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Topic: interesting string problems (Read 7308 times) 

m_aakash
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interesting string problems
« on: Mar 22^{nd}, 2008, 12:18pm » 
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1) Given s string, rearrange characters to form a longest palindrome. If multiple palindromes are possible, return first one in lexicographic order. If no palindrome can be formed then return NULL. 2) Describe an efficient algorithm to find the length of the longest substring that appears both forward and backward in an input string T[1 . n]. The forward and backward must not overlap. Ex: redivide output: 3 (edi)


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Grimbal
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Re: interesting string problems
« Reply #1 on: Mar 22^{nd}, 2008, 4:09pm » 
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1) Order all characters in ascending order. You have a sequence of characters, some repeated. Build the palindrome from the ends while scanning the sorted string from left to right. I.e. start with strings S and E, both empty. For every repeated character, add half of them at the end of S and another half to the front of S. Round the halves down. If any character is left (for an odd number) store the first one you meet as F. When done with the sorted string, your palindrome is S+F+E.


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m_aakash
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Posts: 96


Re: interesting string problems
« Reply #2 on: Mar 22^{nd}, 2008, 8:36pm » 
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on Mar 22^{nd}, 2008, 4:09pm, Grimbal wrote:1) Order all characters in ascending order. You have a sequence of characters, some repeated. Build the palindrome from the ends while scanning the sorted string from left to right. I.e. start with strings S and E, both empty. For every repeated character, add half of them at the end of S and another half to the front of S. Round the halves down. If any character is left (for an odd number) store the first one you meet as F. When done with the sorted string, your palindrome is S+F+E. 
 well done grimbal... can you think of O(n^2) algo for second problem... i think we have to use dp for second one.


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Leo
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Re: interesting string problems
« Reply #3 on: Mar 22^{nd}, 2008, 8:48pm » 
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(b) How about finding the longest common substring of the string and its reverse with an extra condition to check for overlap ... something like If S is the String and S' is its reverse. Then the LCS that comes up should be tested for length(S)  start_index(S')  start_index(S)  (2 * lendth(LCS)) > 0 would this work?


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m_aakash
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Posts: 96


Re: interesting string problems
« Reply #4 on: Mar 22^{nd}, 2008, 9:48pm » 
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on Mar 22^{nd}, 2008, 8:48pm, Leo wrote:(b) How about finding the longest common substring of the string and its reverse with an extra condition to check for overlap ... something like If S is the String and S' is its reverse. Then the LCS that comes up should be tested for length(S)  start_index(S')  start_index(S)  (2 * lendth(LCS)) > 0 would this work? 
 ex: fghqaaaaphgf Longest common substring is aaaa say your test detects overlap even then how do you find the actual result which is fgh


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pscoe2
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Re: interesting string problems
« Reply #5 on: Mar 22^{nd}, 2008, 10:23pm » 
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we can do as Leo suggested just keeping an array with tags which increment if they are counted for LCS Ex: fghqaaaaphgr 000022220000 bcoz this part has been considered for LCS twice for both S and S' in such a situation LCS can be either equal to half (or as the case might be...see the first starting 1 n the ending 1 and take mean) In the above case LCS=4/2=2 but as there are repetitions we hv to run the algo again just this time forbidding the LCS to come out to be the same do this till u get an LCS with all 0's and 1's and the find the longest of them Ex(contd.) in second iteration the array become 111000000111 which gives the value of LCS=3 and we see tht in the second case the LCS is higher and also there are only 0's and 1's in the final array so no more iterations are req. to get the optimal soln.s

« Last Edit: Mar 22^{nd}, 2008, 10:28pm by pscoe2 » 
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Leo
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Re: interesting string problems
« Reply #6 on: Mar 23^{rd}, 2008, 12:09am » 
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here's an implementation in java ... String s = "fghqaaaaphgf"; String bestMatch = ""; String ss = Reverse(s); for (i=0;i<s.length();i++) { for (int k=ss.length()1;k>=i+2;k) { String subString = s.substring(i, k); int index = ss.indexOf(subString); if (index != 1) { //test for overlap here if ((s.length()  index)  i  (2 * (k  i)) > 0) { if (bestMatch.length() < subString.length()) bestMatch = subString; break; } } } }


