How do you solve Super Pow on LeetCode?

Super Pow (LeetCode #372) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using modular exponentiation helps manage large numbers effectively.

What is the brute force solution for Super Pow?

The brute force approach for Super Pow is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach calculates a^b directly by multiplying a by itself b times. However, since b is extremely large and given as an array, this method is impractical for large values. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Super Pow?

Super Pow uses the following concepts: Math, Divide and Conquer. The recommended patterns to study are: Divide and Conquer, Modular Arithmetic.

What are the interview tips for Super Pow?

Always consider edge cases, especially with large numbers. Explain your thought process clearly while coding. Practice modular arithmetic problems to get comfortable with them.

What are common mistakes when solving Super Pow?

Not using modular arithmetic, leading to overflow errors. Failing to handle the array input correctly when calculating the exponent.

#372

Super Pow

Medium

MathDivide and ConquerDivide and ConquerModular Arithmetic

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of converting b into a single integer, we can use properties of modular arithmetic and the fact that we can break down the exponentiation into smaller parts using the modulus operation.

⚙️

Algorithm

3 steps

1Step 1: Define a helper function to compute (a^b) % 1337 using modular exponentiation.
2Step 2: For each digit in b, recursively calculate (a^(b[i]) % 1337) and combine results using the formula: (a^(10 * prev) % 1337) * (a^(b[i]) % 1337).
3Step 3: Return the final result.

solution.py14 lines

1# Full working Python code
2
3def superPow(a, b):
4    def modPow(x, n):
5        if n == 0:
6            return 1
7        half = modPow(x, n // 2)
8        return (half * half * (x if n % 2 else 1)) % 1337
9
10    result = 1
11    for digit in reversed(b):
12        result = result * modPow(a, digit) % 1337
13        a = modPow(a, 10)
14    return result

ℹ

Complexity note: The time complexity is O(n) because we iterate through the array b once and perform constant time operations for each digit.

1Using modular exponentiation helps manage large numbers effectively.
2Breaking down the exponentiation into smaller parts allows for efficient calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.