What is the time complexity of Double Modular Exponentiation?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Double Modular Exponentiation?

This problem uses Array, Math, Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Double Modular Exponentiation?

Insight 1: Modular exponentiation is crucial to prevent integer overflow, especially when dealing with potentially large values of a, b, and c. Applying the modulo operator at each step keeps the intermediate results within manageable bounds. Insight 2: The problem focuses on calculating the expression `((a^b % 10)^c) % m` for each row, highlighting the importance of correctly applying the modulo operator in each step of the exponentiation.

Which companies ask LeetCode 2961. Double Modular Exponentiation in interviews?

This problem is frequently asked at Barclays.

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Double Modular Exponentiation - LeetCode 2961 Solution

Q: How to solve Double Modular Exponentiation?

The solution iterates through each row of the `variables` array. For each row (representing `a`, `b`, `c`, and `m`), it calculates `(a^b % 10)` and then raises the result to the power of `c` and takes the modulo with `m`. If this final result is equal to the `target`, the index `i` of the current row is added to the `good_indices` list. Finally, the list of `good_indices` is returned.

Double Modular Exponentiation - Complete Solution Guide

Double Modular Exponentiation is LeetCode problem 2961, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 0-indexed 2D array variables where variables[i] = [a i , b i , c i, m i ] , and an integer target . An index i is good if the following formula holds: 0 <= i < variables.length ((a i b i % 10) c i ) % m i == target Return an array consisting of good indices in any order . Example 1: Input: variables = [[2,3,3,10],[3,3,3,1],[6,1,1,4]], target = 2 Output: [0,2] Explanation: For each index i in the variables array: 1) For the index 0, variables[0] = [2,3,3,10], (2 3 % 10) 3 % 10 = 2

Detailed Explanation

The problem requires us to iterate through a given 2D array called `variables`. Each row `variables[i]` contains four integers `a`, `b`, `c`, and `m`. For each row, we need to calculate `((a^b % 10)^c) % m`. If this result is equal to a given `target` integer, we consider the index `i` as a 'good' index and add it to a list of good indices. Finally, we return the list of all good indices.

Solution Approach

The solution iterates through each row of the `variables` array. For each row (representing `a`, `b`, `c`, and `m`), it calculates `(a^b % 10)` and then raises the result to the power of `c` and takes the modulo with `m`. If this final result is equal to the `target`, the index `i` of the current row is added to the `good_indices` list. Finally, the list of `good_indices` is returned.

Step-by-Step Algorithm

Step 1: Initialize an empty list `good_indices` to store the indices that satisfy the condition.
Step 2: Iterate through the `variables` array, where each row represents [a, b, c, m].
Step 3: Calculate `first_pow = a^b % 10`. This involves computing `a` to the power of `b`, and then taking the modulo by 10.
Step 4: Calculate `second_pow = first_pow^c % m`. This involves raising `first_pow` to the power of `c`, and then taking the modulo by `m`.
Step 5: Check if `second_pow` is equal to `target`. If it is, add the current index `i` to the `good_indices` list.
Step 6: After iterating through all the rows, return the `good_indices` list.

Key Insights

Insight 1: Modular exponentiation is crucial to prevent integer overflow, especially when dealing with potentially large values of a, b, and c. Applying the modulo operator at each step keeps the intermediate results within manageable bounds.
Insight 2: The problem focuses on calculating the expression `((a^b % 10)^c) % m` for each row, highlighting the importance of correctly applying the modulo operator in each step of the exponentiation.
Insight 3: It is critical to understand the order of operations and apply the modulo operator (%) at the correct stages within nested power calculations to avoid incorrect results. Specifically, (a^b) % 10 must be calculated before raising it to the power of c.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Math, Simulation.

Companies

Asked at: Barclays.

