Count repetitions of an arrayShortest Unique SubstringsWrite a program that finds the most occurring paired...
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Count repetitions of an array
Shortest Unique SubstringsWrite a program that finds the most occurring paired letter in a stringLisp Extraction MissionOrdering words to fit in a given stringReverse Deltas of an ArraySort and Re-apply Deltas of an ArrayTurn an integer n into a list containing it n timesRebuild a rectangular array from a cornerMost Common MultipleLongest Repeating Subsequence of a Single Digit
$begingroup$
You will receive an array and must return the number of integers that occur more than once.
[234, 2, 12, 234, 5, 10, 1000, 2, 99, 234]
This will return 2, since both 234 and 2 are repeated twice.
[234, 2, 12, 234]
[2, 12, 234, 5, 10, 1000, 2]
The list will never be more than 100k integers long, and the integers inside the list will always be in between -100k and 100k.
Integers should be counted if they occur more than once, so if an integer occurs 3 times then it will still only count as one repeated integer.
Test cases
[1, 10, 16, 4, 8, 10, 9, 19, 2, 15, 18, 19, 10, 9, 17, 15, 19, 5, 13, 20] = 4
[11, 8, 6, 15, 9, 19, 2, 2, 4, 19, 14, 19, 13, 12, 16, 13, 0, 5, 0, 8] = 5
[9, 7, 8, 16, 3, 9, 20, 19, 15, 6, 8, 4, 18, 14, 19, 12, 12, 16, 11, 19] = 5
[10, 17, 17, 7, 2, 18, 7, 13, 3, 10, 1, 5, 15, 4, 6, 0, 19, 4, 17, 0] = 5
[12, 7, 17, 13, 5, 3, 4, 15, 20, 15, 5, 18, 18, 18, 4, 8, 15, 13, 11, 13] = 5
[0, 3, 6, 1, 5, 2, 16, 1, 6, 3, 12, 1, 16, 5, 4, 5, 6, 17, 4, 8] = 6
[11, 19, 2, 3, 11, 15, 19, 8, 2, 12, 12, 20, 13, 18, 1, 11, 19, 7, 11, 2] = 4
[6, 4, 11, 14, 17, 3, 17, 11, 2, 16, 14, 1, 2, 1, 15, 15, 12, 10, 11, 13] = 6
[0, 19, 2, 0, 10, 10, 16, 9, 19, 9, 15, 0, 10, 18, 0, 17, 18, 18, 0, 9] = 5
[1, 19, 17, 17, 0, 2, 14, 10, 10, 12, 5, 14, 16, 7, 15, 15, 18, 11, 17, 7] = 5
code-golf array-manipulation
edited 40 mins ago
MickyT
10k21535
asked 1 hour ago
jayko03jayko03
1163
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jayko03 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
You will receive an array and must return the number of integers that occur more than once.
[234, 2, 12, 234, 5, 10, 1000, 2, 99, 234]
This will return 2, since both 234 and 2 are repeated twice.
[234, 2, 12, 234]
[2, 12, 234, 5, 10, 1000, 2]
The list will never be more than 100k integers long, and the integers inside the list will always be in between -100k and 100k.
Integers should be counted if they occur more than once, so if an integer occurs 3 times then it will still only count as one repeated integer.
Test cases
[1, 10, 16, 4, 8, 10, 9, 19, 2, 15, 18, 19, 10, 9, 17, 15, 19, 5, 13, 20] = 4
[11, 8, 6, 15, 9, 19, 2, 2, 4, 19, 14, 19, 13, 12, 16, 13, 0, 5, 0, 8] = 5
[9, 7, 8, 16, 3, 9, 20, 19, 15, 6, 8, 4, 18, 14, 19, 12, 12, 16, 11, 19] = 5
[10, 17, 17, 7, 2, 18, 7, 13, 3, 10, 1, 5, 15, 4, 6, 0, 19, 4, 17, 0] = 5
[12, 7, 17, 13, 5, 3, 4, 15, 20, 15, 5, 18, 18, 18, 4, 8, 15, 13, 11, 13] = 5
[0, 3, 6, 1, 5, 2, 16, 1, 6, 3, 12, 1, 16, 5, 4, 5, 6, 17, 4, 8] = 6
[11, 19, 2, 3, 11, 15, 19, 8, 2, 12, 12, 20, 13, 18, 1, 11, 19, 7, 11, 2] = 4
[6, 4, 11, 14, 17, 3, 17, 11, 2, 16, 14, 1, 2, 1, 15, 15, 12, 10, 11, 13] = 6
[0, 19, 2, 0, 10, 10, 16, 9, 19, 9, 15, 0, 10, 18, 0, 17, 18, 18, 0, 9] = 5
[1, 19, 17, 17, 0, 2, 14, 10, 10, 12, 5, 14, 16, 7, 15, 15, 18, 11, 17, 7] = 5
code-golf array-manipulation
edited 40 mins ago
MickyT
10k21535
asked 1 hour ago
jayko03jayko03
1163
New contributor
jayko03 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
What do you mean byOnce it counts the repetition, don't count again? Also, since we want to find the repetition of a specific integer, how would we know which integer to search for if we are not given it? Lastly, the test cases are a bit confusing; which are output and which are input?
