Check if frequency of each character is equal to its position in English Alphabet
Given string str of lowercase alphabets, the task is to check if the frequency of each distinct characters in the string equals to its position in the English Alphabet. If valid, then print “Yes”, else print “No”.
Examples:
Input: str = “abbcccdddd”
Output: Yes
Explanation:
Since frequency of each distinct character is equals to its position in English Alphabet, i.e.
F(a) = 1,
F(b) = 2,
F(c) = 3, and
F(d) = 4
Hence the output is Yes.Input: str = “w3wiki”
Output: No
Approach:
- Store the frequency of each character in an array of 26, for hashing purpose.
- Now traverse the hash array and check if the frequency of each character at an index i is equal to (i + 1) or not.
- If yes, then print “Yes”, Else print “No”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include "bits/stdc++.h" using namespace std; bool checkValidString(string str) { // Initialise frequency array int freq[26] = { 0 }; // Traverse the string for ( int i = 0; str[i]; i++) { // Update the frequency freq[str[i] - 'a' ]++; } // Check for valid string for ( int i = 0; i < 26; i++) { // If frequency is non-zero if (freq[i] != 0) { // If freq is not equals // to (i+1), then return // false if (freq[i] != i + 1) { return false ; } } } // Return true; return true ; } // Driver Code int main() { // Given string str string str = "abbcccdddd" ; if (checkValidString(str)) cout << "Yes" ; else cout << "No" ; return 0; } |
Java
// Java program for the above approach class GFG{ static boolean checkValidString(String str) { // Initialise frequency array int freq[] = new int [ 26 ]; // Traverse the String for ( int i = 0 ; i < str.length(); i++) { // Update the frequency freq[str.charAt(i) - 'a' ]++; } // Check for valid String for ( int i = 0 ; i < 26 ; i++) { // If frequency is non-zero if (freq[i] != 0 ) { // If freq is not equals // to (i+1), then return // false if (freq[i] != i + 1 ) { return false ; } } } // Return true; return true ; } // Driver Code public static void main(String[] args) { // Given String str String str = "abbcccdddd" ; if (checkValidString(str)) { System.out.print( "Yes" ); } else { System.out.print( "No" ); } } } // This code is contributed by sapnasingh4991 |
Python3
# Python3 program for the # above approach def checkValidString( str ): # Initialise frequency array freq = [ 0 for i in range ( 26 )] # Traverse the string for i in range ( len ( str )): # Update the frequency freq[ ord ( str [i]) - ord ( 'a' )] + = 1 # Check for valid string for i in range ( 26 ): # If frequency is non-zero if (freq[i] ! = 0 ): # If freq is not equals # to (i+1), then return # false if (freq[i] ! = i + 1 ): return False # Return true return True # Driver Code # Given string str str = "abbcccdddd" if (checkValidString( str )): print ( "Yes" ) else : print ( "No" ) # This code is contributed by avanitrachhadiya2155 |
C#
// C# program for the above approach using System; class GFG{ static bool checkValidString(String str) { // Initialise frequency array int []freq = new int [26]; // Traverse the String for ( int i = 0; i < str.Length; i++) { // Update the frequency freq[str[i] - 'a' ]++; } // Check for valid String for ( int i = 0; i < 26; i++) { // If frequency is non-zero if (freq[i] != 0) { // If freq is not equals // to (i+1), then return // false if (freq[i] != i + 1) { return false ; } } } // Return true; return true ; } // Driver Code public static void Main(String[] args) { // Given String str String str = "abbcccdddd" ; if (checkValidString(str)) { Console.Write( "Yes" ); } else { Console.Write( "No" ); } } } // This code is contributed by sapnasingh4991 |
Javascript
<script> // Javascript program for the above approach function checkValidString(str) { // Initialise frequency array var freq = new Array(26).fill(0); // Traverse the String for ( var i = 0; i < str.length; i++) { // Update the frequency freq[str[i].charCodeAt(0) - "a" .charCodeAt(0)]++; } // Check for valid String for ( var i = 0; i < 26; i++) { // If frequency is non-zero if (freq[i] !== 0) { // If freq is not equals // to (i+1), then return // false if (freq[i] !== i + 1) { return false ; } } } // Return true; return true ; } // Driver Code // Given String str var str = "abbcccdddd" ; if (checkValidString(str)) { document.write( "Yes" ); } else { document.write( "No" ); } // This code is contributed by rdtank </script> |
Yes
Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(26)
Method: Using Map
Algorithm :
- Create a HashMap map with character and integer as key and value pairs respectively.
