Tejav
#3622

Check Divisibility by Digit Sum and Product

Easy
MathMathematical OperationsDigit Manipulation
LeetCode ↗

Approaches

Brute ForceOptimal
Complexity Comparison
Brute ForceOptimal Solution★
Time
O(d)
O(d)
Space
O(1)
O(1)
💡

Intuition

Time O(d)Space O(1)

Directly compute the digit sum and product in a single pass through the digits, minimizing operations.

⚙️

Algorithm

3 steps
  1. 1Step 1: Initialize sum and product variables.
  2. 2Step 2: Loop through each digit, updating sum and product.
  3. 3Step 3: Check if n is divisible by the sum of sum and product.
solution.py9 lines
1def check_divisibility(n):
2    digit_sum, digit_product = 0, 1
3    temp = n
4    while temp > 0:
5        digit = temp % 10
6        digit_sum += digit
7        digit_product *= digit
8        temp //= 10
9    return n % (digit_sum + digit_product) == 0
ℹ

Complexity note: The time complexity is O(d) where d is the number of digits in n, achieved by processing each digit once.

  • 1Divisibility checks often involve sums or products of components.
  • 2Single-pass calculations can optimize performance.

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