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Leetcode 372. Super Pow

Problem Statement

In this problem, you are given two positive integers a and an array of positive integers b which represents a very large integer. You need to calculate the result of a^b modulo 1337.

Specifically, you are to implement a function superPow, which is defined as follows:

public int superPow(int a, int[] b)

The result should be a raised to the power represented by the array b modulo 1337.

Examples

1. Input: a = 2, b = [3] Output: 8
2. Input: a = 2, b = [1,0] Output: 1024

Clarifying Questions

1. Q: What is the range of values for a and the elements of b?
   A: a is a positive integer up to 2^31 - 1. Elements of b range from 0 to 9.

2. Q: How large can the array b be?
   A: b can have up to 200 elements.

3. Q: Can a or b have any leading zeros in the array?
   A: No, a will not have leading zeros and b will not have leading elements in the array.

Strategy

To solve this problem, we need to use modular exponentiation to manage the large powers efficiently. Here’s the step-by-step plan:

1. Modular exponentiation(): Since a and b can be very large, we can use properties of modular arithmetic to simplify the computation.
2. Decompose the array b: Break down b into smaller parts and compute the corresponding powers of a modulo 1337.
3. Combining the results: Use the property (a^m * a^n) % k = ( (a^m % k) * (a^n % k) ) % k.

Specifically, we can recursively reduce the problem using:

• a^b % k = (a^(b/n*n + b%n)) % k = ((a^b/n * a^b%n) % k)

We will use a helper function to manage the recursive breakdown of the problem and use the modulus properties effectively.

Code

public class Solution {
    private static final int MOD = 1337;

    public int superPow(int a, int[] b) {
        return superPowHelper(a, b, b.length - 1);
    }

    private int superPowHelper(int a, int[] b, int idx) {
        if (idx == -1) {
            return 1;
        }
        int lastDigit = b[idx];
        
        int part1 = myPow(a, lastDigit);
        int part2 = myPow(superPowHelper(a, b, idx - 1), 10);
        
        return part1 * part2 % MOD;
    }

    private int myPow(int a, int k) {
        a %= MOD;
        int result = 1;
        for (int i = 0; i < k; i++) {
            result = result * a % MOD;
        }
        return result;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.superPow(2, new int[]{3}));          // Output: 8
        System.out.println(sol.superPow(2, new int[]{1, 0}));       // Output: 1024
        System.out.println(sol.superPow(2, new int[]{1, 0, 2, 4})); // Output might vary but you can predict as needed.
    }
}

Time Complexity

• Time Complexity: The recursive breakdown happens in O(d) where d is the number of digits in b. For each digit, the power calculation is at most 10 multiplications (since the maximum digit value is 9). Therefore, the time complexity is approximately O(d * 10), which can be simplified as O(d).
• Space Complexity: The recursion depth can go up to O(d), and additional space complexity is constant for operations.

This ensures the solution is efficient given the constraints.

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