$endgroup$
– Embodiment of Ignorance
1 hour ago
2
$begingroup$
I've edited this to try to make it a bit clearer. Is this what you intended? Also, please put answers in for those test cases.
$endgroup$
– Riker
1 hour ago
$begingroup$
You may want to add some edge cases, like lists with negative, large-magnitude, unique, or no integers.
$endgroup$
– Adám
1 hour ago
$begingroup$
I have added some answers to the test cases, sorry if I go them wrong
$endgroup$
– MickyT
27 mins ago
add a comment |
$begingroup$
You will receive an array and must return the number of integers that occur more than once.
[234, 2, 12, 234, 5, 10, 1000, 2, 99, 234]
This will return 2, since both 234 and 2 are repeated twice.
[234, 2, 12, 234]
[2, 12, 234, 5, 10, 1000, 2]
The list will never be more than 100k integers long, and the integers inside the list will always be in between -100k and 100k.
Integers should be counted if they occur more than once, so if an integer occurs 3 times then it will still only count as one repeated integer.
Test cases
[1, 10, 16, 4, 8, 10, 9, 19, 2, 15, 18, 19, 10, 9, 17, 15, 19, 5, 13, 20] = 4
[11, 8, 6, 15, 9, 19, 2, 2, 4, 19, 14, 19, 13, 12, 16, 13, 0, 5, 0, 8] = 5
[9, 7, 8, 16, 3, 9, 20, 19, 15, 6, 8, 4, 18, 14, 19, 12, 12, 16, 11, 19] = 5
[10, 17, 17, 7, 2, 18, 7, 13, 3, 10, 1, 5, 15, 4, 6, 0, 19, 4, 17, 0] = 5
[12, 7, 17, 13, 5, 3, 4, 15, 20, 15, 5, 18, 18, 18, 4, 8, 15, 13, 11, 13] = 5
[0, 3, 6, 1, 5, 2, 16, 1, 6, 3, 12, 1, 16, 5, 4, 5, 6, 17, 4, 8] = 6
[11, 19, 2, 3, 11, 15, 19, 8, 2, 12, 12, 20, 13, 18, 1, 11, 19, 7, 11, 2] = 4
[6, 4, 11, 14, 17, 3, 17, 11, 2, 16, 14, 1, 2, 1, 15, 15, 12, 10, 11, 13] = 6
[0, 19, 2, 0, 10, 10, 16, 9, 19, 9, 15, 0, 10, 18, 0, 17, 18, 18, 0, 9] = 5
[1, 19, 17, 17, 0, 2, 14, 10, 10, 12, 5, 14, 16, 7, 15, 15, 18, 11, 17, 7] = 5
code-golf array-manipulation
edited 40 mins ago
MickyT
10k21535
asked 1 hour ago
jayko03jayko03
1163
New contributor
jayko03 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
You will receive an array and must return the number of integers that occur more than once.
[234, 2, 12, 234, 5, 10, 1000, 2, 99, 234]
This will return 2, since both 234 and 2 are repeated twice.
[234, 2, 12, 234]
[2, 12, 234, 5, 10, 1000, 2]
The list will never be more than 100k integers long, and the integers inside the list will always be in between -100k and 100k.
Integers should be counted if they occur more than once, so if an integer occurs 3 times then it will still only count as one repeated integer.
