Next Palindrome

Problem Statement: 
                     Given a number find the next Palindrome Number

Solution:
    
In the given problem , we take two pointers one at left and the other on right, we start comparing i and n-i.
The difference of left and right digit (which are multiplied by bases) is added to the number , in case the right digit is greater, we match up the effect introduced by the difference which was added in the previous step.

Time Complexity: O(n)
Space Complexity: O(1)

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/** * Get next pallindrome */ public static int nextPanlindrome(int num) { if(isPalindrome(num)) num++; String temp = num + ""; int end=temp.length() -1; int base=1; for(int start = 0; start < end; start++, end--) { // Gives the right digit int right = Integer.parseInt(temp.charAt(end)+"") * base; // Gives the left digit int left = Integer.parseInt(temp.charAt(start)+"") * base; // add differnce to the number num += left - right ; //------(1) base *=10; if(right > left) { num += base; // for incresing the value of number which got decreased at (1) //number of digits are even, if(start == end - 1) num += base/10; } temp = num + ""; } return num; } /** * Checks if a number is pallindriome * @param num * @return */ public static boolean isPalindrome(int num) { int reverse = 0, temp=num; while( temp != 0 ) { reverse = reverse * 10; reverse = reverse + temp%10; temp = temp/10; } return (num == reverse ? true: false); }

In case of any clarification , suggestion or enhancement of code , please comment.
Happy Coding !! :)