Dada la string str, la string de retorno verdadera sigue el patrón a n b n , es decir, tiene a seguidas de b de modo que el número de a y b es el mismo.
Ejemplos:
C++
// C++ program to check if a string is of
// the form a^nb^n.
#include <iostream>
using namespace std;
// Returns true str is of the form a^nb^n.
bool isAnBn(string str)
{
int n = str.length();
// After this loop 'i' has count of a's
int i;
for (i = 0; i < n; i++)
if (str[i] != 'a')
break;
// Since counts of a's and b's should
// be equal, a should appear exactly
// n/2 times
if (i * 2 != n)
return false;
// Rest of the characters must be all 'b'
int j;
for (j = i; j < n; j++)
if (str[j] != 'b')
return false;
return true;
}
// Driver code
int main()
{
string str = "abab";
// Function call
isAnBn(str) ? cout << "Yes" : cout << "No";
return 0;
}
Java
// Java program to check if a string is of
// the form a^nb^n.
import java.util.*;
import java.lang.*;
import java.io.*;
class CheckPattern {
public static boolean isAnBn(String s)
{
int l = s.length();
// Only even length strings will have same number of
// a's and b's
if (l % 2 == 1) {
return false;
}
// Set two pointers, one from the left and another
// from right
int i = 0;
int j = l - 1;
// Compare the characters till the center
while (i < j) {
if (s.charAt(i) != 'a' || s.charAt(j) != 'b') {
return false;
}
i++;
j--;
}
return true;
}
public static void main(String[] args)
throws java.lang.Exception
{
String s = "abab";
// Function call
boolean value = isAnBn(s);
if (value == true) {
System.out.println("Yes");
}
else {
System.out.println("No");
}
}
}
// Code contributed by Shivani Sanjay Shinde.
Python3
# Python 3program to check if a
# string is of the form a^nb^n.
# Returns true str is of the
# form a^nb^n.
def isAnBn(str):
n = len(str)
# After this loop 'i' has
# count of a's
for i in range(n):
if (str[i] != 'a'):
break
# Since counts of a's and b's should
# be equal, a should appear exactly
# n/2 times
if (i * 2 != n):
return False
# Rest of the characters must
# be all 'b'
for j in range(i, n):
if (str[j] != 'b'):
return False
return True
# Driver code
if __name__ == "__main__":
str = "abab"
print("Yes") if isAnBn(str) else print("No")
# This code is contributed
# by ChitraNayal
C#
// C# program to check if a string
// is of the form a^ nb ^ n.
using System;
class GFG {
// Function returns true str is of the form a^nb^n.
public static bool isAnBn(String s)
{
int l = s.Length;
// Only even length strings will have
// same number of a's and b's
if (l % 2 == 1) {
return false;
}
// Set two pointers, one from the
// left and another from right
int i = 0;
int j = l - 1;
// Compare the characters
// till the center
while (i < j) {
if (s[i] != 'a' || s[j] != 'b') {
return false;
}
i++;
j--;
}
return true;
}
// Driver Code
public static void Main()
{
String s = "abab";
// Function call
bool value = isAnBn(s);
if (value == true) {
Console.Write("Yes");
}
else {
Console.Write("No");
}
}
}
// This code is contributed by Nitin Mittal.
PHP
<?php
// PHP program to check if a string
// is of the form a^nb^n.
// Returns true str is of
// the form a^nb^n.
function isAnBn($str)
{
$n = strlen($str);
// After this loop 'i'
// has count of a's
$i;
for($i = 0; $i < $n; $i++)
if ($str[$i] != 'a')
break;
// Since counts of a's and b's should
// be equal, a should appear exactly
// n/2 times
if ($i * 2 != $n)
return false;
// Rest of the characters
// must be all 'b'
$j;
for($j = $i; $j < $n; $j++)
if ($str[$j] != 'b')
return false;
return true;
}
// Driver code
$str = "abab";
if(isAnBn($str))
echo "Yes";
else
echo "No";
// This code is contributed by nitin mittal.
?>
Javascript
<script>
// Javascript program to check if a string is of
// the form a^nb^n.
function isAnBn(s)
{
let l = s.length;
// Only even length strings will have same number of
// a's and b's
if (l % 2 == 1) {
return false;
}
// Set two pointers, one from the left and another
// from right
let i = 0;
let j = l - 1;
// Compare the characters till the center
while (i < j) {
if (s[i] != 'a' || s[j] != 'b') {
return false;
}
i++;
j--;
}
return true;
}
let s = "abab";
// Function call
let value = isAnBn(s);
if (value == true) {
document.write("Yes");
}
else {
document.write("No");
}
// This code is contributed by ab2127
</script>
C++
// C++ code to check a^nb^n
// pattern
#include <iostream>
using namespace std;
// Returns "Yes" str is of the form a^nb^n.
string isAnBn(string str)
{
int n = str.length();
if (n & 1)
return "No";
// check first half is 'a' and other half is full of 'b'
int i;
for (i = 0; i < n / 2; i++)
if (str[i] != 'a' || str[n - i - 1] != 'b')
return "No";
return "Yes";
}
// Driver code
int main()
{
string str = "ab";
// Function call
cout << isAnBn(str);
return 0;
}
Java
// Java code to check a^nb^n
// pattern
import java.io.*;
class GFG {
// Returns "Yes" str is of the form a^nb^n.
static String isAnBn(String str)
{
int n = str.length();
if ((n & 1) != 0)
return "No";
// check first half is 'a' and other half is full of 'b'
int i;
for (i = 0; i < n / 2; i++)
if (str.charAt(i) != 'a' || str.charAt(n - i - 1) != 'b')
return "No";
return "Yes";
}
// Driver code
public static void main (String[] args)
{
String str = "ab";
// Function call
System.out.println(isAnBn(str));
}
}
// This code is contributed by rag2127
Python3
# Python3 code to check # a^nb^n pattern def isanbn(str): n=len(str) # if length of str is odd return No if n&1: return "No" # check first half is 'a' and other half is full of 'b' for i in range(int(n/2)): if str[i]!='a' or str[n-i-1]!='b': return "No" return "Yes" # Driver code input_str = "ab" # Function call print(isanbn(input_str))
C#
// C# code to check a^nb^n
// pattern
using System;
public class GFG
{
// Returns "Yes" str is of the form a^nb^n.
static string isAnBn(string str)
{
int n = str.Length;
if ((n & 1) != 0)
return "No";
// check first half is 'a' and other half is full of 'b'
int i;
for (i = 0; i < n / 2; i++)
if (str[i] != 'a' || str[n-i-1] != 'b')
return "No";
return "Yes";
}
// Driver code
static public void Main ()
{
string str = "ab";
// Function call
Console.WriteLine(isAnBn(str));
}
}
// This code is contributed by avanitrachhadiya2155
Javascript
<script>
// Javascript code to check a^nb^n
// pattern
// Returns "Yes" str is of the form a^nb^n.
function isAnBn(str)
{
let n = str.length;
if ((n & 1) != 0)
return "No";
// check first half is 'a' and other half is full of 'b'
let i;
for (i = 0; i < n / 2; i++)
if (str[i] != 'a' || str[n - i - 1] != 'b')
return "No";
return "Yes";
}
// Driver code
let str = "ab";
// Function call
document.write(isAnBn(str));
// This code is contributed by unknown2108
</script>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA