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MINMOVE - Minimum Rotations |
English | Vietnamese |
Given a string S[1..n] . A rotation on S is that we move the first character to the right-most of the string. More specific, after a rotation, S becomes T = S[2..n] + S[1].
For example: S = abcaa, then after a rotation we have S = bcaaa.
Find the minimum number of rotations to make S become the smallest lexicographical order string.
Input
A single line contains a string S. S contains only small letters of English alphabet (āaā .. āzā), and the length of S is not more than 100000.
Output
A single line contains an integer which represents the minimum number of rotations.
Example
Input: mississippi Output: 10
Test cases and time limit have been updated. Some accepted solution got TLE.
Added by: | Race with time |
Date: | 2008-12-29 |
Time limit: | 0.100s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ERL JS-RHINO PERL6 |
Resource: | Based on a problem from ACM Central European Programming Contest |
hide comments
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2024-06-30 16:25:59
@coolcoder_iith How did you know it timed out at 12? Finally my O(n*logn) solution got accepted. My previous attempt of O(n*logn*logn) solution gave a TLE Last edit: 2024-06-30 16:41:18 |
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2019-10-03 20:19:13
time out at #12 what to do |
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2019-09-15 17:47:12 yaswanth
Tried n*logn*logn with quite number of optimisations. TLE n*logn went through though |
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2018-08-23 23:15:18
Was getting WA on #17 with prefix hash Made my for loop that compared the current solution with a new candidate run twice instead of once Got AC |
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2016-09-30 13:29:49
learned a lot :D |
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2016-09-11 19:19:06
Those who are using stl, make sure to use stable_sort if using nlogn*logn solution. Costed me few WA |
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2016-04-03 05:42:51 minhthai
"Some accepted solution got TLE." Please, give some care to Java nlogn solution :(( |
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2016-02-05 15:33:28
Same as http://www.spoj.com/problems/BEADS/ . O(nlogn) passed. :) |
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2016-01-22 05:30:20 Dhawal Harkawat
silly mistake costed two WAs at #10. finally AC. |
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2016-01-21 23:37:21 Sumit Vohra
solved using suffix automaton in O(n) |