Check if a string has all of it's parenthesis closed - vb.net

Question

Is there a way to check if a string has all of it's parenthesis closed? So for example it would take as an argument a string like this:
dim ValidOne as string = "This is (good)"
dim ValidOne as string = "This (is (good))"
dim InvalidOne as string = "This is (bad))"
dim InvalidOne as string = "This is (bad"
dim InvalidOne as string = "This is bad)"
And return True or False depending on whether there is a valid number of closed parenthesis.
So it if the string had an open ( and it was not closed, or just a ) that was never opened, it would return false.
I think you can do something like +1 for each open ( and -1 for each ). The rule is that you must end with 0 at the end.

Answer 1

If you want a full versatile and customizable solution then here is my approach:
Output:
Snippet:
''' <summary>
''' Counts the closed and opened pair of chars inside a String.
''' </summary>
''' <param name="PairChars">The pair character.</param>
''' <param name="Input">The string where to count the pair characters.</param>
''' <returns>PairCharacter.</returns>
''' <exception cref="System.Exception">Index of 'PairChar' parameter is out of range.</exception>
Public Function CountPairOfChars(ByVal PairChars As KeyValuePair(Of Char, Char),
ByVal Input As String) As PairOfCharsInfo
If String.IsNullOrEmpty(Input) OrElse String.IsNullOrWhiteSpace(Input) Then
Throw New Exception("'Input' parameter cannot be an empty String.")
End If
Dim CharStack As New Stack(Of Integer)
Dim Result As New PairOfCharsInfo
With Result
.Input = Input
.Characters = New KeyValuePair(Of Char, Char)(PairChars.Key, PairChars.Value)
For i As Integer = 0 To Input.Length - 1
Select Case Input(i)
Case .Characters.Key
CharStack.Push(i)
.OpenedPairsIndex.Add(i)
.CountOpenedPairs += 1
Case .Characters.Value
Select Case CharStack.Count
Case Is = 0
.CountOpenedPairs += 1
.OpenedPairsIndex.Add(i)
Case Else
.CountClosedPairs += 1
.CountOpenedPairs -= 1
.ClosedPairsIndex.Add(Tuple.Create(Of Integer, Integer)(CharStack.Pop, i))
.OpenedPairsIndex.RemoveAt(.OpenedPairsIndex.Count - 1)
End Select '/ CharStack.Count
End Select '/ Input(i)
Next i
.StringHasClosedPairs = .CountClosedPairs <> 0
.StringHasOpenedPairs = .CountOpenedPairs <> 0
End With '/ Result
Return Result
End Function
''' <summary>
''' Stores info about closed and opened pairs of chars in a String.
''' </summary>
Public NotInheritable Class PairOfCharsInfo
''' <summary>
''' Indicates the input string.
''' </summary>
''' <value>The input string.</value>
Public Property Input As String = String.Empty
''' <summary>
''' Indicates the pair of characters.
''' </summary>
''' <value>The pair of characters.</value>
Public Property Characters As KeyValuePair(Of Char, Char) = Nothing
''' <summary>
''' Determines whether the input string contains closed pairs of character.
''' </summary>
''' <value>The closed pairs count.</value>
Public Property StringHasClosedPairs As Boolean = False
''' <summary>
''' Determines whether the input string contains opened pairs of character.
''' </summary>
''' <value>The closed pairs count.</value>
Public Property StringHasOpenedPairs As Boolean = False
''' <summary>
''' Indicates the total amount of closed pairs.
''' </summary>
''' <value>The closed pairs count.</value>
Public Property CountClosedPairs As Integer = 0
''' <summary>
''' Indicates the total amount of opened pairs.
''' </summary>
''' <value>The opened pairs count.</value>
Public Property CountOpenedPairs As Integer = 0
''' <summary>
''' Indicates the closed pairs index position in the string.
''' </summary>
''' <value>The closed pairs positions.</value>
Public Property ClosedPairsIndex As New List(Of Tuple(Of Integer, Integer))
''' <summary>
''' Indicates the opened pairs index position in the string.
''' </summary>
''' <value>The opened pairs positions.</value>
Public Property OpenedPairsIndex As New List(Of Integer)
End Class '/ PairOfCharsInfo
Example Usage:
( The same as I used for the output images above)
Private Sub Test() Handles MyBase.Shown
Dim Inputs As String() =
{
"(This) is (good)",
"This (is (good))",
"This is good",
"This is (bad))",
"This is (bad",
"This is bad)",
"This is bad)("
}
Dim PairChars As New KeyValuePair(Of Char, Char)("(", ")")
For Each s As String In Inputs
Dim Info As PairOfCharsInfo = Me.CountPairOfChars(PairChars, s)
Dim sb As New System.Text.StringBuilder
With sb
.AppendLine(String.Format("Input String: {0}", Info.Input))
.AppendLine(String.Format("Pair of Chars: {0}{1}", Info.Characters.Key, Info.Characters.Value))
.AppendLine()
.AppendLine(String.Format("String has closed pairs?: {0}", Info.StringHasClosedPairs))
.AppendLine(String.Format("String has opened pairs?: {0}", Info.StringHasOpenedPairs))
.AppendLine()
.AppendLine(String.Format("Closed Pairs Count: {0}", Info.CountClosedPairs))
.AppendLine(String.Format("Opened Pairs Count: {0}", Info.CountOpenedPairs))
.AppendLine()
.AppendLine("Closed Pairs Indexes:")
For Each Item As Tuple(Of Integer, Integer) In Info.ClosedPairsIndex
.AppendLine(String.Format("Start Index: {0}, End Index: {1}",
CStr(Item.Item1), CStr(Item.Item2)))
Next Item
.AppendLine()
.AppendLine(String.Format("Opened Pairs Indexes: {0}",
String.Join(", ", Info.OpenedPairsIndex)))
End With '/ sb
MessageBox.Show(sb.ToString, "Count Pair Characters Information",
MessageBoxButtons.OK, MessageBoxIcon.Information)
Next s
End Sub

Answer 2

Something like this should do it:
Public Shared Function TestStringParens(s As String) As Boolean
Dim Ret = 0
For Each C In s
If C = ")"c Then Ret -= 1
If C = "("c Then Ret += 1
'Bail earlier for a closed paren without a matching open
If Ret < 0 Then Return False
Next
Return Ret = 0
End Function
As your post and some other people have said just keep a counter around and walk the string. As #Drico said, any time a negative counter exists then we have a closed parentheses without a corresponding open.
You can test this with:
Dim Tests As New Dictionary(Of String, Boolean)
Tests.Add("This is (good)", True)
Tests.Add("This (is (good))", True)
Tests.Add("This is good", True)
Tests.Add("This is (bad))", False)
Tests.Add("This is (bad", False)
Tests.Add("This is bad)", False)
Tests.Add("This is bad)(", False)
For Each T In Tests
Console.WriteLine(TestStringParens(T.Key) = T.Value)
Next

Answer 3

Here is an example of how you use the mid function to loop through a string
Function checkPar(s As String) As Boolean
Dim i As Integer
Dim parCounter As Integer
parCounter = 0
For i = 1 To Len(s)
If Mid(s, i, 1) = "(" Then
parCounter = parCounter + 1
ElseIf Mid(s, i, 1) = ")" Then
parCounter = parCounter - 1
End If
If parCounter < 0 Then Exit For
Next
If parCounter <> 0 Then
checkPar = False
Else
checkPar = True
End If
End Function