Character whose frequency is equal to the sum of frequencies of other characters of the given string

Given a string str consisting of lowercase English alphabets. The task is to find whether there is any character in the string whose frequency is equal to the sum of the frequencies of other characters of the string. If such character exists then print Yes else print No.

Examples:

Input: str = “hkklkwwwww”
Output: Yes
frequency(w) = frequency(h) + frequency(k) + frequency(l)
4 = 1 + 2 + 1
4 = 4



Input: str = “geeksforgeeks”
Output: No

Approach: If the length of the string is odd then the result will always be No. In case of even length string, calculate the frequency of each of the character of the string and for any character if it’s frequency = half of the length of the string then the result will be Yes else No.

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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 */

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PHP

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<?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
?>

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Output:

No

Time Complexity: O(len) where len is the length of the given string.



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