What is the time complexity of Palindrome Number?

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

What data structures are used to solve Palindrome Number?

This problem uses Math. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Palindrome Number?

Insight 1: Negative numbers are not palindromes because of the minus sign. This allows for an early exit condition. Insight 2: Numbers ending in 0 (except 0 itself) are not palindromes because they would have a leading 0 if reversed. This also helps with early exit.

Which companies ask LeetCode 9. Palindrome Number in interviews?

This problem is frequently asked at Accenture, Adobe, Amazon, Apple, Bloomberg and 27 more companies.

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Palindrome Number - LeetCode 9 Solution

Q: How to solve Palindrome Number?

The provided solution uses a clever iterative approach to check for palindromes without converting the integer to a string. It reverses only the first half of the number and compares it to the remaining part of the original number. This significantly improves efficiency compared to complete reversal.

Palindrome Number - Complete Solution Guide

Palindrome Number is LeetCode problem 9, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an integer x , return true if x is a palindrome , and false otherwise . Example 1: Input: x = 121 Output: true Explanation: 121 reads as 121 from left to right and from right to left. Example 2: Input: x = -121 Output: false Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome. Example 3: Input: x = 10 Output: false Explanation: Reads 01 from right to left. Therefore it is not a palindrome. Constraints: -2 31 <= x <= 2 31 - 1

Detailed Explanation

The problem asks you to determine if an integer is a palindrome. A palindrome is a number that reads the same forwards and backward. For example, 121 is a palindrome, but 123 is not. The input is a single integer `x`, and the output is a boolean value (`true` if it's a palindrome, `false` otherwise). The constraints specify that the input integer will be within the range of a 32-bit signed integer.

Solution Approach

The provided solution uses a clever iterative approach to check for palindromes without converting the integer to a string. It reverses only the first half of the number and compares it to the remaining part of the original number. This significantly improves efficiency compared to complete reversal.

Step-by-Step Algorithm

Step 1: Handle negative numbers and zero. Negative numbers are immediately deemed non-palindromes, and 0 is a palindrome.
Step 2: Check if the number ends in 0 (except for 0). If it does (and it's not 0), it can't be a palindrome.
Step 3: Initialize a `reversed_num` variable to 0. This will store the reversed half of the input number.
Step 4: Iterate while the original number `x` is greater than `reversed_num`. In each iteration:
Step 5: Extract the last digit of `x` using the modulo operator (`% 10`).
Step 6: Append this last digit to `reversed_num` by multiplying it by 10 and adding it to `reversed_num`.
Step 7: Remove the last digit from `x` using integer division (`// 10`).
Step 8: After the loop, check if `x` (the remaining half) is equal to `reversed_num` or if `x` is equal to `reversed_num / 10` (to handle odd-length numbers). If either is true, it's a palindrome.

Key Insights

Insight 1: Negative numbers are not palindromes because of the minus sign. This allows for an early exit condition.
Insight 2: Numbers ending in 0 (except 0 itself) are not palindromes because they would have a leading 0 if reversed. This also helps with early exit.
Insight 3: We only need to compare the first half of the number with its reverse. Reversing the entire number is unnecessary and potentially inefficient for very large numbers.

Complexity Analysis

Time Complexity: O(log(x))

Space Complexity: O(1)

Topics

This problem involves: Math.

Companies

Asked at: Accenture, Adobe, Amazon, Apple, Bloomberg, Capgemini, Capital One, Cognizant, Deloitte, EPAM Systems, FPT, Garmin, HCL, IBM, Infosys, Intel, J.P. Morgan, Meta, Microsoft, Oracle, Qualcomm, Roche, Samsung, Uber, Visa, Walmart Labs, Wipro, Yahoo, Yandex, Zoho, persistent systems, tcs.