Test cases
[1, 10, 16, 4, 8, 10, 9, 19, 2, 15, 18, 19, 10, 9, 17, 15, 19, 5, 13, 20] = 4
[11, 8, 6, 15, 9, 19, 2, 2, 4, 19, 14, 19, 13, 12, 16, 13, 0, 5, 0, 8] = 5
[9, 7, 8, 16, 3, 9, 20, 19, 15, 6, 8, 4, 18, 14, 19, 12, 12, 16, 11, 19] = 5
[10, 17, 17, 7, 2, 18, 7, 13, 3, 10, 1, 5, 15, 4, 6, 0, 19, 4, 17, 0] = 5
[12, 7, 17, 13, 5, 3, 4, 15, 20, 15, 5, 18, 18, 18, 4, 8, 15, 13, 11, 13] = 5
[0, 3, 6, 1, 5, 2, 16, 1, 6, 3, 12, 1, 16, 5, 4, 5, 6, 17, 4, 8] = 6
[11, 19, 2, 3, 11, 15, 19, 8, 2, 12, 12, 20, 13, 18, 1, 11, 19, 7, 11, 2] = 4
[6, 4, 11, 14, 17, 3, 17, 11, 2, 16, 14, 1, 2, 1, 15, 15, 12, 10, 11, 13] = 6
[0, 19, 2, 0, 10, 10, 16, 9, 19, 9, 15, 0, 10, 18, 0, 17, 18, 18, 0, 9] = 5
[1, 19, 17, 17, 0, 2, 14, 10, 10, 12, 5, 14, 16, 7, 15, 15, 18, 11, 17, 7] = 5
code-golf array-manipulation
code-golf array-manipulation
edited 40 mins ago
MickyT
10k21535
asked 1 hour ago
jayko03jayko03
1163
New contributor
jayko03 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 40 mins ago
MickyT
10k21535
asked 1 hour ago
jayko03jayko03
1163
New contributor
jayko03 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 40 mins ago
MickyT
10k21535
edited 40 mins ago
MickyT
10k21535
edited 40 mins ago
MickyT
10k21535
10k21535
asked 1 hour ago
jayko03jayko03
1163
New contributor
jayko03 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
jayko03jayko03
1163
asked 1 hour ago
jayko03jayko03
1163
1163
New contributor
jayko03 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
jayko03 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
jayko03 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
What do you mean byOnce it counts the repetition, don't count again? Also, since we want to find the repetition of a specific integer, how would we know which integer to search for if we are not given it? Lastly, the test cases are a bit confusing; which are output and which are input?
$endgroup$
– Embodiment of Ignorance
1 hour ago
2
$begingroup$
I've edited this to try to make it a bit clearer. Is this what you intended? Also, please put answers in for those test cases.
$endgroup$
– Riker
1 hour ago
$begingroup$
You may want to add some edge cases, like lists with negative, large-magnitude, unique, or no integers.
$endgroup$
– Adám
1 hour ago
$begingroup$
I have added some answers to the test cases, sorry if I go them wrong
$endgroup$
– MickyT
27 mins ago
add a comment |
$begingroup$
What do you mean byOnce it counts the repetition, don't count again? Also, since we want to find the repetition of a specific integer, how would we know which integer to search for if we are not given it? Lastly, the test cases are a bit confusing; which are output and which are input?
$endgroup$
– Embodiment of Ignorance
1 hour ago
2
$begingroup$
I've edited this to try to make it a bit clearer. Is this what you intended? Also, please put answers in for those test cases.
$endgroup$
– Riker
1 hour ago
$begingroup$
You may want to add some edge cases, like lists with negative, large-magnitude, unique, or no integers.
$endgroup$
– Adám
1 hour ago
$begingroup$
I have added some answers to the test cases, sorry if I go them wrong
$endgroup$
– MickyT
27 mins ago
$begingroup$
What do you mean by
Once it counts the repetition, don't count again? Also, since we want to find the repetition of a specific integer, how would we know which integer to search for if we are not given it? Lastly, the test cases are a bit confusing; which are output and which are input?$endgroup$
– Embodiment of Ignorance
1 hour ago
$begingroup$
What do you mean by
Once it counts the repetition, don't count again? Also, since we want to find the repetition of a specific integer, how would we know which integer to search for if we are not given it? Lastly, the test cases are a bit confusing; which are output and which are input?$endgroup$
– Embodiment of Ignorance
1 hour ago
2
2
$begingroup$
I've edited this to try to make it a bit clearer. Is this what you intended? Also, please put answers in for those test cases.
$endgroup$
– Riker
1 hour ago
$begingroup$
I've edited this to try to make it a bit clearer. Is this what you intended? Also, please put answers in for those test cases.
$endgroup$
– Riker
1 hour ago
$begingroup$
You may want to add some edge cases, like lists with negative, large-magnitude, unique, or no integers.
$endgroup$
– Adám
1 hour ago
$begingroup$
You may want to add some edge cases, like lists with negative, large-magnitude, unique, or no integers.
$endgroup$
– Adám
1 hour ago
$begingroup$
I have added some answers to the test cases, sorry if I go them wrong
$endgroup$
– MickyT
27 mins ago
$begingroup$
I have added some answers to the test cases, sorry if I go them wrong
$endgroup$
– MickyT
27 mins ago
add a comment |
10 Answers
10
active
oldest
votes
$begingroup$
APL (Dyalog Unicode), 9 bytesSBCS
Anonymous tacit prefix function.
+/{1<≢⍵}⌸
Try it online!
+/ sum of
{…}⌸ for each unique element:
1< whether 1 is less than
≢⍵ the count of occurrences
answered 1 hour ago
AdámAdám
28.2k274201
$endgroup$
add a comment |
$begingroup$
R, 20 bytes
Is this what you are after? Uses table to count the occurrences of each of the scan input values. Tests if count is > 1 and sums the trues.
sum(table(scan())>1)
Try it online!
answered 1 hour ago
MickyTMickyT
10k21535
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 40 bytes
n=>n.GroupBy(c=>c).Where(x=>x.Count()>1)
Outputs by an IEnumerable<IGrouping<int,int>> with the relevant elements stored in the key.
Try it online!
answered 45 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,120119
$endgroup$
add a comment |
$begingroup$
Bash + coreutils, 18
sort|uniq -d|wc -l
Try it online!
answered 29 mins ago
Digital TraumaDigital Trauma
59.1k787224
$endgroup$
add a comment |
$begingroup$
Haskell, 47 bytes
f[]=0
f(a:b)|x<-filter(/=a)b,x/=b=1+f x|1>0=f b
Try it online!
This is the naïve approach. There is likely something that could be done to improve this.
f[]=0
We return 0 for the empty list
f(a:b)
In the case of a non-empty list starting with a and then b.
|x<-filter(/=a)b,x/=b=1+f x
If filtering a out of b is different from b (that is a is in b) then we return 1 more than f applied to b with the as filtered out.
|1>0=f b
If filtering as doesn't change b then we just run f across the rest.
Here is another similar approach that has the same length:
f[]=0
f(a:b)|elem a b=1+f(filter(/=a)b)|1>0=f b
Try it online!
answered 19 mins ago
Sriotchilism O'ZaicSriotchilism O'Zaic
35.1k10159369
$endgroup$
add a comment |
$begingroup$
J, 11 bytes
1#.(1<#)/.~
Try it online!
Explanation:
/.~ group the list by itself
( ) for each group
1<# is the length greater than 1
1#. sum by base-1 conversion
answered 18 mins ago
Galen IvanovGalen Ivanov
6,80711033
$endgroup$
add a comment |
$begingroup$
JavaScript (ES6), 40 bytes
a=>a.map(o=x=>n+=(o[x]=-~o[x])==2,n=0)|n
Try it online!
answered 10 mins ago
ArnauldArnauld
76.7k693322
$endgroup$
add a comment |
$begingroup$
Python 3, 93 bytes
import collections as c;print(sum(filter(lambda x:x>1,[y-1 for y in c.Counter(i).values()])))
Try it online!
answered 7 mins ago
Mike ShlantaMike Shlanta
357310
$endgroup$
add a comment |
$begingroup$
Ruby, 40 bytes
->a{a.select{|x|a.count(x)>1}.uniq.size}
Try it online!
answered 36 secs ago
Kirill L.Kirill L.
