Check if the frequency of any character is more than half the length of the string
Given a string str, the task is to check if the frequency of any character is more than half the length of the given string. The characters can be lowercase or uppercase alphabets, digits and special characters.
Examples:
Input: str = “AAa*2AAAA”
Output: Yes
The frequency of ‘A’ is more than half the length of the string.Input: str = “abB@2a”
Output: No
Approach: The problem can be easily solved by using a frequency array of length 28 i.e. 256 as there are 256 different characters. Iterate through the string and increase the count of the character by one in the frequency array every time it is encountered. Finally, iterate through the frequency array to check if the frequency of any character is more than half the length of the string.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define MAXN 256 // Function that returns true if the frequency // of any character is more than half the // length of the string bool checkHalfFrequency(string str) { // Length of the string int L = str.size(); // Initialize a frequency array int fre[MAXN] = { 0 }; // Iterate the string and update the // frequency of each of the character for ( int i = 0; i < L; i++) fre[str[i]]++; // Iterate the frequency array for ( int i = 0; i < MAXN; i++) // If condition is satisfied if (fre[i] > L / 2) return true ; return false ; } // Driver code int main() { string str = "GeeksforGeeks" ; if (checkHalfFrequency(str)) cout << "Yes" ; else cout << "No" ; return 0; } |
Java
// Java implementation of the approach class GFG { static int MAXN = 256 ; // Function that returns true if the frequency // of any character is more than half the // length of the string static boolean checkHalfFrequency(String str) { // Length of the string int L = str.length(); // Initialize a frequency array int fre[] = new int [MAXN]; // Iterate the string and update the // frequency of each of the character for ( int i = 0 ; i < L; i++) { fre[str.charAt(i)]++; } // Iterate the frequency array for ( int i = 0 ; i < MAXN; i++) // If condition is satisfied { if (fre[i] > L / 2 ) { return true ; } } return false ; } // Driver code public static void main(String[] args) { String str = "GeeksforGeeks" ; if (checkHalfFrequency(str)) { System.out.println( "Yes" ); } else { System.out.println( "No" ); } } } /* This code contributed by PrinciRaj1992 */ |
Python3
# Python3 implementation of the approach MAXN = 256 # Function that returns true if the frequency # of any character is more than half the # length of the String def checkHalfFrequency( Str ): # Length of the String L = len ( Str ) # Initialize a frequency array fre = [ 0 for i in range (MAXN)] # Iterate the and update the # frequency of each of the character for i in range (L): fre[ ord ( Str [i])] + = 1 # Iterate the frequency array for i in range (MAXN): # If condition is satisfied if (fre[i] > L / / 2 ): return True return False # Driver code Str = "GeeksforGeeks" if (checkHalfFrequency( Str )): print ( "Yes" ) else : print ( "No" ) # This code is contributed by mohit kumar |
C#
// C# implementation of the approach using System; class GFG { static int MAXN = 256; // Function that returns true if the frequency // of any character is more than half the // length of the string static bool checkHalfFrequency( string str) { // Length of the string int L = str.Length; // Initialize a frequency array int [] fre = new int [MAXN]; // Iterate the string and update the // frequency of each of the character for ( int i = 0; i < L; i++) { fre[str[i]]++; } // Iterate the frequency array for ( int i = 0; i < MAXN; i++) // If condition is satisfied { if (fre[i] > L / 2) { return true ; } } return false ; } // Driver code public static void Main() { string str = "GeeksforGeeks" ; if (checkHalfFrequency(str)) { Console.WriteLine( "Yes" ); } else { Console.WriteLine( "No" ); } } } /* This code contributed by Code_Mech */ |
PHP
<?php // PHP implementation of the approach // Function that returns true if the frequency // of any character is more than half the // length of the string function checkHalfFrequency( $str ) { $MAXN = 256; // Length of the string $L = strlen ( $str ); // Initialize a frequency array $fre = array_fill (0, $MAXN , 0 ); // Iterate the string and update the // frequency of each of the character for ( $i = 0; $i < $L ; $i ++) { $fre [ord( $str [ $i ])]++; } // Iterate the frequency array for ( $i = 0; $i < $MAXN ; $i ++) // If condition is satisfied { if ( $fre [ $i ] > (int)( $L / 2)) { return true; } } return false; } // Driver code $str = "GeeksforGeeks" ; if (checkHalfFrequency( $str )) { echo ( "Yes" ); } else { echo ( "No" ); } // This code is contributed by Code_Mech |
Javascript
<script> // Javascript implementation of the approach var MAXN = 256 // Function that returns true if the frequency // of any character is more than half the // length of the string function checkHalfFrequency(str) { // Length of the string var L = str.length; // Initialize a frequency array var fre = Array(MAXN); // Iterate the string and update the // frequency of each of the character for ( var i = 0; i < L; i++) fre[str[i]]++; // Iterate the frequency array for ( var i = 0; i < MAXN; i++) // If condition is satisfied if (fre[i] > L / 2) return true ; return false ; } // Driver code var str = "GeeksforGeeks" ; if (checkHalfFrequency(str)) document.write( "Yes" ); else document.write( "No" ); // This code is contributed by itsok </script> |
No
Time Complexity: O(N), N = length of the string
Auxiliary Space: O(1)