Super Pow Solution in Java

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Problem Description

Welcome to the world of Super Pow, where numbers are raised to the power of other numbers, and you get to do it in a way that would make even a calculator sweat! The problem is simple: given an integer a and an array of integers b, you need to compute a^{b[0] × 10^{k} + b[1] × 10^{k-1} + ... + b[k]} mod 1337.

Now, if you’re scratching your head wondering why we need to do this modulo 1337, just think of it as a quirky way to keep our numbers from getting too big. After all, who wants to deal with numbers larger than the population of the universe?

Imagine you’re trying to calculate how many times you can raise a pizza to the power of toppings. If you have 3 toppings and you want to know how many pizzas you can make, you might end up with a number that’s just too large to handle. So, we use modulo to keep it manageable.

Code Solution


class Solution {
  public int superPow(int a, int[] b) {
    int ans = 1;

    a %= kMod;
    for (final int i : b)
      ans = modPow(ans, 10) * modPow(a, i) % kMod;

    return ans;
  }

  private static final int kMod = 1337;

  private int modPow(int x, int n) {
    if (n == 0)
      return 1;
    if (n % 2 == 1)
      return x * modPow(x % kMod, (n - 1)) % kMod;
    return modPow(x * x % kMod, (n / 2)) % kMod;
  }
}

Approach

The approach here is as elegant as a cat walking on a tightrope. We first reduce a modulo 1337 to keep our calculations manageable. Then, for each digit in the array b, we calculate the power of a raised to that digit while also considering the powers of 10. The modPow function is a recursive helper that efficiently computes powers using the method of exponentiation by squaring.

Time and Space Complexity

Time Complexity

O(log n) for each call to modPow, where n is the exponent. Since we call modPow for each digit in b, the overall complexity is O(k log n), where k is the number of digits in b.

Space Complexity

O(log n) due to the recursion stack in modPow.

Real-World Example

Let’s say you’re a baker trying to figure out how many cupcakes you can make with a certain number of ingredients. If you have a as the number of ingredients and b as the number of cupcakes you want to make, the Super Pow problem helps you calculate the maximum number of cupcakes you can bake without running out of ingredients. Just remember to keep it under control with modulo 1337!