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    <title>DEV Community: Ganthier Didier</title>
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      <title>DEV Community: Ganthier Didier</title>
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    <item>
      <title>How to Determine if an Integer is a Palindrome on LeetCode</title>
      <dc:creator>Ganthier Didier</dc:creator>
      <pubDate>Fri, 17 May 2024 22:07:55 +0000</pubDate>
      <link>https://dev.to/didierganthier/how-to-determine-if-an-integer-is-a-palindrome-on-leetcode-1md5</link>
      <guid>https://dev.to/didierganthier/how-to-determine-if-an-integer-is-a-palindrome-on-leetcode-1md5</guid>
      <description>&lt;h2&gt;
  
  
  Introduction
&lt;/h2&gt;

&lt;p&gt;In this article, we will explore a straightforward method to determine if a given integer is a palindrome. This problem is a common algorithmic challenge that you might encounter on platforms like LeetCode. We will use JavaScript to implement our solution.&lt;/p&gt;

&lt;h2&gt;
  
  
  Thought Process
&lt;/h2&gt;

&lt;p&gt;A palindrome is a number that reads the same backward as forward. For example, &lt;code&gt;121&lt;/code&gt; is a palindrome, while &lt;code&gt;123&lt;/code&gt; is not. To solve this problem, we can leverage the properties of strings to reverse the digits of the number and then compare the reversed version with the original.&lt;/p&gt;

&lt;h2&gt;
  
  
  Approach
&lt;/h2&gt;

&lt;p&gt;Our approach can be broken down into the following steps:&lt;/p&gt;

&lt;ol&gt;
&lt;li&gt;&lt;p&gt;&lt;strong&gt;Handle Negative Numbers&lt;/strong&gt;: Negative numbers cannot be palindromes due to the presence of the negative sign. Thus, if the input number is negative, we can immediately return &lt;code&gt;false&lt;/code&gt;.&lt;/p&gt;&lt;/li&gt;
&lt;li&gt;
&lt;p&gt;&lt;strong&gt;Convert to String and Reverse&lt;/strong&gt;: &lt;/p&gt;

&lt;ul&gt;
&lt;li&gt;Convert the number to a string using the &lt;code&gt;toString()&lt;/code&gt; method.&lt;/li&gt;
&lt;li&gt;Split the string into an array of characters.&lt;/li&gt;
&lt;li&gt;Reverse the array using the &lt;code&gt;reverse()&lt;/code&gt; method.&lt;/li&gt;
&lt;li&gt;Join the reversed array back into a string.&lt;/li&gt;
&lt;/ul&gt;
&lt;/li&gt;
&lt;li&gt;&lt;p&gt;&lt;strong&gt;Comparison&lt;/strong&gt;: Convert the reversed string back to an integer using &lt;code&gt;parseInt()&lt;/code&gt; and compare it with the original number. If they are equal, the number is a palindrome; otherwise, it is not.&lt;/p&gt;&lt;/li&gt;
&lt;/ol&gt;

&lt;h2&gt;
  
  
  Complexity
&lt;/h2&gt;

&lt;ul&gt;
&lt;li&gt;
&lt;strong&gt;Time Complexity&lt;/strong&gt;: The time complexity is ( O(n) ), where ( n ) is the number of digits in the number. This is because converting the number to a string, reversing the string, and joining the string all require linear time relative to the number of digits.&lt;/li&gt;
&lt;li&gt;
&lt;strong&gt;Space Complexity&lt;/strong&gt;: The space complexity is ( O(n) ), because we create a new string (and an intermediate array) that is proportional in size to the number of digits in the number.&lt;/li&gt;
&lt;/ul&gt;

&lt;h2&gt;
  
  
  Code
&lt;/h2&gt;

&lt;p&gt;Here is the implementation of the above approach in JavaScript:&lt;br&gt;
&lt;/p&gt;

&lt;div class="highlight js-code-highlight"&gt;
&lt;pre class="highlight javascript"&gt;&lt;code&gt;&lt;span class="cm"&gt;/**
 * @param {number} x
 * @return {boolean}
 */&lt;/span&gt;
&lt;span class="kd"&gt;var&lt;/span&gt; &lt;span class="nx"&gt;isPalindrome&lt;/span&gt; &lt;span class="o"&gt;=&lt;/span&gt; &lt;span class="kd"&gt;function&lt;/span&gt;&lt;span class="p"&gt;(&lt;/span&gt;&lt;span class="nx"&gt;x&lt;/span&gt;&lt;span class="p"&gt;)&lt;/span&gt; &lt;span class="p"&gt;{&lt;/span&gt;
    &lt;span class="k"&gt;if &lt;/span&gt;&lt;span class="p"&gt;(&lt;/span&gt;&lt;span class="nx"&gt;x&lt;/span&gt; &lt;span class="o"&gt;&amp;lt;&lt;/span&gt; &lt;span class="mi"&gt;0&lt;/span&gt;&lt;span class="p"&gt;)&lt;/span&gt; &lt;span class="p"&gt;{&lt;/span&gt;
        &lt;span class="k"&gt;return&lt;/span&gt; &lt;span class="kc"&gt;false&lt;/span&gt;&lt;span class="p"&gt;;&lt;/span&gt;
    &lt;span class="p"&gt;}&lt;/span&gt;
    &lt;span class="c1"&gt;// Convert the number to a string, reverse it, and compare with the original&lt;/span&gt;
    &lt;span class="k"&gt;return&lt;/span&gt; &lt;span class="nf"&gt;parseInt&lt;/span&gt;&lt;span class="p"&gt;(&lt;/span&gt;&lt;span class="nx"&gt;x&lt;/span&gt;&lt;span class="p"&gt;.&lt;/span&gt;&lt;span class="nf"&gt;toString&lt;/span&gt;&lt;span class="p"&gt;().&lt;/span&gt;&lt;span class="nf"&gt;split&lt;/span&gt;&lt;span class="p"&gt;(&lt;/span&gt;&lt;span class="dl"&gt;""&lt;/span&gt;&lt;span class="p"&gt;).&lt;/span&gt;&lt;span class="nf"&gt;reverse&lt;/span&gt;&lt;span class="p"&gt;().&lt;/span&gt;&lt;span class="nf"&gt;join&lt;/span&gt;&lt;span class="p"&gt;(&lt;/span&gt;&lt;span class="dl"&gt;""&lt;/span&gt;&lt;span class="p"&gt;))&lt;/span&gt; &lt;span class="o"&gt;===&lt;/span&gt; &lt;span class="nx"&gt;x&lt;/span&gt;&lt;span class="p"&gt;;&lt;/span&gt;
&lt;span class="p"&gt;};&lt;/span&gt;
&lt;/code&gt;&lt;/pre&gt;

&lt;/div&gt;



&lt;h2&gt;
  
  
  Conclusion
&lt;/h2&gt;

&lt;p&gt;This solution is a simple and effective way to check if an integer is a palindrome. By leveraging JavaScript's string manipulation methods, we can achieve this in a few lines of code. This method ensures clarity and efficiency, making it a great approach for solving this problem on coding platforms like LeetCode.&lt;/p&gt;

&lt;p&gt;Feel free to experiment with this code and try solving other similar problems to enhance your understanding of palindromes and string manipulations. Happy coding!&lt;/p&gt;

</description>
      <category>leetcode</category>
      <category>javascript</category>
      <category>algorithms</category>
      <category>interviewprep</category>
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