Solution: Happy Number

Explore how to identify happy numbers through the fast and slow pointers technique, detecting cycles efficiently without extra space. Learn to implement a helper function for digit square sums, apply cycle detection concepts, and analyze algorithmic time and space complexity for optimized coding interview solutions.

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Statement
Solution
- Naive approach
- Optimized approach using Fast and Slow Pointers pattern

Statement

Write an algorithm to determine if a number $n$ is a happy number.

We use the following process to check if a given number is a happy number:

Starting with the given number $n$ , replace the number with the sum of the squares of its digits.
Repeat the process until:
- The number equals $1$ , which will depict that the given number $n$ is a happy number.
- It enters a cycle, which will depict that the given number $n$ is not a happy number.

Return TRUE if $n$ is a happy number, and FALSE if not.

Constraints

$1 \leq$ $n$ $\leq 2^{31} - 1$

Solution

So far, you have probably brainstormed some approaches and have an idea of how to solve this problem. Let’s explore some of these approaches and figure out which one to follow based on considerations such as time complexity and any implementation constraints.

Naive approach

The brute force approach is to repeatedly calculate the squared sum of digits of the input number and store the computed sum in a hash set. For every calculation, we check if the sum is already present in the set. If yes, we've detected a cycle and should return FALSE. Otherwise, we add it to our hash set and continue further. If our sum converges to $1$ , we've found a happy number.

While this approach works well for small numbers, we might have to perform several computations for larger numbers to get the required result. So, it might get infeasible for such cases. The time complexity is $O(\log n)$ because finding the next value requires processing each digit, which takes $O(\log n)$ time as the number of digits in nn is approximately $\log n$ . Once the value drops below $243$ , it can take, at most, $243$ more steps to terminate, which is a constant and, therefore, negligible. For numbers greater than ...