How do you solve Double Modular Exponentiation on LeetCode?

Double Modular Exponentiation (LeetCode #2961) 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: Understanding modular arithmetic can simplify calculations.

What is the brute force solution for Double Modular Exponentiation?

The brute force approach for Double Modular Exponentiation is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach checks each index in the variables array and computes the formula for each one. If the result matches the target, we consider that index as good. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Double Modular Exponentiation?

Double Modular Exponentiation uses the following concepts: Array, Math, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Double Modular Exponentiation?

Explain your thought process clearly while coding. Test your code with edge cases before finalizing. Be prepared to discuss the time and space complexity of your solution.

What are common mistakes when solving Double Modular Exponentiation?

Forgetting to apply the modulo operation correctly. Not handling edge cases like m=1 or very large values.

#2961

Double Modular Exponentiation

Medium

ArrayMathSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

This approach leverages the properties of modular arithmetic to compute results more efficiently. Instead of recalculating powers multiple times, we can simplify our calculations.

⚙️

Algorithm

6 steps

1Step 1: Initialize an empty list to store good indices.
2Step 2: Loop through each index i in the variables array.
3Step 3: For each index, extract a, b, c, m from variables[i].
4Step 4: Compute the value using the formula (pow(a, b, 10) ** c) % m directly.
5Step 5: If the computed value equals the target, add i to the list of good indices.
6Step 6: Return the list of good indices.

solution.py15 lines

1# Full working Python code
2
3def find_good_indices(variables, target):
4    good_indices = []
5    for i in range(len(variables)):
6        a, b, c, m = variables[i]
7        value = (pow(a, b, 10) ** c) % m
8        if value == target:
9            good_indices.append(i)
10    return good_indices
11
12# Example usage
13variables = [[2,3,3,10],[3,3,3,1],[6,1,1,4]]
14target = 2
15print(find_good_indices(variables, target))

ℹ

Complexity note: The time complexity remains O(n) since we still iterate through the list of variables once. The space complexity is O(1) as we are using a fixed amount of space for the output.

1Understanding modular arithmetic can simplify calculations.
2Recognizing that we can reduce the number of computations by using properties of exponents.

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