LeetCode 202. Happy Number

Solving LeetCode 202, Happy Number. Click here and try it out your self!

LeetCode Problem Statement

Write an algorithm to determine if a number n 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. 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. Return true if n is a happy number, and false if not.

Examples:

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

Input: n = 2
Output: false

Initial Thoughts

Happy Number. I didn’t realize numbers could feel emotions…

HA!

I thought I would have to keep track of my solutions. If a solution was equal to 1, then it would be a happy number. If I saw it before, then it would be a sad number.

This solution would require extra space though… can we do better?

Could we solve with with O(1) space?

Indeed we can by using slow and fast pointers. Lets see how!

The Algorithm Plan

The key insight here is that we are searching for a cycle. Either a cycle that gets stuck in a number 1, or a cycle that doesn’t. Lets keep that in mind as we go through this Algorithm plan.

Approach Overview

Set up two pointers, a slow one and a fast one
- Slow = number
- Fast = getNextSum(number)
While fast != 1 && slow != fast
Slow pointer will solve the problem for the current number
Fast pointer will solve the problem for two steps ahead
When the loops break
- return fast === 1
- If this is true, the number is happy, else it’s sad

getNextSum implementation

Accepts number
Declare sum variable and set to zero
While number > 0
- Get the leading digit with modulo operator number % 10
- Update the sum with the square value sum = digit * digit
- Remove the leading digit from the number number = Math.floor(number/10)
Return the sum

That’s our plan. Overall this algorithm is fairly easy to implement. Nothing fancy here.

Once you realize that you are searching for cycles, this solution follows nicely. That’s the key insight we need to uncover.

Lets code it up!

Code


const isHappy = (number) => {
    let slow = number;
    let fast = getNextSum(number);
    while(fast !== 1 && slow !== fast) {
        slow = getNextSum(slow);
        fast = getNextSum(getNextSum(fast));
    }
    return fast === 1;
}

const getNextSum = (number) => {
    let sum = 0;
    while(number > 0) {
        let digit = number % 10;
        sum += digit * digit;
        number = Math.floor(number / 10);
    }
    return sum;
}