What is the brute force approach and complexity for Happy Number?

Store every number produced; a repeat before 1 means not happy with time complexity O(log N) and space complexity O(log N).

What is the optimal approach and complexity for Happy Number?

slow = next(slow), fast = next(next(fast)); stop at 1 or when they meet with optimal time complexity O(log N) and space complexity O(1).

How do you identify the pattern for Happy Number?

Look for key signals in the prompt: "repeat the process" indicates Each value produces exactly one next value → a hidden linked list.; "loops endlessly in a cycle" indicates The statement itself says the walk can cycle → cycle detection.; "reaches 1" indicates 1 maps to itself, so "reaches 1" is just "the cycle it ends in is {1}"..

Happy Number

Easy·LeetCode #202 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

Start with a positive integer n. Replace it with the sum of the squares of its digits, and keep doing that.

If this process ever reaches 1, the number is happy. If it never reaches 1 (it keeps going round the same numbers forever), it is not happy.

Return true if n is happy, otherwise false.

Input

A positive integer n.

Output

true if repeating "sum of squared digits" reaches 1, otherwise false.

Example 1

Input: n = 19

Output: true

1² + 9² = 82 → 8² + 2² = 68 → 6² + 8² = 100 → 1² + 0² + 0² = 1. We reached 1.

Example 2

Input: n = 2

Output: false

2 → 4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4 → … The numbers repeat from 4 and never reach 1.

Constraints

• 1 ≤ n ≤ 2^31 − 1

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“repeat the process”

→ Each value produces exactly one next value → a hidden linked list.

“loops endlessly in a cycle”

→ The statement itself says the walk can cycle → cycle detection.

“reaches 1”

→ 1 maps to itself, so "reaches 1" is just "the cycle it ends in is {1}".

Approach Complexity Ladder

Brute Force — set of seen numbersO(log N)O(log N)

Store every number produced; a repeat before 1 means not happy

Optimal — Floyd's cycle detection on numbersO(log N)O(1)

slow = next(slow), fast = next(next(fast)); stop at 1 or when they meet

Brute Force Intuition

Keep producing the next number. Put each one into a hash set. If we reach 1, return true. If we produce a number that is already in the set, the sequence will repeat the same numbers forever without reaching 1, so return false.

Recognition rule for this sub-pattern:

Fast moves 2, slow moves 1. If there is a loop, fast catches slow. Reset one to head to find the loop start.

Remember this · Linked List — Fast & Slow

Trigger

Cycle detection, middle of list, "n-th from end", two lists meeting, repeating sequences.

Think

If one pointer moves faster or starts earlier, where will they meet?

Key question

What speed or gap makes the meeting point tell me the answer?

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