Palindrome Number | LeetCode 9

Easy

Math

Time: O(log(x))Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean isPalindrome(int x) {
        if (x < 0) {
            return false;
        }
        if (x == 0) {
            return true;
        }
        if (x % 10 == 0) {
            return false;
        }

        int reversedNum = 0;
        while (x > reversedNum) {
            reversedNum = reversedNum * 10 + x % 10;
            x /= 10;
        }

        return x == reversedNum || x == reversedNum / 10;
    }
}

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Problem Info

Description

Problem 9: Palindrome Number - Detailed explanation and solution approaches.

Topics

Math

Companies

AccentureAdobeAmazonAppleBloombergCapgeminiCapital OneCognizant

Stats

Difficulty:Easy

Languages:4

Companies:32

Problem #:9

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

8 steps

1
Step 1: Handle negative numbers and zero. Negative numbers are immediately deemed non-palindromes, and 0 is a palindrome.
2
Step 2: Check if the number ends in 0 (except for 0). If it does (and it's not 0), it can't be a palindrome.
3
Step 3: Initialize a `reversed_num` variable to 0. This will store the reversed half of the input number.
4
Step 4: Iterate while the original number `x` is greater than `reversed_num`. In each iteration:
5
Step 5: Extract the last digit of `x` using the modulo operator (`% 10`).
6
Step 6: Append this last digit to `reversed_num` by multiplying it by 10 and adding it to `reversed_num`.
7
Step 7: Remove the last digit from `x` using integer division (`// 10`).
8
Step 8: After the loop, check if `x` (the remaining half) is equal to `reversed_num` or if `x` is equal to `reversed_num / 10` (to handle odd-length numbers). If either is true, it's a palindrome.

💡

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(log(x))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is logarithmic because the number of iterations in the `while` loop is proportional to the number of digits in the input integer `x`. The number of digits is approximately log10(x). The space complexity is constant because only a few integer variables are used, regardless of the input size.

📖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

Solution Approaches

String Conversion

Convert to string and check if it reads the same forwards and backwards

Time: O(log x)Space: O(log x)

Mathematical Reversal

Reverse half the number mathematically

Time: O(log x)Space: O(1)

Alternative Approaches

String Conversion

Convert the integer to a string, reverse the string, and compare the original and reversed strings.

Time: O(n), where n is the number of digitsSpace: O(n), to store the reversed string

Trade-offs: Simpler to understand, but less efficient than the provided solution, especially for very large numbers.

Recursive Approach

Recursively reverse the number and compare.

Time: O(log10(x))Space: O(log10(x)) due to recursive calls

Trade-offs: Less efficient in space compared to the iterative solution due to recursive call stack.

✨

Key Insights

1Insight 1: Negative numbers are not palindromes because of the minus sign. This allows for an early exit condition.
2Insight 2: Numbers ending in 0 (except 0 itself) are not palindromes because they would have a leading 0 if reversed. This also helps with early exit.
3Insight 3: We only need to compare the first half of the number with its reverse. Reversing the entire number is unnecessary and potentially inefficient for very large numbers.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: Negative numbers. The code correctly handles this by immediately returning `false`.
Edge case 2: Numbers ending in 0 (except 0). These are correctly identified as not palindromes.
Edge case 3: Numbers with an odd number of digits. The condition `x == reversed_num / 10` handles this correctly by comparing the remaining single digit with the reversed half's integer division.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle negative numbers and numbers ending in 0.

✗Common mistake 2: Incorrectly handling the case of odd-length numbers. Failing to account for the middle digit in comparison.

✗Common mistake 3: Inefficiently reversing the entire number instead of just the first half.

✅

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: Validating user inputs, such as credit card numbers or social security numbers (though these often have additional validation rules).

🏦Real-world scenario 2: Data processing tasks involving numerical palindromes, potentially in cryptography or pattern recognition.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: June 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadMay 29

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92May 22

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

@AlgoMasterMay 15

❓

Frequently Asked Questions

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Palindrome Number"?

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

🤖 Need help with this problem?