Happy Number Explained

Problem Statement

A happy number is a number that, when you repeatedly replace it with the sum of the squares of its digits, eventually reaches 1. If it loops endlessly in a cycle that does not include 1, it is not happy. Given an integer n, determine if it is a happy number. This problem tests cycle detection in iterative processes.

Example

Input: n = 19

Output: true

Explanation: 1^2 + 9^2 = 82, 8^2 + 2^2 = 68, 6^2 + 8^2 = 100, 1^2 + 0^2 + 0^2 = 1.

Code

public class Solution {
    public boolean isHappy(int n) {
        Set seen = new HashSet<>();
        while (n != 1 && !seen.contains(n)) {
            seen.add(n);
            int sum = 0;
            while (n > 0) {
                int digit = n % 10;
                sum += digit * digit;
                n /= 10;
            }
            n = sum;
        }
        return n == 1;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.isHappy(19)); // true
    }
}
            

def is_happy(n):
    seen = set()
    while n != 1 and n not in seen:
        seen.add(n)
        sum_squares = 0
        while n > 0:
            digit = n % 10
            sum_squares += digit * digit
            n //= 10
        n = sum_squares
    return n == 1

# Example usage
print(is_happy(19))  # True
            

function isHappy(n) {
    const seen = new Set();
    while (n !== 1 && !seen.has(n)) {
        seen.add(n);
        let sum = 0;
        while (n > 0) {
            let digit = n % 10;
            sum += digit * digit;
            n = Math.floor(n / 10);
        }
        n = sum;
    }
    return n === 1;
}

// Example usage
console.log(isHappy(19)); // true
            

Explanation

• Use a set to track numbers seen during the process.
• For each number, compute the sum of the squares of its digits.
• Update n to this sum and repeat.
• If n becomes 1, it’s a happy number.
• If n repeats (cycle detected), it’s not happy.

Note