- Loop through each character of the string str and count the frequency.
- Loop through each character in the string str again and check if its frequency is equal to its position in the English alphabet.
- If any character fails to satisfy the condition, set the boolean variable isEqual to false and break out of the loop.
- Print the result based on the value of isEqual.
C++
#include <iostream> #include <unordered_map> using namespace std; bool check(string str) { unordered_map< char , int > map; // Count frequency of each character in the string for ( char c : str) { if (map.find(c) != map.end()) { map++; } else { map = 1; } } // Check if frequency of each character is equal to its position in English alphabet bool isEqual = true ; for ( char c : str) { int frequency = map; int position = c - 'a' + 1; if (frequency != position) { isEqual = false ; break ; } } return isEqual; } int main() { string str = "abbcccdddd" ; bool ans = check(str); if (ans) { cout << "Yes" << endl; } else { cout << "No" << endl; } return 0; } //This code is contributed by aeroabrar_31 |
Java
import java.util.*; public class Main { public static void main(String[] args) { String str = "abbcccdddd" ; boolean ans = check(str); if (ans) { System.out.println( "Yes" ); } else { System.out.println( "No" ); } } public static boolean check(String str) { Map<Character, Integer> map = new HashMap<>(); // Count frequency of each character in the string for ( char c : str.toCharArray()) { if (map.containsKey(c)) { map.put(c, map.get(c) + 1 ); } else { map.put(c, 1 ); } } // Check if frequency of each character is equal to // its position in English alphabet boolean isEqual = true ; for ( char c : str.toCharArray()) { int frequency = map.get(c); int position = c - 'a' + 1 ; if (frequency != position) { isEqual = false ; break ; } } return isEqual; } } //This code is contributed by aeroabrar_31 |
Python3
def check(s): char_count = {} # Use a dictionary to store character frequencies # Count the frequency of each character in the string for char in s: if char in char_count: char_count[char] + = 1 else : char_count[char] = 1 # Check if the frequency of each character is equal to its position in the English alphabet is_equal = True for char in s: frequency = char_count[char] position = ord (char) - ord ( 'a' ) + 1 if frequency ! = position: is_equal = False break return is_equal if __name__ = = "__main__" : input_str = "abbcccdddd" result = check(input_str) if result: print ( "Yes" ) else : print ( "No" ) #This code is contributed by aeroabrar_31 |
C#
using System; using System.Collections.Generic; public class GFG { public static void Main( string [] args) { string str = "abbcccdddd" ; bool ans = Check(str); if (ans) { Console.WriteLine( "Yes" ); } else { Console.WriteLine( "No" ); } } public static bool Check( string str) { Dictionary< char , int > map = new Dictionary< char , int >(); // Count frequency of each character in the string foreach ( char c in str) { if (map.ContainsKey(c)) { map++; } else { map = 1; } } // Check if frequency of each character is equal to // its position in English alphabet bool isEqual = true ; foreach ( char c in str) { int frequency = map; int position = c - 'a' + 1; if (frequency != position) { isEqual = false ; break ; } } return isEqual; } } //This code is contributed by aeroabrar_31 |
Javascript
function check(str) { const map = new Map(); // Count frequency of each character in the string for (const c of str) { if (map.has(c)) { map.set(c, map.get(c) + 1); } else { map.set(c, 1); } } // Check if frequency of each character is equal to its position in English alphabet let isEqual = true ; for (const c of str) { const frequency = map.get(c); const position = c.charCodeAt(0) - 'a' .charCodeAt(0) + 1; if (frequency !== position) { isEqual = false ; break ; } } return isEqual; } const str = "abbcccdddd" ; const ans = check(str); if (ans) { console.log( "Yes" ); } else { console.log( "No" ); } |
Yes
Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(1)
Note: The size of the map never exceeds 26 here as the lower case English alphabets are only 26, so the Auxiliary space is 1 (constant).
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