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Jigsaw
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Re: interesting string problems
« Reply #7 on: Jun 18^{th}, 2008, 4:22am » 
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on Mar 22^{nd}, 2008, 12:18pm, m_aakash wrote:1) Given s string, rearrange characters to form a longest palindrome. If multiple palindromes are possible, return first one in lexicographic order. If no palindrome can be formed then return NULL. 2) Describe an efficient algorithm to find the length of the longest substring that appears both forward and backward in an input string T[1 . n]. The forward and backward must not overlap. Ex: redivide output: 3 (edi) 
 int n=str.size()1; //1 based string for(int i=1;i<=n;i++) for(int j=n;j>i;j++) ls[i][j]=((str[i]==str[j])?ls[i1][j+1]+1:0); The code is self explanatory I guess. The longest substring ending at i1 from one direction and at j+1 from other direction can be extended by 1 if a[i] and b[j] are equal. Once you calculate the lcs matrix, if ls[i][j] contains the largest value, i gives the position in string and ls[i][j] gives the length of the substring which has the stated properties. [edit: replaced all lcs with ls]

« Last Edit: Jun 23^{rd}, 2008, 9:05am by jagatsastry » 
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curt_cobain
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Posts: 33


Re: interesting string problems
« Reply #8 on: Jun 21^{st}, 2008, 12:14am » 
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1) Order all characters in ascending order. You have a sequence of characters, some repeated. Build the palindrome from the ends while scanning the sorted string from left to right. I.e. start with strings S and E, both empty. For every repeated character, add half of them at the end of S and another half to the front of S. Round the halves down. If any character is left (for an odd number) store the first one you meet as F. When done with the sorted string, your palindrome is S+F+E. [hide][/hide] CAN U TELL ME WHAT IS IN E ALSO CAN U ELABORATE UR LOGIC WITH THE EAXAMPLE


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tuxilogy
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Re: interesting string problems
« Reply #9 on: Jun 23^{rd}, 2008, 1:33am » 
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@ Jigsaw, Can you please xplain your LCS matrix logic, i could not get it


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Jigsaw
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Re: interesting string problems
« Reply #10 on: Jun 23^{rd}, 2008, 9:02am » 
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on Jun 23^{rd}, 2008, 1:33am, tuxilogy wrote:@ Jigsaw, Can you please xplain your LCS matrix logic, i could not get it 
 Yeah, sure. Say str[ik. . .i1] and rev(str[j+1. . .k+j]) are equal(i.e. has the desired property) and str[ik...i1] is the longest substring(as asked) ending at i1 and is of length k, then the length of the longest substring ending at i is of length k+1 if str[i] == str[j]. Otherwise it(ls[i][j]) would be of length 0. E.g.: str=xyzzy ls[2][5]=1 (str[2](y)==str[5](y)) ls[3][4]=ls[2][5]+1=2 since str[3](z)==str[4](z) and ls[3][5]=0 since str[3](z)!=str[5](y) After constructing the whole matrix, find the indices i and j such that ls[i][j] is maximum. i is the position where the required substring ends and ls[i][j] is its length. Here comes the code(I've just coded and tested) Code: string getLcs(const string& str) { int n=str.size(); vector<vector<int> > ls(n+2, vector<int>(n+2,0)); pair<int, int> max(0, 1); //Max value, index for(int i=1;i<=n;i++) for(int j=n;j>i;j) { ls[i][j]=(str[i1]==str[j1])?ls[i1][j+1]+1:0; //Indices adjusted for 0 based string indices if(ls[i][j]>max.first) max.first=ls[i][j], max.second=i; } if(max.second==1) return ""; else return string(str, max.secondmax.first, max.first); } 
 [edit: replaced all lcs with "ls"]

« Last Edit: Jun 23^{rd}, 2008, 9:07am by jagatsastry » 
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manita23
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Re: interesting string problems
« Reply #11 on: Jul 22^{nd}, 2014, 9:54pm » 
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Here is java method to find longest Palindrome string[] public static String longestPalindromeString(String s) { if (s == null) return null; String longest = s.substring(0, 1); for (int i = 0; i < s.length()  1; i++) { //odd cases like 121 String palindrome = intermediatePalindrome(s, i, i); if (palindrome.length() > longest.length()) { longest = palindrome; } //even cases like 1221 palindrome = intermediatePalindrome(s, i, i + 1); if (palindrome.length() > longest.length()) { longest = palindrome; } } return longest; }

« Last Edit: Aug 13^{th}, 2014, 6:59am by Grimbal » 
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