Happy Number

Problem

A number is "happy" if repeatedly summing the squares of its digits eventually reaches 1. Otherwise it cycles. Return true iff n is happy.

Constraints

• 1 ≤ n ≤ 2³¹ − 1

Approach — Hashing & Counting

This is a Hashing & Counting problem. The idea: store what you've seen in a hash map for O(1) lookups, trading a little space to avoid a nested scan. Work through the reference code below line by line, then re-derive it yourself in the editor — that's how the pattern sticks.

Complexity: O(n) time, O(n) space.

Solution code

Python

class Solution:
    def isHappy(self, n):
        seen = set()
        while n != 1 and n not in seen:
            seen.add(n)
            n = sum(int(d) ** 2 for d in str(n))
        return n == 1

Java

import java.util.*;

class Solution {
    public boolean isHappy(int n) {
        // Hash-set cycle detection: the digit-square sequence either
        // reaches 1 or revisits a number it has already seen (a cycle).
        Set<Integer> seen = new HashSet<>();
        while (n != 1 && seen.add(n)) n = next(n);
        return n == 1;
    }

    private int next(int n) {
        int s = 0;
        while (n > 0) { int d = n % 10; s += d * d; n /= 10; }
        return s;
    }

    // O(1)-space alternative (the Fast & Slow track's approach):
    //   int slow = n, fast = next(n);
    //   while (fast != 1 && slow != fast) {
    //       slow = next(slow);
    //       fast = next(next(fast));
    //   }
    //   return fast == 1;
}

Practice it

Reading a solution isn't the same as being able to write it under pressure. Open this problem in the in-browser editor, hide the solution, and solve it from scratch — your code runs against real test cases instantly.