# Solving Leetcode Interviews in Seconds with AI: Super Pow
# Introduction
In this blog post, we will explore how to solve the LeetCode problem "372" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.
# Problem Statement
> Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array. Example 1: Input: a = 2, b = [3] Output: 8 Example 2: Input: a = 2, b = [1,0] Output: 1024 Example 3: Input: a = 1, b = [4,3,3,8,5,2] Output: 1 Constraints: 1 <= a <= 231 - 1 1 <= b.length <= 2000 0 <= b[i] <= 9 b does not contain leading zeros.
# Explanation
Here's the breakdown of the solution:
* **Modular Exponentiation:** Utilize the property (a * b) mod m = ((a mod m) * (b mod m)) mod m and a(x+y) mod m = (ax mod m * ay mod m) mod m. This avoids calculating huge intermediate results.
* **Iterate through b:** Process the array `b` digit by digit. For each digit, update the result by raising it to the power of 10 and then multiplying by adigit.
* **Optimization:** Use the modulo operator (%) after each multiplication to keep the intermediate results within the range [0, 1336].
* **Runtime & Storage Complexity:** O(N), where N is the length of the array b, and O(1) storage complexity.
# Code
```python
def superPow(a: int, b: list[int]) -> int:
"""
Calculates a^b mod 1337 where b is represented as an array of digits.
Args:
a: The base integer.
b: The exponent represented as a list of digits.
Returns:
The result of a^b mod 1337.
"""
MOD = 1337
result = 1
a %= MOD # Reduce a modulo MOD to avoid large intermediate values
for digit in b:
result = pow(result, 10, MOD) * pow(a, digit, MOD) % MOD
return result
```