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# Check if frequency of each character is equal to its position in English Alphabet

• Last Updated : 13 May, 2021

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++

 `// 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

 ``
Output:
`Yes`

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

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