Double Modular Exponentiation | LeetCode 2961

Medium

Array Math Simulation

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<Integer> getGoodIndices(int[][] variables, int target) {
        List<Integer> good_indices = new ArrayList<>();
        for (int i = 0; i < variables.length; i++) {
            int a = variables[i][0];
            int b = variables[i][1];
            int c = variables[i][2];
            int m = variables[i][3];

            long firstPow = 1;
            for (int j = 0; j < b; j++) {
                firstPow = (firstPow * a) % 10;
            }
            
            long secondPow = 1;
            for (int j = 0; j < c; j++) {
                secondPow = (secondPow * firstPow) % m;
            }

            if (secondPow == target) {
                good_indices.add(i);
            }
        }
        return good_indices;
    }
}

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Problem Info

Description

Problem 2961: Double Modular Exponentiation - Detailed explanation and solution approaches.

Topics

Array Math Simulation

Companies

Barclays

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2961

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize an empty list `good_indices` to store the indices that satisfy the condition.
2
Step 2: Iterate through the `variables` array, where each row represents [a, b, c, m].
3
Step 3: Calculate `first_pow = a^b % 10`. This involves computing `a` to the power of `b`, and then taking the modulo by 10.
4
Step 4: Calculate `second_pow = first_pow^c % m`. This involves raising `first_pow` to the power of `c`, and then taking the modulo by `m`.
5
Step 5: Check if `second_pow` is equal to `target`. If it is, add the current index `i` to the `good_indices` list.
6
Step 6: After iterating through all the rows, return the `good_indices` list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the nested loop structure, specifically the iteration through `variables` and the inner power calculations that are done via iteration in Java, C++, and C. The space complexity is derived from the size of the `good_indices` list, which can grow linearly with the size of the input `variables` in the worst-case scenario.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using precomputed powers

Instead of calculating a^b % 10 and (a^b % 10)^c % m within the loop for each variable set, we could precompute all possible powers for each 'a' and store them in a hash map or array. This way, we reduce the exponentiation computation within the loop. However, this approach requires extra space to store the precomputed results.

Time: O(n), assuming the precomputation is done only once and stored.Space: O(max(a)*max(b) + max((a^b % 10))*max(c)). Significant space overhead for large max(b) and max(c).

Trade-offs: Useful when 'a', 'b' and 'c' values have a limited range and are used multiple times across different variable sets. If the range is large or the variables are only used once, the precomputation overhead outweighs the benefit.

✨

Key Insights

1Insight 1: Modular exponentiation is crucial to prevent integer overflow, especially when dealing with potentially large values of a, b, and c. Applying the modulo operator at each step keeps the intermediate results within manageable bounds.
2Insight 2: The problem focuses on calculating the expression `((a^b % 10)^c) % m` for each row, highlighting the importance of correctly applying the modulo operator in each step of the exponentiation.
3Insight 3: It is critical to understand the order of operations and apply the modulo operator (%) at the correct stages within nested power calculations to avoid incorrect results. Specifically, (a^b) % 10 must be calculated before raising it to the power of c.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: `m` being equal to 1. When `m` is 1, the result of the modulo operation `% m` will always be 0. Therefore, the `target` must also be 0 for the index to be considered a good index.
Edge case 2: `b` or `c` being very large. The modular exponentiation is designed to handle large exponents efficiently by calculating the modulo in each step to avoid integer overflow.
Edge case 3: `a` being zero. If `a` is zero, `a^b` will be zero, and the final result depends on `c` and `m`. Proper handling of these values is necessary in the calculation.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer Overflow. Failing to apply the modulo operator at each step can lead to integer overflow, resulting in incorrect results.

✗Common mistake 2: Incorrect order of operations. Applying the modulo operator at the wrong step can change the entire outcome of the expression, and thus, returning a wrong result.

✗Common mistake 3: Ignoring the edge case where m is 1. This always results in a modulo 0, so the target must also be 0 to be considered good.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Cryptography. Modular exponentiation is a core component of many cryptographic algorithms, such as RSA, for encryption and decryption. This problem highlights the fundamental calculation involved.

🏦Real-world scenario 2: Hashing. Hash functions often use modular arithmetic to map large data sets to smaller, fixed-size values. Understanding the modulo operation is key to designing effective hash functions.

🎮Real-world scenario 3: Digital signatures. Algorithms for creating digital signatures rely heavily on modular exponentiation to ensure the authenticity and integrity of digital documents.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 21, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Double Modular Exponentiation"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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