MSUBSTR - Mirror Strings !!!
As we all know Utkarsh is very good at solving number based problems, this time Arpit thinks smartly and gives Utkarsh to solve a problem on Strings. Arpit gives Utkarsh a string and challenges him to find the length of largest substring that have its mirror string same as its original one and number of such substrings. Now Utkarsh is busy at preparing Avishkar papers so he asks you to help him in doing this task.
E.g. for mirror string: Consider string "lalit" then its mirror string will be "tilal".
Input
There are t numbers of test cases (t <= 200) followed t lines where each line contains a character string of lower case characters (a-z) of length l (1 <= l <= 3000).
Output
There will be two integers per line separated by space indicating the length of largest substring which have its mirror string same and number of such substrings.
Example
Input: 3 lalit abedcdetr abcde Output: 3 1 5 1 1 5
hide comments
trung_chu:
2024-06-12 18:07:07
I have a test string "tilit".
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cjn2007:
2021-10-28 13:06:06
AC in one go with O(n). |
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rks14:
2019-09-21 12:48:53
QL^2logL passes, weak test cases. |
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suvro_coder:
2017-06-18 23:50:47
AC in 2 goes. Used maacher's algorithm.
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Gaurav Dahima:
2016-08-31 20:01:27
@Rahul Dubey O(n*n) is giving TLE
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Ayush Mishra:
2015-06-17 22:01:18
Did it without manchester algorithm. Used polynomial hashing and binary search. Complexity per testcase is O(n log n). |
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FoolForCS:
2015-05-19 21:32:25
Simple and so elegant. Wow-ed by this algorithm! Last edit: 2015-05-19 21:33:42 |
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vip_yadav:
2015-01-11 13:26:50
very nice problem !!! finally AC>>>!!!! |
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The Arrow:
2014-08-19 19:27:30
@admin please delete spoiler comment
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nitesh kumar:
2014-07-17 14:20:26
manacher algorithm
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Added by: | ! include(L.ppt) |
Date: | 2012-09-21 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
Resource: | MNNIT OPC 21-09-2012 |