palindrome

So on Project Euler the Problem 4 states the following: A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99. Find the largest palindrome made from the product of two 3-digit numbers. I have tried the following: #include <stdio.h> #include <stdlib.h> int check(int result) { char b[7]; sprintf(b, "%d", result); if (b[0] == b[5] && b[1] == b[4] && b[2] == b[3]) { return 1; } else { return 0; } } int main () { int i; int g; int final; for (i = 999; i > 99; i--) { for (g = 999; g > 99; g--) { if (check(g*i) == 1) { final

I am aware of solutions that uses the bottom up dynamic programing approach to solve this problem in O(n^2). I am specifically looking for a top down dp approach. Is it possible to achieve longest palindromic substring using a recursive solution? Here is what I have tried but it fails for certain cases, but I feel I am almost on the right track. #include <iostream> #include <string> using namespace std; string S; int dp[55][55]; int solve(int x,int y,int val) { if(x>y)return val; int &ret = dp[x][y]; if(ret!=0){ret = val + ret;return ret;} //cout<<"x: "<<x<<" y: "<<y<<" val: "<<val<<endl; if(S

Folks, I was recently interviewed and got a question on Palindrome. Given a string ( which might represent a date ), check if it's a palindrome or not using Stack. I tried to come up with solution, but he didn't like that. Can anyone show me the code snippet for it in Java ? Thanks PS : This is not a homework, actual interview question. import java.util.Stack; public class PalindromeTest { public static void main(String[] args) { String input = "test"; Stack<Character> stack = new Stack<Character>(); for (int i = 0; i < input.length(); i++) { stack.push(input.charAt(i)); } String reverseInput

I have the program made up until the point where it has to ignore and punctuations and spaces in the thread and I was wondering if anyone could help me with the coding for that? What I've been trying out doesn't seem to be working. Here is what I have so far: import java.util.Scanner; public class PalindromeTester { public static void main (String[] args) { String str, another = "y"; int left, right; char charLeft, charRight; Scanner scan = new Scanner (System.in); while (another.equalsIgnoreCase("y")) // allows y or Y { System.out.println ("Enter a potential palindrome: "); str = scan

After reading polygenelubricants 's series of articles on advanced regular expressions techniques (particularly How does this Java regex detect palindromes? ), I decided to attempt to create my own PCRE regex to parse a palindrome, using recursion (in PHP). What I came up with was: ^(([a-z])(?1)\2|[a-z]?)$ My understanding of this expression is that it should either match zero or one characters (every string of less than 2 characters is implicitly a palindrome, as well as to account for palindromes with odd lengths in the recursion), or two of the same character separated by a recursion of the

Is there any suggestion on solving next higher prime and palindrome number from a given int. Here is the snippet I am trying but its a kind of slow, please suggest if you ve any good algorithm that i can test. #!/usr/bin/python def next_higher(n): while True: s = str(n) if not any([n % i == 0 \ for i in range(2, int(n**0.5))]) and s == s[::-1]: return n n = n + 1 print next_higher(2004) print next_higher(20) Output: 10201 101 Updated code testing for palindrome before prime. much faster than my previous code. I am implementing the suggestion from user2357112. #!/usr/bin/python def next_higher

Using Python, I am trying to solve problem #4 of the Project Euler problems. Can someone please tell me what I am doing incorrectly? The problem is to Find the largest palindrome made from the product of two 3-digit numbers . Here is what I have thus far. import math def main(): for z in range(100, 1000): for y in range(100, 1000): for x in range(1, 1000000): x = str(x) if x == x[::-1] and x == z*y: print x if __name__ == '__main__': main() Some efficiency issues: start at the top (since we can use this in skipping a lot of calculations) don't double-calculate def is_palindrome(n): s = str(n)