Dada una string str que consta de alfabetos ingleses en minúsculas. La tarea es encontrar si hay algún carácter en la string cuya frecuencia sea igual a la suma de las frecuencias de otros caracteres de la string. Si tal carácter existe, imprima Sí , de lo contrario, imprima No.
Ejemplos:
Entrada: str = “hkklkwwwww”
Salida: Sí
frecuencia(w) = frecuencia(h) + frecuencia(k) + frecuencia(l)
4 = 1 + 2 + 1
4 = 4
Entrada: str = “geeksforgeeks”
Salida: No
Enfoque: si la longitud de la string es impar, el resultado siempre será No . En el caso de una string de longitud uniforme, calcule la frecuencia de cada uno de los caracteres de la string y para cualquier carácter si su frecuencia = la mitad de la longitud de la string, el resultado será Sí , de lo contrario , No.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns true if some character
// exists in the given string whose frequency
// is equal to the sum frequencies of
// other characters of the string
bool isFrequencyEqual(string str, int len)
{
// If string is of odd length
if (len % 2 == 1)
return false;
// To store the frequency of each
// character of the string
int i, freq[26] = { 0 };
// Update the frequencies of the characters
for (i = 0; i < len; i++)
freq[str[i] - 'a']++;
for (i = 0; i < 26; i++)
if (freq[i] == len / 2)
return true;
// No such character exists
return false;
}
// Driver code
int main()
{
string str = "geeksforgeeks";
int len = str.length();
if (isFrequencyEqual(str, len))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java implementation of the above approach.
class GFG
{
// Function that returns true if some character
// exists in the given string whose frequency
// is equal to the sum frequencies of
// other characters of the string
static boolean isFrequencyEqual(String str, int len)
{
// If string is of odd length
if (len % 2 == 1)
{
return false;
}
// To store the frequency of each
// character of the string
int i, freq[] = new int[26];
// Update the frequencies of the characters
for (i = 0; i < len; i++)
{
freq[str.charAt(i) - 'a']++;
}
for (i = 0; i < 26; i++)
{
if (freq[i] == len / 2)
{
return true;
}
}
// No such character exists
return false;
}
// Driver code
public static void main(String[] args)
{
String str = "geeksforgeeks";
int len = str.length();
if (isFrequencyEqual(str, len))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code contributed by Rajput-Ji
Python3
# Python3 implementation of the approach
# Function that returns true if some character
# exists in the given string whose frequency
# is equal to the sum frequencies of
# other characters of the string
def isFrequencyEqual(string, length):
# If string is of odd length
if length % 2 == 1:
return False
# To store the frequency of each
# character of the string
freq = [0] * 26
# Update the frequencies of
# the characters
for i in range(0, length):
freq[ord(string[i]) - ord('a')] += 1
for i in range(0, 26):
if freq[i] == length // 2:
return True
# No such character exists
return False
# Driver code
if __name__ == "__main__":
string = "geeksforgeeks"
length = len(string)
if isFrequencyEqual(string, length):
print("Yes")
else:
print("No")
# This code is contributed by Rituraj Jain
C#
// C# implementation of the above approach.
using System;
class GFG
{
// Function that returns true if some character
// exists in the given string whose frequency
// is equal to the sum frequencies of
// other characters of the string
static bool isFrequencyEqual(String str, int len)
{
// If string is of odd length
if (len % 2 == 1)
{
return false;
}
// To store the frequency of each
// character of the string
int i;
int []freq = new int[26];
// Update the frequencies of the characters
for (i = 0; i < len; i++)
{
freq[str[i] - 'a']++;
}
for (i = 0; i < 26; i++)
{
if (freq[i] == len / 2)
{
return true;
}
}
// No such character exists
return false;
}
// Driver code
public static void Main()
{
String str = "geeksforgeeks";
int len = str.Length;
if (isFrequencyEqual(str, len))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
/* This code contributed by PrinciRaj1992 */
PHP
<?php
// PHP implementation of the approach
// Function that returns true if some character
// exists in the given string whose frequency
// is equal to the sum frequencies of
// other characters of the string
function isFrequencyEqual($str, $len)
{
// If string is of odd length
if ($len % 2 == 1)
return false;
// To store the frequency of each
// character of the string
$freq = array();
for($i = 0; $i < 26 ; $i++)
$freq[$i] = 0;
// Update the frequencies of the characters
for($i = 0; $i < $len ; $i++)
$freq[ord($str[$i]) - 97]++;
for($i = 0; $i < 26 ; $i++)
if ($freq[$i] == $len / 2)
return true;
// No such character exists
return false;
}
// Driver code
$str = "geeksforgeeks";
$len = strlen($str);
if (isFrequencyEqual($str, $len))
echo "Yes";
else
echo "No";
// This code is contributed by ihritik
?>
Javascript
<script>
// Javascript implementation of the approach
// Function that returns true if some character
// exists in the given string whose frequency
// is equal to the sum frequencies of
// other characters of the string
function isFrequencyEqual(str, len)
{
// If string is of odd length
if (len % 2 == 1)
return false;
// To store the frequency of each
// character of the string
var i, freq=Array(26).fill(0);
// Update the frequencies of the characters
for (i = 0; i < len; i++)
freq[str[i] - 'a']++;
for (i = 0; i < 26; i++)
if (freq[i] == parseInt(len / 2))
return true;
// No such character exists
return false;
}
// Driver code
var str = "geeksforgeeks";
var len = str.length;
if (isFrequencyEqual(str, len))
document.write("Yes");
else
document.write("No");
// This code is contributed by noob2000.
</script>
Producción:
No
Complejidad de tiempo: O(len) donde len es la longitud de la string dada.