Find first repeating Char in String
Given a string, find the first repeating character in it.
Examples:
firstUnique("Vikrant")→None
firstUnique("VikrantVikrant")→Some(V)
Scala implementation:
object FirstUniqueChar extends App {
def firstUnique(s: String): Option[Char] = {
val countMap = (s groupBy (c=>c)) mapValues(_.length)
def checkOccurence(s1: String ): Option[Char] = {
if (countMap(s1.head) > 1) Some(s1.head)
else if (s1.length == 1) None
else checkOccurence(s1.tail)
}
checkOccurence(s)
}
println(firstUnique("abcdebC"))
println(firstUnique("abcdef"))
}
I also have a followup question. What is the recommended way if I do not want to solve this problem with recursion? Instead of using the checkOccurence method I can traverse through the string and break when I find the first element with a count more than 1. But that will require a break, which is discouraged in Scala.
strings recursion interview-questions functional-programming scala
edited 6 hours ago
200_success
128k15152414
asked yesterday
vikrantvikrant
828
add a comment |
Given a string, find the first repeating character in it.
Examples:
firstUnique("Vikrant")→None
firstUnique("VikrantVikrant")→Some(V)
Scala implementation:
object FirstUniqueChar extends App {
def firstUnique(s: String): Option[Char] = {
val countMap = (s groupBy (c=>c)) mapValues(_.length)
def checkOccurence(s1: String ): Option[Char] = {
if (countMap(s1.head) > 1) Some(s1.head)
else if (s1.length == 1) None
else checkOccurence(s1.tail)
}
checkOccurence(s)
}
println(firstUnique("abcdebC"))
println(firstUnique("abcdef"))
}
I also have a followup question. What is the recommended way if I do not want to solve this problem with recursion? Instead of using the checkOccurence method I can traverse through the string and break when I find the first element with a count more than 1. But that will require a break, which is discouraged in Scala.
strings recursion interview-questions functional-programming scala
edited 6 hours ago
200_success
128k15152414
asked yesterday
vikrantvikrant
828
1
Well there seems to be a problem with either the problem-statement or with the examples.
– Heslacher
22 hours ago
add a comment |
Given a string, find the first repeating character in it.
Examples:
firstUnique("Vikrant")→None
firstUnique("VikrantVikrant")→Some(V)
Scala implementation:
object FirstUniqueChar extends App {
def firstUnique(s: String): Option[Char] = {
val countMap = (s groupBy (c=>c)) mapValues(_.length)
def checkOccurence(s1: String ): Option[Char] = {
if (countMap(s1.head) > 1) Some(s1.head)
else if (s1.length == 1) None
else checkOccurence(s1.tail)
}
checkOccurence(s)
}
println(firstUnique("abcdebC"))
println(firstUnique("abcdef"))
}
I also have a followup question. What is the recommended way if I do not want to solve this problem with recursion? Instead of using the checkOccurence method I can traverse through the string and break when I find the first element with a count more than 1. But that will require a break, which is discouraged in Scala.
strings recursion interview-questions functional-programming scala
edited 6 hours ago
200_success
128k15152414
asked yesterday
vikrantvikrant
828
Given a string, find the first repeating character in it.
Examples:
firstUnique("Vikrant")→None
firstUnique("VikrantVikrant")→Some(V)
Scala implementation:
object FirstUniqueChar extends App {
def firstUnique(s: String): Option[Char] = {
val countMap = (s groupBy (c=>c)) mapValues(_.length)
def checkOccurence(s1: String ): Option[Char] = {
if (countMap(s1.head) > 1) Some(s1.head)
else if (s1.length == 1) None
else checkOccurence(s1.tail)
}
checkOccurence(s)
}
println(firstUnique("abcdebC"))
println(firstUnique("abcdef"))
}
I also have a followup question. What is the recommended way if I do not want to solve this problem with recursion? Instead of using the checkOccurence method I can traverse through the string and break when I find the first element with a count more than 1. But that will require a break, which is discouraged in Scala.
strings recursion interview-questions functional-programming scala
strings recursion interview-questions functional-programming scala
edited 6 hours ago
200_success
128k15152414
asked yesterday
vikrantvikrant
828
edited 6 hours ago
200_success
128k15152414
asked yesterday
vikrantvikrant
828
edited 6 hours ago
200_success
128k15152414
edited 6 hours ago
200_success
128k15152414
edited 6 hours ago
200_success
128k15152414
128k15152414
asked yesterday
vikrantvikrant
828
asked yesterday
vikrantvikrant
828
asked yesterday
vikrantvikrant
828
828
1
Well there seems to be a problem with either the problem-statement or with the examples.
– Heslacher
22 hours ago
add a comment |
1
Well there seems to be a problem with either the problem-statement or with the examples.
– Heslacher
22 hours ago
1
1
Well there seems to be a problem with either the problem-statement or with the examples.
– Heslacher
22 hours ago
Well there seems to be a problem with either the problem-statement or with the examples.
– Heslacher
22 hours ago
add a comment |
1 Answer
1
active
oldest
votes
Your checkOccurrence(s) is just a clumsy way to write s.find(countMap(_) > 1).
You can significantly simplify the solution by taking advantage of .distinct.
def firstUnique(s: String): Option[Char] =
s.zipAll(s.distinct, 'u0000', 'u0000')
.collectFirst({ case ab if ab._1 != ab._2 => ab._1 })
answered 6 hours ago
200_success200_success
128k15152414
thanks for this wonderful answer.
– vikrant
2 hours ago
1
Slightly more concise syntax for a very nice answer:collectFirst{case (a,b) if a != b => a}
– jwvh
2 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your checkOccurrence(s) is just a clumsy way to write s.find(countMap(_) > 1).
You can significantly simplify the solution by taking advantage of .distinct.
def firstUnique(s: String): Option[Char] =
s.zipAll(s.distinct, 'u0000', 'u0000')
.collectFirst({ case ab if ab._1 != ab._2 => ab._1 })
answered 6 hours ago
200_success200_success
128k15152414
thanks for this wonderful answer.
– vikrant
2 hours ago
1
Slightly more concise syntax for a very nice answer:collectFirst{case (a,b) if a != b => a}
– jwvh
2 hours ago
add a comment |
Your checkOccurrence(s) is just a clumsy way to write s.find(countMap(_) > 1).
You can significantly simplify the solution by taking advantage of .distinct.
def firstUnique(s: String): Option[Char] =
s.zipAll(s.distinct, 'u0000', 'u0000')
.collectFirst({ case ab if ab._1 != ab._2 => ab._1 })
answered 6 hours ago
200_success200_success
128k15152414
thanks for this wonderful answer.
– vikrant
2 hours ago
1
Slightly more concise syntax for a very nice answer:collectFirst{case (a,b) if a != b => a}
– jwvh
2 hours ago
add a comment |
Your checkOccurrence(s) is just a clumsy way to write s.find(countMap(_) > 1).
You can significantly simplify the solution by taking advantage of .distinct.
def firstUnique(s: String): Option[Char] =
s.zipAll(s.distinct, 'u0000', 'u0000')
.collectFirst({ case ab if ab._1 != ab._2 => ab._1 })
answered 6 hours ago
200_success200_success
128k15152414
Your checkOccurrence(s) is just a clumsy way to write s.find(countMap(_) > 1).
You can significantly simplify the solution by taking advantage of .distinct.
def firstUnique(s: String): Option[Char] =
s.zipAll(s.distinct, 'u0000', 'u0000')
.collectFirst({ case ab if ab._1 != ab._2 => ab._1 })
answered 6 hours ago
200_success200_success
128k15152414
edited 6 hours ago
answered 6 hours ago
200_success200_success
128k15152414
answered 6 hours ago
200_success200_success
128k15152414
answered 6 hours ago
200_success200_success
128k15152414
128k15152414
thanks for this wonderful answer.
– vikrant
2 hours ago
1
Slightly more concise syntax for a very nice answer:collectFirst{case (a,b) if a != b => a}
– jwvh
2 hours ago
add a comment |
thanks for this wonderful answer.
– vikrant
2 hours ago
1
Slightly more concise syntax for a very nice answer:collectFirst{case (a,b) if a != b => a}
– jwvh
2 hours ago
thanks for this wonderful answer.
– vikrant
2 hours ago
thanks for this wonderful answer.
– vikrant
2 hours ago
1
1
Slightly more concise syntax for a very nice answer:
collectFirst{case (a,b) if a != b => a}– jwvh
2 hours ago
Slightly more concise syntax for a very nice answer:
collectFirst{case (a,b) if a != b => a}– jwvh
2 hours ago
add a comment |
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1
Well there seems to be a problem with either the problem-statement or with the examples.
– Heslacher
22 hours ago