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 a 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 = “neveropen”
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++
// 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 stringbool 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 codeint main(){ string str = "neveropen"; 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 = "neveropen"; 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 = "neveropen" 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 = "neveropen"; 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 stringfunction 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 = "neveropen";$len = strlen($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 stringfunction 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 codevar str = "neveropen";var len = str.length;if (isFrequencyEqual(str, len)) document.write("Yes");else document.write("No");// This code is contributed by noob2000.</script> |
Output:
No
Time Complexity: O(len) where len is the length of the given string.
Auxiliary Space: O(26) ? O(1), no extra space is required, so it is a constant.
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