4,5251523
$endgroup$
add a comment |
$begingroup$
05AB1E, 4 bytes
Ù¢≠O
Try it online!
or as a Test Suite
Explanation
O # sum
≠ # the false values
¢ # in the count
Ù # of unique digits in input
answered 5 mins ago
EmignaEmigna
46.5k432142
$endgroup$
add a comment |
Your Answer
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10 Answers
10
active
oldest
votes
10 Answers
10
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
APL (Dyalog Unicode), 9 bytesSBCS
Anonymous tacit prefix function.
+/{1<≢⍵}⌸
Try it online!
+/ sum of
{…}⌸ for each unique element:
1< whether 1 is less than
≢⍵ the count of occurrences
answered 1 hour ago
AdámAdám
28.2k274201
$endgroup$
add a comment |
$begingroup$
APL (Dyalog Unicode), 9 bytesSBCS
Anonymous tacit prefix function.
+/{1<≢⍵}⌸
Try it online!
+/ sum of
{…}⌸ for each unique element:
1< whether 1 is less than
≢⍵ the count of occurrences
answered 1 hour ago
AdámAdám
28.2k274201
$endgroup$
add a comment |
$begingroup$
APL (Dyalog Unicode), 9 bytesSBCS
Anonymous tacit prefix function.
+/{1<≢⍵}⌸
Try it online!
+/ sum of
{…}⌸ for each unique element:
1< whether 1 is less than
≢⍵ the count of occurrences
answered 1 hour ago
AdámAdám
28.2k274201
$endgroup$
APL (Dyalog Unicode), 9 bytesSBCS
Anonymous tacit prefix function.
+/{1<≢⍵}⌸
Try it online!
+/ sum of
{…}⌸ for each unique element:
1< whether 1 is less than
≢⍵ the count of occurrences
answered 1 hour ago
AdámAdám
28.2k274201
edited 1 hour ago
answered 1 hour ago
AdámAdám
28.2k274201
answered 1 hour ago
AdámAdám
28.2k274201
answered 1 hour ago
AdámAdám
28.2k274201
28.2k274201
add a comment |
add a comment |
$begingroup$
R, 20 bytes
Is this what you are after? Uses table to count the occurrences of each of the scan input values. Tests if count is > 1 and sums the trues.
sum(table(scan())>1)
Try it online!
answered 1 hour ago
MickyTMickyT
10k21535
$endgroup$
add a comment |
$begingroup$
R, 20 bytes
Is this what you are after? Uses table to count the occurrences of each of the scan input values. Tests if count is > 1 and sums the trues.
sum(table(scan())>1)
Try it online!
answered 1 hour ago
MickyTMickyT
10k21535
$endgroup$
add a comment |
$begingroup$
R, 20 bytes
Is this what you are after? Uses table to count the occurrences of each of the scan input values. Tests if count is > 1 and sums the trues.
sum(table(scan())>1)
Try it online!
answered 1 hour ago
MickyTMickyT
10k21535
$endgroup$
R, 20 bytes
Is this what you are after? Uses table to count the occurrences of each of the scan input values. Tests if count is > 1 and sums the trues.
sum(table(scan())>1)
Try it online!
answered 1 hour ago
MickyTMickyT
10k21535
answered 1 hour ago
MickyTMickyT
10k21535
answered 1 hour ago
MickyTMickyT
10k21535
answered 1 hour ago
MickyTMickyT
10k21535
10k21535
add a comment |
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 40 bytes
n=>n.GroupBy(c=>c).Where(x=>x.Count()>1)
Outputs by an IEnumerable<IGrouping<int,int>> with the relevant elements stored in the key.
Try it online!
answered 45 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,120119
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 40 bytes
n=>n.GroupBy(c=>c).Where(x=>x.Count()>1)
Outputs by an IEnumerable<IGrouping<int,int>> with the relevant elements stored in the key.
Try it online!
answered 45 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,120119
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 40 bytes
n=>n.GroupBy(c=>c).Where(x=>x.Count()>1)
Outputs by an IEnumerable<IGrouping<int,int>> with the relevant elements stored in the key.
Try it online!
answered 45 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,120119
$endgroup$
C# (Visual C# Interactive Compiler), 40 bytes
n=>n.GroupBy(c=>c).Where(x=>x.Count()>1)
Outputs by an IEnumerable<IGrouping<int,int>> with the relevant elements stored in the key.
