Find first Unique Char in String
Given a string, find the first non-repeating character in it.
Example
i/p - Vikrant
o/p - None
i/p - VikrantVikrant
o/p - 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 17 mins ago
Jamal♦
30.3k11116226
asked 2 hours ago
vikrantvikrant
807
add a comment |
Given a string, find the first non-repeating character in it.
Example
i/p - Vikrant
o/p - None
i/p - VikrantVikrant
o/p - 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 17 mins ago
Jamal♦
30.3k11116226
asked 2 hours ago
vikrantvikrant
807
add a comment |
Given a string, find the first non-repeating character in it.
Example
i/p - Vikrant
o/p - None
i/p - VikrantVikrant
o/p - 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 17 mins ago
Jamal♦
30.3k11116226
asked 2 hours ago
vikrantvikrant
807
Given a string, find the first non-repeating character in it.
Example
i/p - Vikrant
o/p - None
i/p - VikrantVikrant
o/p - 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 17 mins ago
Jamal♦
30.3k11116226
asked 2 hours ago
vikrantvikrant
807
edited 17 mins ago
Jamal♦
30.3k11116226
asked 2 hours ago
vikrantvikrant
807
edited 17 mins ago
Jamal♦
30.3k11116226
edited 17 mins ago
Jamal♦
30.3k11116226
edited 17 mins ago
Jamal♦
30.3k11116226
30.3k11116226
asked 2 hours ago
vikrantvikrant
807
asked 2 hours ago
vikrantvikrant
807
asked 2 hours ago
vikrantvikrant
807
807
add a comment |
add a comment |
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