3536. Maximum Product of Two Digits

Math
Greedy

TIME

O(d)

SPACE

O(1)

Intuition

Digits are non-negative, so the largest product of any two of them is simply the product of the two largest digits (using the same position isn't allowed, but a repeated value is, and tracking the top two by value handles that automatically since n >= 10 guarantees at least two digits).

Approach

Scan the digits of n, maintaining the largest (max1) and second-largest (max2) digit values seen.
Return max1 * max2.

Complexity

Time: O(d) where d = number of digits (at most 10)
Space: O(1)

Edge Cases

Repeated largest digit (e.g. 22) → max1 and max2 both that value → squared
At least two digits always exist (n >= 10)
A 0 digit just never wins the top two unless forced; products stay correct

Code

class Solution:
    def maxProduct(self, n: int) -> int:
        """
        3536. Maximum Product of Two Digits
        Time: O(d)
        Space: O(1)
        """
        max1 = max2 = -1  # two largest digit values
        for ch in str(n):
            d = int(ch)
            if d >= max1:
                max2 = max1
                max1 = d
            elif d > max2:
                max2 = d
        return max1 * max2

/**
 * 3536. Maximum Product of Two Digits
 * Time: O(d)
 * Space: O(1)
 */
class Solution {
    public int maxProduct(int n) {
        int max1 = -1, max2 = -1; // two largest digit values
        while (n > 0) {
            int d = n % 10;
            if (d >= max1) {
                max2 = max1;
                max1 = d;
            } else if (d > max2) {
                max2 = d;
            }
            n /= 10;
        }
        return max1 * max2;
    }
}