• Question
• Solution

Question

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

For example:

Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.

Follow up:

Could you do it without any loop/recursion in O(1) runtime?

Hint:

1. A naive implementation of the above process is trivial. Could you come up with other methods? Show More Hint

Solution

Result: Accepted Time: 4 ms

Here should be some explanations.

int addDigits(int num) {
    return (num - 1)%9 + 1;
}

Complexity Analytics

• Time Complexity: $O(1)$
• Space Complexity: $O(1)$

Result: Accepted Time: 8 ms

int addDigits(int num) {
    int ans = 0;
    do
    {
        while(num)
        {
            ans += num%10;
            num /=10;
        }
        num = ans;
        ans  = 0;
    }while(num > 9);
    return num;
}

Complexity Analytics

• Time Complexity: $O(log(n))$
• Space Complexity: $O(1)$