Try it online!
answered 45 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,120119
answered 45 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,120119
answered 45 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,120119
answered 45 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,120119
1,120119
add a comment |
add a comment |
$begingroup$
Bash + coreutils, 18
sort|uniq -d|wc -l
Try it online!
answered 29 mins ago
Digital TraumaDigital Trauma
59.1k787224
$endgroup$
add a comment |
$begingroup$
Bash + coreutils, 18
sort|uniq -d|wc -l
Try it online!
answered 29 mins ago
Digital TraumaDigital Trauma
59.1k787224
$endgroup$
add a comment |
$begingroup$
Bash + coreutils, 18
sort|uniq -d|wc -l
Try it online!
answered 29 mins ago
Digital TraumaDigital Trauma
59.1k787224
$endgroup$
Bash + coreutils, 18
sort|uniq -d|wc -l
Try it online!
answered 29 mins ago
Digital TraumaDigital Trauma
59.1k787224
answered 29 mins ago
Digital TraumaDigital Trauma
59.1k787224
answered 29 mins ago
Digital TraumaDigital Trauma
59.1k787224
answered 29 mins ago
Digital TraumaDigital Trauma
59.1k787224
59.1k787224
add a comment |
add a comment |
$begingroup$
Haskell, 47 bytes
f[]=0
f(a:b)|x<-filter(/=a)b,x/=b=1+f x|1>0=f b
Try it online!
This is the naïve approach. There is likely something that could be done to improve this.
f[]=0
We return 0 for the empty list
f(a:b)
In the case of a non-empty list starting with a and then b.
|x<-filter(/=a)b,x/=b=1+f x
If filtering a out of b is different from b (that is a is in b) then we return 1 more than f applied to b with the as filtered out.
|1>0=f b
If filtering as doesn't change b then we just run f across the rest.
Here is another similar approach that has the same length:
f[]=0
f(a:b)|elem a b=1+f(filter(/=a)b)|1>0=f b
Try it online!
answered 19 mins ago
Sriotchilism O'ZaicSriotchilism O'Zaic
35.1k10159369
$endgroup$
add a comment |
$begingroup$
Haskell, 47 bytes
f[]=0
f(a:b)|x<-filter(/=a)b,x/=b=1+f x|1>0=f b
Try it online!
This is the naïve approach. There is likely something that could be done to improve this.
f[]=0
We return 0 for the empty list
f(a:b)
In the case of a non-empty list starting with a and then b.
|x<-filter(/=a)b,x/=b=1+f x
If filtering a out of b is different from b (that is a is in b) then we return 1 more than f applied to b with the as filtered out.
|1>0=f b
If filtering as doesn't change b then we just run f across the rest.
Here is another similar approach that has the same length:
f[]=0
f(a:b)|elem a b=1+f(filter(/=a)b)|1>0=f b
Try it online!
answered 19 mins ago
Sriotchilism O'ZaicSriotchilism O'Zaic
35.1k10159369
$endgroup$
add a comment |
$begingroup$
Haskell, 47 bytes
f[]=0
f(a:b)|x<-filter(/=a)b,x/=b=1+f x|1>0=f b
Try it online!
This is the naïve approach. There is likely something that could be done to improve this.
f[]=0
We return 0 for the empty list
f(a:b)
In the case of a non-empty list starting with a and then b.
|x<-filter(/=a)b,x/=b=1+f x
If filtering a out of b is different from b (that is a is in b) then we return 1 more than f applied to b with the as filtered out.
|1>0=f b
If filtering as doesn't change b then we just run f across the rest.
Here is another similar approach that has the same length:
f[]=0
f(a:b)|elem a b=1+f(filter(/=a)b)|1>0=f b
Try it online!
answered 19 mins ago
Sriotchilism O'ZaicSriotchilism O'Zaic
35.1k10159369
$endgroup$
Haskell, 47 bytes
f[]=0
f(a:b)|x<-filter(/=a)b,x/=b=1+f x|1>0=f b
Try it online!
This is the naïve approach. There is likely something that could be done to improve this.
f[]=0
We return 0 for the empty list
f(a:b)
In the case of a non-empty list starting with a and then b.
|x<-filter(/=a)b,x/=b=1+f x
If filtering a out of b is different from b (that is a is in b) then we return 1 more than f applied to b with the as filtered out.
|1>0=f b
If filtering as doesn't change b then we just run f across the rest.
