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 = “geeksforgeeks”
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 Codeint main(){ // Given string str string str = "abbcccdddd"; if (checkValidString(str)) cout << "Yes"; else cout << "No"; return 0;} |
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Java
// Java program for the above approachclass 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 Codepublic 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
# Python3 program for the # above approachdef 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#
// C# program for the above approachusing 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 Codepublic 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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