Happy Number.md

Algorithm

This is similar to LeetCode #141 - Linked List Cycle. We will reuse the algorithm from that problem to detect cycles.

1. Create slow. It will move 1 step at a time.
2. Create fast. It will move 2 steps at a time.
3. We will keep generating numbers until either
   1. We generate 1 (we have a "happy number"), or
   2. slow and fast collide (we don't have a "happy number").

Solution

class Solution {
    public boolean isHappy(int n) {
        if (n <= 0) {
            return false;
        }
        int slow = n;
        int fast = digitSquareSum(n);  // setting fast to n would cause an immediate `slow == fast` collision

        while (fast != 1 && slow != fast) {
            slow = digitSquareSum(slow);
            fast = digitSquareSum(fast);
            fast = digitSquareSum(fast);
        }

        return fast == 1;
    }

    private int digitSquareSum(int n) {
        int sum = 0;
        while (n > 0) {
            int digit = n % 10;
            sum += digit * digit;
            n /= 10;
        }
        return sum;
    }
}

Time Complexity

If number of bits in int n is capped at 32 by Java

O(1)

If number of bits in int n is allowed to grow

• digitSquareSum() will be O(log n)
• When (or if) while (fast != 1 && slow != fast) terminates as n increases needs a mathematical proof that won't be necessary in an interview.
• This gives us O(log n) < time complexity < ???

Space Complexity

O(1)

Links

• github.com/RodneyShag