Here is another similar approach that has the same length:
f[]=0
f(a:b)|elem a b=1+f(filter(/=a)b)|1>0=f b
Try it online!
answered 19 mins ago
Sriotchilism O'ZaicSriotchilism O'Zaic
35.1k10159369
answered 19 mins ago
Sriotchilism O'ZaicSriotchilism O'Zaic
35.1k10159369
answered 19 mins ago
Sriotchilism O'ZaicSriotchilism O'Zaic
35.1k10159369
answered 19 mins ago
Sriotchilism O'ZaicSriotchilism O'Zaic
35.1k10159369
35.1k10159369
add a comment |
add a comment |
$begingroup$
J, 11 bytes
1#.(1<#)/.~
Try it online!
Explanation:
/.~ group the list by itself
( ) for each group
1<# is the length greater than 1
1#. sum by base-1 conversion
answered 18 mins ago
Galen IvanovGalen Ivanov
6,80711033
$endgroup$
add a comment |
$begingroup$
J, 11 bytes
1#.(1<#)/.~
Try it online!
Explanation:
/.~ group the list by itself
( ) for each group
1<# is the length greater than 1
1#. sum by base-1 conversion
answered 18 mins ago
Galen IvanovGalen Ivanov
6,80711033
$endgroup$
add a comment |
$begingroup$
J, 11 bytes
1#.(1<#)/.~
Try it online!
Explanation:
/.~ group the list by itself
( ) for each group
1<# is the length greater than 1
1#. sum by base-1 conversion
answered 18 mins ago
Galen IvanovGalen Ivanov
6,80711033
$endgroup$
J, 11 bytes
1#.(1<#)/.~
Try it online!
Explanation:
/.~ group the list by itself
( ) for each group
1<# is the length greater than 1
1#. sum by base-1 conversion
answered 18 mins ago
Galen IvanovGalen Ivanov
6,80711033
answered 18 mins ago
Galen IvanovGalen Ivanov
6,80711033
answered 18 mins ago
Galen IvanovGalen Ivanov
6,80711033
answered 18 mins ago
Galen IvanovGalen Ivanov
6,80711033
6,80711033
add a comment |
add a comment |
$begingroup$
JavaScript (ES6), 40 bytes
a=>a.map(o=x=>n+=(o[x]=-~o[x])==2,n=0)|n
Try it online!
answered 10 mins ago
ArnauldArnauld
76.7k693322
$endgroup$
add a comment |
$begingroup$
JavaScript (ES6), 40 bytes
a=>a.map(o=x=>n+=(o[x]=-~o[x])==2,n=0)|n
Try it online!
answered 10 mins ago
ArnauldArnauld
76.7k693322
$endgroup$
add a comment |
$begingroup$
JavaScript (ES6), 40 bytes
a=>a.map(o=x=>n+=(o[x]=-~o[x])==2,n=0)|n
Try it online!
answered 10 mins ago
ArnauldArnauld
76.7k693322
$endgroup$
JavaScript (ES6), 40 bytes
a=>a.map(o=x=>n+=(o[x]=-~o[x])==2,n=0)|n
Try it online!
answered 10 mins ago
ArnauldArnauld
76.7k693322
answered 10 mins ago
ArnauldArnauld
76.7k693322
answered 10 mins ago
ArnauldArnauld
76.7k693322
answered 10 mins ago
ArnauldArnauld
76.7k693322
76.7k693322
add a comment |
add a comment |
$begingroup$
Python 3, 93 bytes
import collections as c;print(sum(filter(lambda x:x>1,[y-1 for y in c.Counter(i).values()])))
Try it online!
answered 7 mins ago
Mike ShlantaMike Shlanta
357310
$endgroup$
add a comment |
$begingroup$
Python 3, 93 bytes
import collections as c;print(sum(filter(lambda x:x>1,[y-1 for y in c.Counter(i).values()])))
Try it online!
answered 7 mins ago
Mike ShlantaMike Shlanta
357310
$endgroup$
add a comment |
$begingroup$
Python 3, 93 bytes
import collections as c;print(sum(filter(lambda x:x>1,[y-1 for y in c.Counter(i).values()])))
Try it online!
