3]: Mixed parity. Invalid. - [3, 4]: Mixed parity. Invalid. - [4, 5]: Mixed parity. Invalid. - [5, 2]: Mixed parity. Invalid. 4. Wait, let's re-evaluate. The cycle is 2->3->4->5->2. - Subsequence [2]: Even. - Subsequence [3]: Odd. - Subsequence [4]: Even. - Subsequence [5]: Odd. - Are there any adjacent nodes with same parity? No. 2(e), 3(o), 4(e), 5(o), 2(e). - Therefore, no contiguous subsequence of length > 1 has consistent parity. - Result: 1.

Constraints

The linked list contains at least 3 nodes.
The linked list is guaranteed to have a cycle.
1 <= Node.val <= 10^9
The cycle length is at most 10^5.
All node values are positive integers.

Optimal Approach & Strategy

Apply Floyd’s algorithm to locate the cycle entry in O(n) time, then perform a single pass around the cycle to compute the longest monotonic parity subsequence, achieving O(n) time and O(1) extra space.

Brute Force Approach

Repeatedly traverse the list from the head to detect the cycle entry for each node, then for every possible sub‑segment of the cycle recompute parity monotonicity, leading to O(n²) time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums) {
   if (nums.length === 0) return 0;
   if (nums.length === 1) return nums[0];
   let result = nums[0];
   for (let i = 1; i < nums.length; i++) {
       if (nums[i] > nums[i - 1]) {
           result += nums[i];
       }
   }
   return result;
}

class Solution {
   public:
       int solution(vector<int>& nums) {
           if (nums.size() == 0) return 0;
           if (nums.size() == 1) return nums[0];
           int result = nums[0];
           for (int i = 1; i < nums.size(); i++) {
               if (nums[i] > nums[i - 1]) {
                   result += nums[i];
               }
           }
           return result;
       }
   };

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

def solution(nums):
   if len(nums) == 0:
       return 0
   if len(nums) == 1:
       return nums[0]
   result = nums[0]
   for i in range(1, len(nums)):
       if nums[i] > nums[i - 1]:
           result += nums[i]
   return result

function solution(nums) {
   if (nums.length === 0) return 0;
   if (nums.length === 1) return nums[0];
   let result = nums[0];
   for (let i = 1; i < nums.length; i++) {
       if (nums[i] > nums[i - 1]) {
           result += nums[i];
       }
   }
   return result;
}

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Monotonic Parity Sequence Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Linked List

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Monotonic Parity Sequence — 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