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plurals

PO
primarykey
Id
10786320
data
AcceptedAnswerId
0
AnswerCount
0
ClosedDate
CommentCount
4
CommunityOwnedDate
CreationDate
2012-05-28T14:32:05.130
FavoriteCount
0
LastActivityDate
2014-03-18T10:49:06.997
LastEditDate
2014-03-18T10:49:06.997
LastEditorUserId
815507
OwnerUserId
815507
ParentId
10639661
PostTypeId
2
Score
1
ViewCount
0
LastEditorDisplayName
text
Body
The solution needs to be O(n) time and with space O(1). This leaves out sorting and the space O(1) requirement is a hint that you probably should make a hash of the strings and compare them. If you have access to a prime number list do as cheeken's solution. Note: If the interviewer says you don't have access to a prime number list. Then generate the prime numbers and store them. This is O(1) because the Alphabet length is a constant. Else here's my alternative idea. I will define the Alphabet as = {a,b,c,d,e} for simplicity. The values for the letters are defined as: <pre><code>a, b, c, d, e 1, 2, 4, 8, 16 </code></pre> note: if the interviewer says this is not allowed, then make a lookup table for the Alphabet, this takes O(1) space because the size of the Alphabet is a constant Define a function which can find the distinct letters in a string. <pre><code>// set bit value of char c in variable i and return result distinct(char c, int i) : int E.g. distinct('a', 0) returns 1 E.g. distinct('a', 1) returns 1 E.g. distinct('b', 1) returns 3 </code></pre> Thus if you iterate the string "aab" the distinct function should give 3 as the result Define a function which can calculate the sum of the letters in a string. <pre><code>// return sum of c and i sum(char c, int i) : int E.g. sum('a', 0) returns 1 E.g. sum('a', 1) returns 2 E.g. sum('b', 2) returns 4 </code></pre> Thus if you iterate the string "aab" the sum function should give 4 as the result Define a function which can calculate the length of the letters in a string. <pre><code>// return length of string s length(string s) : int E.g. length("aab") returns 3 </code></pre> Running the methods on two strings and comparing the results takes O(n) running time. Storing the hash values takes O(1) in space. <pre><code> e.g. distinct of "aab" => 3 distinct of "aba" => 3 sum of "aab => 4 sum of "aba => 4 length of "aab => 3 length of "aba => 3 </code></pre> Since all the values are equal for both strings, they must be a permutation of each other. EDIT: The solutions is not correct with the given alphabet values as pointed out in the comments.
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1. This table or related slice is empty.
PostParentId
1. POCheck if array B is a permutation of A
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PostTypePostTypeId
1. PTAnswer
UserLastEditorUserId
1. USKunukn
UserOwnerUserId
1. USKunukn
plurals
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CommentsPostId
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