Dada una string str , la tarea es encontrar el recuento de todas las substrings de longitud cuatro cuyos caracteres se pueden reorganizar para formar la palabra «clap» .
Ejemplos:
Entrada: str = “clapc”
Salida: 2
“clap” y “lapc” son las substrings requeridas
Entrada: str = “abcd”
Salida: 0
Enfoque: para cada substring de longitud cuatro, cuente las ocurrencias de los caracteres de la palabra «clap» . Si cada carácter tiene la ocurrencia exactamente uno en la substring, incremente el conteo . Imprime el conteo al final.
A continuación se muestra la implementación del enfoque anterior:
C++
// CPP implementation of the approach
#include<bits/stdc++.h>
using namespace std;
// Function to return the count of
// required occurrence
int countOcc(string s)
{
// To store the count of occurrences
int cnt = 0;
// Check first four characters from ith position
for (int i = 0; i < s.length() - 3; i++)
{
// Variables for counting the required characters
int c = 0, l = 0, a = 0, p = 0;
// Check the four contiguous characters which
// can be reordered to form 'clap'
for (int j = i; j < i + 4; j++)
{
switch (s[j])
{
case 'c':
c++;
break;
case 'l':
l++;
break;
case 'a':
a++;
break;
case 'p':
p++;
break;
}
}
// If all four contiguous characters are present
// then increment cnt variable
if (c == 1 && l == 1 && a == 1 && p == 1)
cnt++;
}
return cnt;
}
// Driver code
int main()
{
string s = "clapc";
transform(s.begin(), s.end(), s.begin(), ::tolower);
cout << (countOcc(s));
}
// This code is contributed by
// Surendra_Gangwar
Java
// Java implementation of the approach
class GFG {
// Function to return the count of
// required occurrence
static int countOcc(String s)
{
// To store the count of occurrences
int cnt = 0;
// Check first four characters from ith position
for (int i = 0; i < s.length() - 3; i++) {
// Variables for counting the required characters
int c = 0, l = 0, a = 0, p = 0;
// Check the four contiguous characters which
// can be reordered to form 'clap'
for (int j = i; j < i + 4; j++) {
switch (s.charAt(j)) {
case 'c':
c++;
break;
case 'l':
l++;
break;
case 'a':
a++;
break;
case 'p':
p++;
break;
}
}
// If all four contiguous characters are present
// then increment cnt variable
if (c == 1 && l == 1 && a == 1 && p == 1)
cnt++;
}
return cnt;
}
// Driver code
public static void main(String args[])
{
String s = "clapc";
System.out.print(countOcc(s.toLowerCase()));
}
}
Python3
# Python3 implementation of the approach # Function to return the count of # required occurrence def countOcc(s): # To store the count of occurrences cnt = 0 # Check first four characters from ith position for i in range(0, len(s) - 3): # Variables for counting the required characters c, l, a, p = 0, 0, 0, 0 # Check the four contiguous characters # which can be reordered to form 'clap' for j in range(i, i + 4): if s[j] == 'c': c += 1 elif s[j] == 'l': l += 1 elif s[j] == 'a': a += 1 elif s[j] == 'p': p += 1 # If all four contiguous characters are # present then increment cnt variable if c == 1 and l == 1 and a == 1 and p == 1: cnt += 1 return cnt # Driver code if __name__ == "__main__": s = "clapc" print(countOcc(s.lower())) # This code is contributed by Rituraj Jain
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count of
// required occurrence
static int countOcc(string s)
{
// To store the count of occurrences
int cnt = 0;
// Check first four characters
// from ith position
for (int i = 0; i < s.Length - 3; i++)
{
// Variables for counting the
// required characters
int c = 0, l = 0, a = 0, p = 0;
// Check the four contiguous characters
// which can be reordered to form 'clap'
for (int j = i; j < i + 4; j++)
{
switch (s[j])
{
case 'c':
c++;
break;
case 'l':
l++;
break;
case 'a':
a++;
break;
case 'p':
p++;
break;
}
}
// If all four contiguous characters are
// present then increment cnt variable
if (c == 1 && l == 1 && a == 1 && p == 1)
cnt++;
}
return cnt;
}
// Driver code
public static void Main()
{
string s = "clapc";
Console.Write(countOcc(s.ToLower()));
}
}
// This code is contributed by Akanksha Rai
PHP
<?php
// PHP implementation of the approach
// Function to return the count of
// required occurrence
function countOcc($s)
{
// To store the count of occurrences
$cnt = 0;
// Check first four characters
// from ith position
for ($i = 0; $i < strlen($s) - 3; $i++)
{
// Variables for counting the
// required characters
$c = 0; $l = 0; $a = 0; $p = 0;
// Check the four contiguous characters
// which can be reordered to form 'clap'
for ($j = $i; $j < $i + 4; $j++)
{
switch ($s[$j])
{
case 'c':
$c++;
break;
case 'l':
$l++;
break;
case 'a':
$a++;
break;
case 'p':
$p++;
break;
}
}
// If all four contiguous characters are present
// then increment cnt variable
if ($c == 1 && $l == 1 &&
$a == 1 && $p == 1)
$cnt++;
}
return $cnt;
}
// Driver code
$s = "clapc";
echo countOcc(strtolower($s));
// This code is contributed by Ryuga
?>
Javascript
<script>
// Javascript implementation of the approach
// Function to return the count of
// required occurrence
function countOcc(s)
{
// To store the count of occurrences
var cnt = 0;
// Check first four characters from ith position
for (var i = 0; i < s.length - 3; i++)
{
// Variables for counting the required characters
var c = 0, l = 0, a = 0, p = 0;
// Check the four contiguous characters which
// can be reordered to form 'clap'
for (var j = i; j < i + 4; j++)
{
switch (s[j])
{
case 'c':
c++;
break;
case 'l':
l++;
break;
case 'a':
a++;
break;
case 'p':
p++;
break;
}
}
// If all four contiguous characters are present
// then increment cnt variable
if (c == 1 && l == 1 && a == 1 && p == 1)
cnt++;
}
return cnt;
}
// Driver code
var s = "clapc";
s = s.toLowerCase();
document.write(countOcc(s));
</script>
Producción:
2
Publicación traducida automáticamente
Artículo escrito por facebookruppal y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA