Happy Number

• LeetCode: Happy Number

• Problem:

Write an algorithm to determine if a number is "happy".

A happy number is a number defined by the following process: 

Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. 

Those numbers for which this process ends in 1 are happy numbers.

• Note:

19 is a happy number

12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1

• Analysis:

From the problem description, it loops endlessly in a cycle which does not include 1.

However, we can store the unhappy numbers in a container, when we match the number again.

• Code - Java:

class Solution {
    public boolean isHappy(int n) {
        Set<Integer> unHappy = new HashSet<Integer>();
        while (n != 1) {
            if (unHappy.contains(n)) {
                return false;
            }
            unHappy.add(n);
            int temp = 0;
            while (n > 0) {
                int val = n % 10;
                temp += val * val;
                n /= 10;
            }
            n = temp;
        }
        return true;
    }
}

• Code - C++:

class Solution {
public:
    bool isHappy(int n) {
        set<int> unHappy;
        while (n != 1) {
            if (unHappy.find(n) != unHappy.end()) {
                return false;
            }
            unHappy.insert(n);
            int temp = 0;
            while (n > 0) {
                int val = n % 10;
                temp += val * val;
                n /= 10;
            }
            n = temp;
        }
        return true;
    }
};

• Code - Python:

class Solution(object):
    def isHappy(self, n):
        """
        :type n: int
        :rtype: bool
        """
        unHappy = set()
        while n != 1:
            if n in unHappy:
                return False
            unHappy.add(n)
            temp = 0
            while n > 0:
                val = n % 10
                temp += val ** 2
                n /= 10
            n = temp
        return True

• Time Complexity: O(N), N is the rounds.

• Space Complexity: O(N), N is the rounds.