Dada la string str , la tarea es verificar si cada grupo de aes consecutivos es seguido por un grupo de bs consecutivos de la misma longitud. Si la condición es verdadera para cada grupo, imprima 1 ; de lo contrario, imprima 0 .
Ejemplos:
Entrada: str = “ababaabb”
Salida: 1
ab, ab, aabb. Todos los grupos son válidos
Entrada: str = “aabbabb”
Salida: 0
aabb, abb (Una sola ‘a’ seguida de 2 ‘b’)
Acercarse:
- Por cada a en la string, incremente el conteo .
- A partir de la primera b , disminuya la cuenta para cada b .
- Si al final del ciclo anterior, count != 0 , devuelve false .
- De lo contrario, repita los dos primeros pasos para el resto de la string.
- Devuelve verdadero si la condición se cumple para todos los ciclos, de lo contrario imprime 0 .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to match whether there are always n consecutive b's
// followed by n consecutive a's throughout the string
int matchPattern(string s)
{
int count = 0;
int n = s.length();
// Traverse through the string
int i = 0;
while (i < n) {
// Count a's in current segment
while (i < n && s[i] == 'a') {
count++;
i++;
}
// Count b's in current segment
while (i < n && s[i] == 'b') {
count--;
i++;
}
// If both counts are not same.
if (count != 0)
return false;
}
return true;
}
// Driver code
int main()
{
string s = "bb";
if (matchPattern(s) == true)
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java implementation of the above approach
public class GFG{
// Function to match whether there are always n consecutive b's
// followed by n consecutive a's throughout the string
static boolean matchPattern(String s)
{
int count = 0;
int n = s.length();
// Traverse through the string
int i = 0;
while (i < n) {
// Count a's in current segment
while (i < n && s.charAt(i) == 'a') {
count++;
i++;
}
// Count b's in current segment
while (i < n && s.charAt(i) == 'b') {
count--;
i++;
}
// If both counts are not same.
if (count != 0)
return false;
}
return true;
}
// Driver code
public static void main(String []args)
{
String s = "bb";
if (matchPattern(s) == true)
System.out.println("Yes");
else
System.out.println("No");
}
// This code is contributed by Ryuga
}
Python3
# Python 3 implementation of the approach
# Function to match whether there are
# always n consecutive b's followed by
# n consecutive a's throughout the string
def matchPattern(s):
count = 0;
n = len(s);
# Traverse through the string
i = 0;
while (i < n) :
# Count a's in current segment
while (i < n and s[i] == 'a'):
count += 1 ;
i =+ 1;
# Count b's in current segment
while (i < n and s[i] == 'b'):
count -= 1 ;
i += 1;
# If both counts are not same.
if (count != 0):
return False;
return True;
# Driver code
s = "bb";
if (matchPattern(s) == True):
print("Yes");
else:
print("No");
# This code is contributed
# by Akanksha Rai
C#
// C# implementation of the above approach
using System;
public class GFG{
// Function to match whether there are always n consecutive b's
// followed by n consecutive a's throughout the string
static bool matchPattern(string s)
{
int count = 0;
int n = s.Length;
// Traverse through the string
int i = 0;
while (i < n) {
// Count a's in current segment
while (i < n && s[i] == 'a') {
count++;
i++;
}
// Count b's in current segment
while (i < n && s[i] == 'b') {
count--;
i++;
}
// If both counts are not same.
if (count != 0)
return false;
}
return true;
}
// Driver code
public static void Main()
{
string s = "bb";
if (matchPattern(s) == true)
Console.Write("Yes");
else
Console.Write("No");
}
}
PHP
<?php
//PHP implementation of the approach
// Function to match whether there are always n consecutive b's
// followed by n consecutive a's throughout the string
function matchPattern($s)
{
$count = 0;
$n = strlen($s);
// Traverse through the string
$i = 0;
while ($i < $n) {
// Count a's in current segment
while ($i < $n && $s[$i] == 'a') {
$count++;
$i++;
}
// Count b's in current segment
while ($i < $n && $s[$i] == 'b') {
$count--;
$i++;
}
// If both counts are not same.
if ($count != 0)
return false;
}
return true;
}
// Driver code
$s = "bb";
if (matchPattern($s) == true)
echo "Yes";
else
echo "No";
// This code is contributed by ajit
?>
Javascript
<script>
// Javascript implementation of
// the above approach
// Function to match whether there are
// always n consecutive b's
// followed by n consecutive a's
// throughout the string
function matchPattern(s)
{
let count = 0;
let n = s.length;
// Traverse through the string
let i = 0;
while (i < n)
{
// Count a's in current segment
while (i < n && s[i] == 'a')
{
count++;
i++;
}
// Count b's in current segment
while (i < n && s[i] == 'b')
{
count--;
i++;
}
// If both counts are not same.
if (count != 0)
return false;
}
return true;
}
let s = "bb";
if (matchPattern(s) == true)
document.write("Yes");
else
document.write("No");
</script>
Producción:
No
Complejidad de tiempo : O( | s | ) ,donde | s | es la longitud de la string dada s.
Complejidad espacial: O (1), ya que no estamos usando ningún espacio adicional
Publicación traducida automáticamente
Artículo escrito por Devyansh_Ojha y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA