Approach: Single pass over digits

def subtractProductAndSum(n):
    product, total = 1, 0
    while n > 0:
        d = n % 10
        product *= d
        total += d
        n //= 10
    return product - total

Complexity: O(log n) time, O(1) space.

Full Solution — all languages

class Solution:
    def subtractProductAndSum(self, n):
        product, total = 1, 0
        while n > 0:
            d = n % 10
            product *= d
            total += d
            n //= 10
        return product - total

import java.util.*;
class Solution {
    public int subtractProductAndSum(int n) {
        int product = 1, total = 0;
        while (n > 0) {
            int d = n % 10;
            product *= d;
            total += d;
            n /= 10;
        }
        return product - total;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int subtractProductAndSum(int n) {
        int product = 1, total = 0;
        while (n > 0) {
            int d = n % 10;
            product *= d;
            total += d;
            n /= 10;
        }
        return product - total;
    }
};

/**
 * @param {number} n
 * @return {number}
 */ var subtractProductAndSum = function(n) {
    let product = 1, total = 0;
    while (n > 0) {
        const d = n % 10;
        product *= d;
        total += d;
        n = Math.floor(n / 10);
    }
    return product - total;
};

Subtract the Product and Sum of Digits of an Integer

Example 1

Example 2

Constraints