What is the time complexity of Sum of k-Mirror Numbers?

The optimal solution has a time complexity of O(n * k^(len_k/2) * log_k(k^(len_k/2))).

What is the space complexity of Sum of k-Mirror Numbers?

The space complexity is O(log_k(k^(len_k/2))).

What data structures are used to solve Sum of k-Mirror Numbers?

This problem uses Math, Enumeration. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Sum of k-Mirror Numbers?

Insight 1: Generating palindromes in base k is more efficient than checking all numbers for being k-mirror numbers. This is because the number of potential palindromes is significantly less than the total number of integers. Insight 2: Constructing palindromes can be done by generating the first half of the palindrome and then mirroring it to create the full palindrome. Special consideration is needed for odd vs. even length palindromes.

Which companies ask LeetCode 2081. Sum of k-Mirror Numbers in interviews?

This problem is frequently asked at Cisco.

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Sum of k-Mirror Numbers - LeetCode 2081 Solution

Q: How to solve Sum of k-Mirror Numbers?

The solution constructs k-mirror numbers by generating palindromes in base `k`. It iterates through different lengths of palindromes, creating the first half of the palindrome in base `k`, mirroring it to create the full palindrome in base `k`, converting the base `k` palindrome to base 10, and finally checking if the base 10 number is also a palindrome. The process continues until `n` k-mirror numbers have been found.

Sum of k-Mirror Numbers - Complete Solution Guide

Sum of k-Mirror Numbers is LeetCode problem 2081, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A k-mirror number is a positive integer without leading zeros that reads the same both forward and backward in base-10 as well as in base-k. For example, 9 is a 2-mirror number. The representation of 9 in base-10 and base-2 are 9 and 1001 respectively, which read the same both forward and backward. On the contrary, 4 is not a 2-mirror number. The representation of 4 in base-2 is 100 , which does not read the same both forward and backward. Given the base k and the number n , return the sum of th

Detailed Explanation

The problem asks us to find the sum of the first `n` k-mirror numbers. A k-mirror number is a number that is a palindrome (reads the same forwards and backward) in both base 10 and base k. The input includes the base `k` (2 <= k <= 9) and the number `n` (1 <= n <= 30), which represents how many k-mirror numbers we need to sum. We need to return the sum as a long integer.

Solution Approach

The solution constructs k-mirror numbers by generating palindromes in base `k`. It iterates through different lengths of palindromes, creating the first half of the palindrome in base `k`, mirroring it to create the full palindrome in base `k`, converting the base `k` palindrome to base 10, and finally checking if the base 10 number is also a palindrome. The process continues until `n` k-mirror numbers have been found.

Step-by-Step Algorithm

Step 1: Initialize `total_sum` to 0 and `found_count` to 0. These variables track the sum of the k-mirror numbers found so far and the number of k-mirror numbers found, respectively.
Step 2: Initialize `len_k` to 1. This represents the length of the k-base palindrome we are trying to find.
Step 3: While `found_count` is less than `n`, iterate through potential palindromes of length `len_k`.
Step 4: Calculate `half_len` as `(len_k + 1) // 2`. This gives the length of the first half of the palindrome.
Step 5: Iterate from `k**(half_len - 1)` to `k**half_len` (exclusive). These are the possible values for the first half of the palindrome represented in base k.
Step 6: Convert the current number `i` (which is the first half of our palindrome in base k expressed as base 10) to a base `k` string `first_half`.
Step 7: Construct the second half of the base `k` palindrome (`second_half`) by reversing the `first_half`. If `len_k` is odd, exclude the last character of `first_half` before reversing. If `len_k` is even, reverse all of `first_half`.
Step 8: Concatenate `first_half` and `second_half` to form the complete base `k` palindrome string (`palindrome_k_str`).
Step 9: Convert `palindrome_k_str` from base `k` to base 10 to get `num_base10`.
Step 10: Check if `num_base10` is a palindrome in base 10. If it is, increment `total_sum` by `num_base10` and increment `found_count`.
Step 11: If `found_count` equals `n`, return `total_sum`.
Step 12: Increment `len_k` to search for the next length of k-mirror numbers.

Key Insights

Insight 1: Generating palindromes in base k is more efficient than checking all numbers for being k-mirror numbers. This is because the number of potential palindromes is significantly less than the total number of integers.
Insight 2: Constructing palindromes can be done by generating the first half of the palindrome and then mirroring it to create the full palindrome. Special consideration is needed for odd vs. even length palindromes.
Insight 3: Since `n` is relatively small (<= 30), we can generate palindromes in base k iteratively, stopping when we have found the required number of k-mirror numbers.

Complexity Analysis

Time Complexity: O(n * k^(len_k/2) * log_k(k^(len_k/2)))

Space Complexity: O(log_k(k^(len_k/2)))

Topics

This problem involves: Math, Enumeration.

Companies

Asked at: Cisco.

