LeetCode //C - 202. Happy Number

202. Happy Number

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 .

Example 1:

Input: n = 19
Output: true
Explanation:
1 2 + 9 2 = 82 1^2 + 9^2 = 82 12+92=82
8 2 + 2 2 = 68 8^2 + 2^2 = 68 82+22=68
6 2 + 8 2 = 100 6^2 + 8^2 = 100 62+82=100
1 2 + 0 2 + 0 2 = 1 1^2 + 0^2 + 0^2 = 1 12+02+02=1

Example 2:

Input: n = 2
Output: false

Constraints:

• 1 < = n < = 2 31 − 1 1 <= n <= 2^{31} - 1 1<=n<=231−1

From: LeetCode

Link: 202. Happy Number

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Solution:

Ideas：

Here, the sumOfSquares function calculates the sum of the squares of the digits of the given number. The isHappy function uses the Floyd's Tortoise and Hare (Cycle Detection) algorithm to detect if there is a loop in the sequence. If a cycle is detected, the number is not a happy number. If the sequence reaches 1, the number is a happy number.

Code:

bool isHappy(int n) {
    int slow = n, fast = n; // Two pointers for cycle detection

    do {
        slow = sumOfSquares(slow); // Move slow one step
        fast = sumOfSquares(sumOfSquares(fast)); // Move fast two steps
    } while (slow != fast && fast != 1);

    return fast == 1;
}

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