2961. Double Modular Exponentiation

Problem Description

In this problem, you are given a two-dimensional array named variables where each element is a list of four integers [a_i, b_i, c_i, m_i]. In addition, you are provided with an integer target. An index i in the variables array is considered good if it satisfies a certain mathematical condition based on the values of a_i, b_i, c_i, and m_i. The condition is that the expression ((a_i^b_i % 10)^c_i) % m_i must be equal to the target. Here the ^ represents exponentiation, and % is the modulo operation.

The task is to identify all the indices that are good and return them as an array in any order. An index is good if the above formula matches the target when evaluated with the values at that index in the variables array. You need to consider each index i from 0 to variables.length - 1 and check the formula for each.

Intuition

The intuition behind the solution lies in understanding how modulo arithmetic can be used to simplify calculations and the properties of exponents. Since the final expression involves taking a_i to the power of b_i, then to the power of c_i, and applying modulo m_i to check if it equals the target, we can take advantage of the following properties:

1. Exponentiation by Squaring (Fast Power Method): This method allows us to compute a^b efficiently by squaring and reducing the number of multiplications we need to perform. This is especially useful when b is very large.

2. Modulo Properties: We can use the property (a * b) % m = ((a % m) * (b % m)) % m, which states that the result of a product modulo m is the same regardless of whether we perform the modulo before or after multiplication. We can also apply this principle to exponentiation.

By applying these optimizations, we can effectively reduce the calculations needed to verify whether an index i satisfies the condition. Hence, the process for each index i is to apply the modulo operation as soon as possible during the exponentiation steps to keep the numbers manageable and prevent overflow. This is done by calculating a^b % 10 and then ((a^b % 10)^c) % m_i.

The solution applies the Python pow function, which takes an additional modulo argument, making these operations straightforward and efficient. By doing so, we loop through each variables[i] and apply the property (a^b % 10)^c % m_i and compare it to the target. If it matches, we add the index i to our output list. This approach is both clean and efficient, allowing us to handle even large powers effectively.

Learn more about Math patterns.

Solution Approach

The implementation of the solution uses a relatively simple approach aligned with the properties of modulo arithmetic and exponentiation. Here's how it works:

1. Iteration: We iterate through each index i of the input variables array using a for-loop.

2. Exponentiation: For every tuple (a, b, c, m) at index i, we need to calculate the expression ((a^b % 10)^c) % m and check if it equals the target.

3. Exponentiation with Modulo: The Python pow function is perfect for this task because it allows us to efficiently calculate powers with a modulo. The regular power operation a^b may lead to very large numbers, especially when b and c are large, which is computationally expensive and can cause integer overflows. However, by using the pow function with a modulo, we can calculate a^b % 10 much more quickly and safely.

   • The first power operation is pow(a, b, 10) which computes a^b % 10. We modulo by 10 because exponentiation modulo 10 gives us the last digit of a^b, which is all we need for the next power.
   • The resulting number then needs to be exponentiated again with c and taken modulo m. This is continued with another pow operation - pow(result, c, m) where result is a^b % 10.

4. Check Against Target: The final result of the nested pow function is compared to the target. If they match, the index i is considered good.

5. Building the Output List: The indices where the condition holds true are gathered into a list using a list comprehension. This list is returned as the final output.

The process uses a direct simulation of the problem's formula. The fast power method (also known as exponentiation by squaring) is implicitly applied through Python's built-in pow function with mod argument, as this is a known efficient way to compute powers in modular arithmetic.

Here's a breakdown of the algorithm based on the solution code:

1class Solution:
2    def getGoodIndices(self, variables: List[List[int]], target: int) -> List[int]:
3        return [
4            i
5            for i, (a, b, c, m) in enumerate(variables)
6            if pow(pow(a, b, 10), c, m) == target
7        ]

In the provided code, enumerate(variables) gives us both the index i and the list [a, b, c, m]. For each index and list, if pow(pow(a, b, 10), c, m) == target evaluates the condition. If the condition is true, the i is included in the output list. This is an elegant and effective way to solve the problem with minimal code and high readability.

Example Walkthrough

Let's work through a small example to illustrate the solution approach.

Assume we have the following variables array and the target value:

1variables = [
2    [2, 3, 1, 4],  # Index 0
3    [3, 2, 2, 3],  # Index 1
4    [2, 5, 2, 1],  # Index 2
5]
6
7target = 1

We need to find all indices where the condition ((a_i^b_i % 10)^c_i) % m_i == target holds true.

Step 1: We start by iterating over each index in the variables array.

For index 0: The tuple is [2, 3, 1, 4], representing a=2, b=3, c=1, m=4.

• We calculate 2^3 % 10, which is 8 % 10 giving us 8.
• Next, we calculate (8^1) % 4, which is simply 8 % 4 giving us 0.
• The result is 0, which does not match the target of 1. Hence, index 0 is not good.

For index 1: The tuple is [3, 2, 2, 3], representing a=3, b=2, c=2, m=3.

• We calculate 3^2 % 10, which is 9 % 10 giving us 9.
• Next, we calculate (9^2) % 3, which is 81 % 3 giving us 0.
• The result is 0, which does not match the target of 1. Hence, index 1 is not good.

For index 2: The tuple is [2, 5, 2, 1], representing a=2, b=5, c=2, m=1.

• We calculate 2^5 % 10, which is 32 % 10 giving us 2.
• Next, we calculate (2^2) % 1, which is 4 % 1 giving us 0.
• However, since any number modulo 1 is 0, this does not match our target of 1. Hence, index 2 is not good.

At the end of the iteration, unfortunately, no indices satisfy the condition for our small example, and therefore the resulting array of good indices would be empty - [].

The solution code using this approach would be:

1class Solution:
2    def getGoodIndices(self, variables: List[List[int]], target: int) -> List[int]:
3        return [
4            i
5            for i, (a, b, c, m) in enumerate(variables)
6            if pow(pow(a, b, 10), c, m) == target
7        ]

The code works efficiently through the properties of exponents and modulo arithmetic, using Python's pow function to simplify the computations and avoid overflow, while checking each index against the target.

Solution Implementation

Time and Space Complexity

The given Python code in the Solution class contains a method called getGoodIndices which computes good indices based on the power operation condition.

The time complexity of this code is primarily dependent on the enumerate function that iterates through the variables list and the pow function used inside the list comprehension. The pow function in Python has a time complexity of O(log n), where n is the exponent.

Here, the pow function is used twice; the first call executes pow(a, b, 10), and the second call executes pow(result_of_first_call, c, m). Since b and c are the exponents used in each call, the time complexity for each iteration is controlled by these values.

Considering M as the maximum value among all values of b_i and c_i, and with M being limited to 10^3 in the problem, the time complexity per iteration is O(log M). Since there are n elements in variables, where n is the length of variables, the total time complexity is O(n * log M).

The space complexity of the code is O(1) because the extra space required is constant and does not depend on the input size. There are no additional data structures that grow with the size of variables. The list comprehension simply produces the final list of indices, which the function returns and does not count towards extra space as it is the required output.

Learn more about how to find time and space complexity quickly using problem constraints.