Sum of k-Mirror Numbers | LeetCode 2081

Hard

Math Enumeration

Time: O(n * k^(len_k/2) * log_k(k^(len_k/2)))Space: O(log_k(k^(len_k/2)))

Solution Code

Available in:

class Solution {
    public long kMirror(int k, int n) {
        long total_sum = 0;
        int found_count = 0;

        int len_k = 1;
        while (found_count < n) {
            int half_len = (len_k + 1) / 2;

            long start = (long)Math.pow(k, half_len - 1);
            long end = (long)Math.pow(k, half_len);

            for (long i = start; i < end; i++) {
                String first_half = "";
                long temp_i = i;
                while (temp_i > 0) {
                    first_half = (temp_i % k) + first_half;
                    temp_i /= k;
                }

                String second_half;
                if (len_k % 2 == 1) {
                    second_half = new StringBuilder(first_half.substring(0, first_half.length() - 1)).reverse().toString();
                } else {
                    second_half = new StringBuilder(first_half).reverse().toString();
                }

                String palindrome_k_str = first_half + second_half;

                long num_base10 = Long.parseLong(palindrome_k_str, k);

                if (is_palindrome(String.valueOf(num_base10))) {
                    total_sum += num_base10;
                    found_count++;
                    if (found_count == n) {
                        return total_sum;
                    }
                }
            }

            len_k++;
        }

        return total_sum;
    }

    private boolean is_palindrome(String s) {
        return s.equals(new StringBuilder(s).reverse().toString());
    }
}

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Problem Info

Description

Problem 2081: Sum of k-Mirror Numbers - Detailed explanation and solution approaches.

Topics

Math Enumeration

Companies

Cisco

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2081

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

12 steps

1
Step 1: Initialize `total_sum` to 0 and `found_count` to 0. These variables track the sum of the k-mirror numbers found so far and the number of k-mirror numbers found, respectively.
2
Step 2: Initialize `len_k` to 1. This represents the length of the k-base palindrome we are trying to find.
3
Step 3: While `found_count` is less than `n`, iterate through potential palindromes of length `len_k`.
4
Step 4: Calculate `half_len` as `(len_k + 1) // 2`. This gives the length of the first half of the palindrome.
5
Step 5: Iterate from `k**(half_len - 1)` to `k**half_len` (exclusive). These are the possible values for the first half of the palindrome represented in base k.
6
Step 6: Convert the current number `i` (which is the first half of our palindrome in base k expressed as base 10) to a base `k` string `first_half`.
7
Step 7: Construct the second half of the base `k` palindrome (`second_half`) by reversing the `first_half`. If `len_k` is odd, exclude the last character of `first_half` before reversing. If `len_k` is even, reverse all of `first_half`.
8
Step 8: Concatenate `first_half` and `second_half` to form the complete base `k` palindrome string (`palindrome_k_str`).
9
Step 9: Convert `palindrome_k_str` from base `k` to base 10 to get `num_base10`.
10
Step 10: Check if `num_base10` is a palindrome in base 10. If it is, increment `total_sum` by `num_base10` and increment `found_count`.
11
Step 11: If `found_count` equals `n`, return `total_sum`.
12
Step 12: Increment `len_k` to search for the next length of k-mirror numbers.

💡

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 * k^(len_k/2) * log_k(k^(len_k/2)))

💾

Space Complexity

O(log_k(k^(len_k/2)))

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The time complexity is determined by the process of generating base `k` palindromes and checking if their base 10 equivalents are also palindromes. The outer loop finds n numbers. The most significant operation is converting numbers to base k (log_k(number)) and checking if a number is a palindrome, along with string manipulation when forming k-base palindrome candidates. The space complexity depends on the length of the base-k string, which is logarithmic to the number.

📖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

Brute-Force Approach

Iterate through all numbers from 1 onwards. For each number, convert it to base k and check if it's a palindrome in both base 10 and base k. If it is, add it to the sum and increment the count. Stop when the count reaches n.

Time: O(M * (log_k(M) + log(M))), where M is the largest k-mirror number found. This can be very large depending on n and k.Space: O(log_k(M) + log(M)) for storing base k and base 10 representations.

Trade-offs: The brute-force approach is simpler to implement but significantly less efficient than the palindrome generation approach. It is not practical for larger values of n.

✨

Key Insights

1Insight 1: Generating palindromes in base k is more efficient than checking all numbers for being k-mirror numbers. This is because the number of potential palindromes is significantly less than the total number of integers.
2Insight 2: Constructing palindromes can be done by generating the first half of the palindrome and then mirroring it to create the full palindrome. Special consideration is needed for odd vs. even length palindromes.
3Insight 3: Since `n` is relatively small (<= 30), we can generate palindromes in base k iteratively, stopping when we have found the required number of k-mirror numbers.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: k=2, n=1. The smallest k-mirror number is 1. The solution correctly identifies and returns this case.
Edge case 2: Finding a large number of k-mirror numbers may require considering base-k palindromes of significant length. The solution handles large numbers using `long` (Java, C++) or `int` (Python) type.
Edge case 3: The problem statement guarantees that `k` will be between 2 and 9 inclusive and n will be between 1 and 30, meaning no invalid input checks are strictly necessary in most languages.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle the different cases for constructing the second half of the base `k` palindrome string when `len_k` is even or odd. This leads to generating incorrect palindromes.

✗Common mistake 2: Incorrectly converting between base `k` and base 10. Pay attention to the correct use of place values and modulo operations.

✗Common mistake 3: Not using a `long` (Java, C++) data type to store the intermediate `total_sum`. The sum of the first 30 k-mirror numbers can easily exceed the range of an `int`.

✅

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: This problem explores number base conversions and palindrome generation. While not directly applicable, the concepts of base conversion are important in cryptography, data compression, and network protocols.

🏦Real-world scenario 2: Palindrome detection and generation, while not directly used for k-mirror numbers, finds applications in bioinformatics (detecting palindromic sequences in DNA), data validation (ensuring data integrity), and recreational mathematics.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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: September 10, 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

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Sum of k-Mirror Numbers"?

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

🤖 Need help with this problem?