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Leetcode 372. Super Pow

Problem Statement

You are given two integers a and an array of integers b that represents an extremely large integer. You need to calculate a raised to the power of this large integer (a^b) modulo 1337.

Example:

Constraints:

Clarifying Questions

  1. Q: Can b contain leading zeros? A: No, b does not contain leading zeros.

  2. Q: What is the modulo value we need to compute? A: The modulo value is 1337.

  3. Q: What is the maximum length of array b? A: The maximum length of the array b is 2000.

  4. Q: Should the solution handle extremely large integers efficiently? A: Yes, use appropriate methods to handle the extremely large power efficiently.

Strategy

Given the constraints, we can’t directly compute a^b as b can be very large. Instead, we can utilize properties of modular arithmetic, particularly Euler’s theorem and modular exponentiation.

Steps:

  1. Modular Exponentiation:
    • We’ll use the function powmod(a, k, mod) that computes (a^k) % mod using iterative squaring for efficiency.
  2. Breaking the problem:
    • We will process each digit in the array b in reverse, treating it like a big integer.
  3. Combining results using properties:
    • We need to compute the result in parts and combine using (a^b) % mod = (a^(b1*10^k + b2*10^(k-1) + ... + bn)) % mod.

Time Complexity

The solution will be efficient with time complexity approximately O(N * log(M)), where N is the length of b and M is the largest value of a.

Code

Here’s the implementation in C++:

class Solution {
private:
    const int MOD = 1337;

    int powmod(int a, int k, int mod) {
        a %= mod;
        int result = 1;
        while (k > 0) {
            if (k % 2 == 1) {
                result = (result * a) % mod;
            }
            a = (a * a) % mod;
            k /= 2;
        }
        return result;
    }

public:
    int superPow(int a, vector<int>& b) {
        int result = 1;
        for (int i = 0; i < b.size(); ++i) {
            result = powmod(result, 10, MOD) * powmod(a, b[i], MOD) % MOD;
        }
        return result;
    }
};

Explanation:

  1. powmod function:
    • Efficiently calculates (a^k) % mod using iterative squaring with time complexity O(log k).
  2. superPow function:
    • Iterates through each element of b (considering it as a large number).
    • Combines the results effectively using properties of modular exponentiation.

This strategy ensures that even for large values of b, the computation remains feasible within constraints.

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