answered 7 mins ago
Mike ShlantaMike Shlanta
357310
$endgroup$
Python 3, 93 bytes
import collections as c;print(sum(filter(lambda x:x>1,[y-1 for y in c.Counter(i).values()])))
Try it online!
answered 7 mins ago
Mike ShlantaMike Shlanta
357310
answered 7 mins ago
Mike ShlantaMike Shlanta
357310
answered 7 mins ago
Mike ShlantaMike Shlanta
357310
answered 7 mins ago
Mike ShlantaMike Shlanta
357310
357310
add a comment |
add a comment |
$begingroup$
Ruby, 40 bytes
->a{a.select{|x|a.count(x)>1}.uniq.size}
Try it online!
answered 36 secs ago
Kirill L.Kirill L.
4,5251523
$endgroup$
add a comment |
$begingroup$
Ruby, 40 bytes
->a{a.select{|x|a.count(x)>1}.uniq.size}
Try it online!
answered 36 secs ago
Kirill L.Kirill L.
4,5251523
$endgroup$
add a comment |
$begingroup$
Ruby, 40 bytes
->a{a.select{|x|a.count(x)>1}.uniq.size}
Try it online!
answered 36 secs ago
Kirill L.Kirill L.
4,5251523
$endgroup$
Ruby, 40 bytes
->a{a.select{|x|a.count(x)>1}.uniq.size}
Try it online!
answered 36 secs ago
Kirill L.Kirill L.
4,5251523
answered 36 secs ago
Kirill L.Kirill L.
4,5251523
answered 36 secs ago
Kirill L.Kirill L.
4,5251523
answered 36 secs ago
Kirill L.Kirill L.
4,5251523
4,5251523
add a comment |
add a comment |
$begingroup$
05AB1E, 4 bytes
Ù¢≠O
Try it online!
or as a Test Suite
Explanation
O # sum
≠ # the false values
¢ # in the count
Ù # of unique digits in input
answered 5 mins ago
EmignaEmigna
46.5k432142
$endgroup$
add a comment |
$begingroup$
05AB1E, 4 bytes
Ù¢≠O
Try it online!
or as a Test Suite
Explanation
O # sum
≠ # the false values
¢ # in the count
Ù # of unique digits in input
answered 5 mins ago
EmignaEmigna
46.5k432142
$endgroup$
add a comment |
$begingroup$
05AB1E, 4 bytes
Ù¢≠O
Try it online!
or as a Test Suite
Explanation
O # sum
≠ # the false values
¢ # in the count
Ù # of unique digits in input
answered 5 mins ago
EmignaEmigna
46.5k432142
$endgroup$
05AB1E, 4 bytes
Ù¢≠O
Try it online!
or as a Test Suite
Explanation
O # sum
≠ # the false values
¢ # in the count
Ù # of unique digits in input
answered 5 mins ago
EmignaEmigna
46.5k432142
edited 12 secs ago
answered 5 mins ago
EmignaEmigna
46.5k432142
answered 5 mins ago
EmignaEmigna
46.5k432142
answered 5 mins ago
EmignaEmigna
46.5k432142
46.5k432142
add a comment |
add a comment |
jayko03 is a new contributor. Be nice, and check out our Code of Conduct.
jayko03 is a new contributor. Be nice, and check out our Code of Conduct.
jayko03 is a new contributor. Be nice, and check out our Code of Conduct.
jayko03 is a new contributor. Be nice, and check out our Code of Conduct.
If this is an answer to a challenge…
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$begingroup$
What do you mean by
Once it counts the repetition, don't count again? Also, since we want to find the repetition of a specific integer, how would we know which integer to search for if we are not given it? Lastly, the test cases are a bit confusing; which are output and which are input?$endgroup$
– Embodiment of Ignorance
1 hour ago
2
$begingroup$
I've edited this to try to make it a bit clearer. Is this what you intended? Also, please put answers in for those test cases.
$endgroup$
– Riker
1 hour ago
$begingroup$
You may want to add some edge cases, like lists with negative, large-magnitude, unique, or no integers.
$endgroup$
– Adám
1 hour ago
$begingroup$
I have added some answers to the test cases, sorry if I go them wrong
$endgroup$
– MickyT
27 mins ago