Given a string S, the task is to design a Deterministic Finite Automata (DFA) for accepting the language L = C (A + B)+. If the given string is accepted by DFA, then print "Yes". Otherwise, print "No".
Examples:
Input: S = "CABABABAB"
Output: Yes
Explanation: The given string is of the form C(A + B)+ as the first character is C and it is followed by A or B.Input: S = "ABAB"
Output: No
Approach: The idea is to interpret the given language L = C (A + B)+ and for the regular expression of the form C(A + B)+, the following is the DFA State Transition Diagram:
Follow the steps below to solve the problem:
- If the given string is of length less than equal to 1, then print "No".
- If the first character is always C, then traverse the remaining string and check if any of the characters is A or B.
- If there exists any character other than A or B while traversing in the above step, then print "No".
- Otherwise, print "Yes".
Below is the implementation of the above approach:
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find whether the given
// string is Accepted by the DFA
void DFA(string str, int N)
{
// If n <= 1, then print No
if (N <= 1) {
cout << "No";
return;
}
// To count the matched characters
int count = 0;
// Check if the first character is C
if (str[0] == 'C') {
count++;
// Traverse the rest of string
for (int i = 1; i < N; i++) {
// If character is A or B,
// increment count by 1
if (str[i] == 'A' || str[i] == 'B')
count++;
else
break;
}
}
else {
// If the first character
// is not C, print -1
cout << "No";
return;
}
// If all characters matches
if (count == N)
cout << "Yes";
else
cout << "No";
}
// Driver Code
int main()
{
string str = "CAABBAAB";
int N = str.size();
DFA(str, N);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to find whether the given
// String is Accepted by the DFA
static void DFA(String str, int N)
{
// If n <= 1, then print No
if (N <= 1)
{
System.out.print("No");
return;
}
// To count the matched characters
int count = 0;
// Check if the first character is C
if (str.charAt(0) == 'C')
{
count++;
// Traverse the rest of String
for (int i = 1; i < N; i++)
{
// If character is A or B,
// increment count by 1
if (str.charAt(i) == 'A' ||
str.charAt(i) == 'B')
count++;
else
break;
}
}
else
{
// If the first character
// is not C, print -1
System.out.print("No");
return;
}
// If all characters matches
if (count == N)
System.out.print("Yes");
else
System.out.print("No");
}
// Driver Code
public static void main(String[] args)
{
String str = "CAABBAAB";
int N = str.length();
DFA(str, N);
}
}
// This code is contributed by 29AjayKumar
# Python3 program for the above approach
# Function to find whether the given
# is Accepted by the DFA
def DFA(str, N):
# If n <= 1, then print No
if (N <= 1):
print("No")
return
# To count the matched characters
count = 0
# Check if the first character is C
if (str[0] == 'C'):
count += 1
# Traverse the rest of string
for i in range(1, N):
# If character is A or B,
# increment count by 1
if (str[i] == 'A' or str[i] == 'B'):
count += 1
else:
break
else:
# If the first character
# is not C, print -1
print("No")
return
# If all characters matches
if (count == N):
print("Yes")
else:
print("No")
# Driver Code
if __name__ == '__main__':
str = "CAABBAAB"
N = len(str)
DFA(str, N)
# This code is contributed by mohit kumar 29.
// C# program for the above approach
using System;
class GFG
{
// Function to find whether the given
// String is Accepted by the DFA
static void DFA(string str, int N)
{
// If n <= 1, then print No
if (N <= 1)
{
Console.Write("No");
return;
}
// To count the matched characters
int count = 0;
// Check if the first character is C
if (str[0] == 'C') {
count++;
// Traverse the rest of String
for (int i = 1; i < N; i++) {
// If character is A or B,
// increment count by 1
if (str[i] == 'A'
|| str[i] == 'B')
count++;
else
break;
}
}
else {
// If the first character
// is not C, print -1
Console.Write("No");
return;
}
// If all characters matches
if (count == N)
Console.Write("Yes");
else
Console.Write("No");
}
// Driver Code
static public void Main()
{
string str = "CAABBAAB";
int N = str.Length;
DFA(str, N);
}
}
// This code is contributed by Dharanendra L V
<script>
// Javascript program for the above approach
// Function to find whether the given
// String is Accepted by the DFA
function DFA(str,N) {
// If n <= 1, then print No
if (N <= 1)
{
document.write("No");
return;
}
// To count the matched characters
let count = 0;
// Check if the first character is C
if (str[0] == 'C')
{
count++;
// Traverse the rest of String
for (let i = 1; i < N; i++)
{
// If character is A or B,
// increment count by 1
if (str[i] == 'A' ||
str[i] == 'B')
count++;
else
break;
}
}
else
{
// If the first character
// is not C, print -1
document.write("No");
return;
}
// If all characters matches
if (count == N)
document.write("Yes");
else
document.write("No");
}
// Driver Code
let str = "CAABBAAB";
let N = str.length;
DFA(str, N);
// This code is contributed by patel2127
</script>
Output:
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
