Happy Number

Question: 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.

Example:

Input: 19
Output: true
Explanation: 
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1

You may view the full question here.

Approach 1: Using HashMaps to store the evaluated values to detect repetitions.

//Approach 1:
//Runtime: 2ms
//Memory usage: 33MB
class Solution {
    
    public boolean isHappy(int n) {
        HashSet<Integer> set = new HashSet();
        set.add(n);
        int sum = n;
        while(sum!=1){
            sum = getDigitSqrSum(sum);
            if(set.contains(sum)){
                return false;
            } else {
                set.add(sum);
            }
        }
        return true;
    }
    
    private int getDigitSqrSum(int n){
        int sum = 0;
        while(n!=0){
            int d = n%10;
            sum+=(d*d);
            n=n/10;
        }
        return sum;
    }
}

Approach 2: Using cycle detection algorithms such as the Floyd’s Tortoise and the Hare algorithm. You may look up more such algorithms here.

//Approach 2:
//Runtime: 1ms
//Memory usage: 32.6MB 
class Solution {
    //Using Floyd Cycle detection algorithm
    public boolean isHappy(int n) {
        int slow, fast;
        slow = fast = n;
        do {
            slow = digitSquareSum(slow);
            fast = digitSquareSum(fast);
            fast = digitSquareSum(fast);
        } while(slow != fast);
        if (slow == 1) return true;
        else return false;
    }
    
    private int digitSquareSum(int n) {
    int sum = 0, tmp;
    while (n>0) {
        tmp = n % 10;
        sum += tmp * tmp;
        n /= 10;
    }
    return sum;
}
}

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Cheers & Chao!