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Check if frequency of each character is equal to its position in English Alphabet
  • Last Updated : 16 Oct, 2020

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 = “geeksforgeeks” 
Output: No 

Approach: 

  1. Store the frequency of each character in an array of 26, for hashing purpose.
  2. Now traverse the hash array and check if the frequency of each character at an index i is equal to (i + 1) or not.
  3. If yes, then print “Yes”, Else print “No”.

Below is the implementation of the above approach: 
 

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 

Yes


 

Time Complexity: O(N), where N is the length of the string. 
Auxiliary Space: O(26)

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