3754. Concatenate Non-Zero Digits and Multiply by Sum I

Math
String
Simulation

TIME

O(d)

SPACE

O(1)

Intuition

Directly follow the definition: filter out zero digits, rebuild the number, sum its digits, multiply.

Approach

Iterate over the digits of n (left to right).
Keep the non-zero digits: append each to x (as x = x * 10 + d) and add d to sum.
Return x * sum. If there were no non-zero digits, x and sum stay 0, giving 0.

Complexity

Time: O(d) where d = number of digits (at most 10)
Space: O(1) (or O(d) if using a string)

Edge Cases

n = 0 → no non-zero digits → answer 0
All-zero-except-one (e.g. 1000) → x = 1, sum = 1
Order matters: x preserves the original digit order
Max value x * sum fits comfortably in 64-bit (x <= 10^9, sum <= 81)

Code

class Solution:
    def concatenateAndMultiply(self, n: int) -> int:
        """
        3754. Concatenate Non-Zero Digits and Multiply by Sum I
        Time: O(d)
        Space: O(1)
        """
        x = 0
        total = 0
        for ch in str(n):
            d = int(ch)
            if d != 0:
                x = x * 10 + d
                total += d
        return x * total

/**
 * 3754. Concatenate Non-Zero Digits and Multiply by Sum I
 * Time: O(d)
 * Space: O(1)
 */
class Solution {
    public long concatenateAndMultiply(int n) {
        long x = 0;
        long total = 0;
        for (char ch : Integer.toString(n).toCharArray()) {
            int d = ch - '0';
            if (d != 0) {
                x = x * 10 + d;
                total += d;
            }
        }
        return x * total;
    }
}