Python Happy Number Program

What Is a Happy Number?

A positive integer is a happy number if repeatedly replacing it with the sum of the squares of its digits eventually reaches 1. If the process loops without hitting 1, the number is unhappy.

n is happy  ⇔  digit-square chain reaches 1

Example: 19
  1² + 9² = 82
  8² + 2² = 68
  6² + 8² = 100
  1² + 0² + 0² = 1   ← happy

Unhappy numbers enter:
  4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4

How to Check a Happy Number in Python

Four clear steps — then pick a visited set or Floyd’s tortoise and hare.

1. Require n > 0 (positive integers only).
2. Define f(n) = sum of squares of digits of n.
3. Keep applying f until you reach 1 (happy) or detect a repeat (unhappy).
4. Detect repeats with a set, or use Floyd cycle detection for O(1) extra space.

Pseudocode

function SumOfSquares(n):
    total = 0
    while n > 0:
        d = n mod 10
        total = total + d * d
        n = n / 10
    return total

function IsHappy(n):
    if n <= 0:
        return false

    seen = empty set
    while n != 1 and n not in seen:
        add n to seen
        n = SumOfSquares(n)

    return n == 1

Cheat Sheet

Keep these facts handy while writing the program.

Live Preview

Enter a positive integer to see the digit-square chain and the happy/unhappy verdict.

n = 19
Chain: 19 → 82 → 68 → 100 → 1
Verdict: HAPPY NUMBER

Worked Walkthrough — n = 19

Square each digit, sum, and replace until you hit 1.

Step 1: 1² + 9² = 1 + 81 = 82
Step 2: 8² + 2² = 64 + 4  = 68
Step 3: 6² + 8² = 36 + 64 = 100
Step 4: 1² + 0² + 0² = 1

Reached 1 → 19 is happy

Python Programs

Start with a visited set so the stop condition is obvious. Then use Floyd’s cycle detection when you want O(1) extra space.

def sum_of_squares(n):
    total = 0
    while n > 0:
        d = n % 10
        total += d * d
        n //= 10
    return total


def is_happy(n):
    if n <= 0:
        return False

    seen = set()
    while n != 1 and n not in seen:
        seen.add(n)
        n = sum_of_squares(n)
    return n == 1


number = 19
if is_happy(number):
    print(f"{number} is a Happy Number.")
else:
    print(f"{number} is not a Happy Number.")

19 is a Happy Number.

def sum_of_squares(n):
    total = 0
    while n > 0:
        d = n % 10
        total += d * d
        n //= 10
    return total


def is_happy(n):
    if n <= 0:
        return False

    slow = n
    fast = sum_of_squares(n)

    while fast != 1 and slow != fast:
        slow = sum_of_squares(slow)
        fast = sum_of_squares(sum_of_squares(fast))

    return fast == 1


print("19 is happy" if is_happy(19) else "19 is unhappy")
print("2 is happy" if is_happy(2) else "2 is unhappy")
print("7 is happy" if is_happy(7) else "7 is unhappy")

19 is happy
2 is unhappy
7 is happy

Time and Space Complexity

Each square-sum step costs O(digits). Values shrink quickly into a small range, so k (chain length until 1 or a cycle) stays small in practice.

Key Takeaways