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:12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1
Example 2:
Input: n = 2
Output: false
Constraints:
- 1 <= n <= 231 - 1
My Solution
We can solve this problem by using a HashSet to store all the numbers we see. If we ever see the same number again, we can return false. If we encounter a 1, we return true. We compute the next number by extracting all the digits, squaring them, and adding them together.
Time complexity is O(1) because we converge in very few iterations, no matter how big the number is. Space complexity is also O(1) for the same reason.
class Solution {
public boolean isHappy(int n) {
if (n == 1) {
return true;
}
Set<Integer> all = new HashSet<>();
while (n != 1 && !all.contains(n)) {
all.add(n);
n = next(n);
}
return !all.contains(n);
}
public int next(int n) {
int out = 0;
while (n > 0) {
int digit = n % 10;
out += digit*digit;
n /= 10;
}
return out;
}
}