Given a string S consisting of N lowercase alphabets, the task is to remove the characters from the string whose frequency is not a power of 2 and then sort the string in ascending order.
Examples:
Input: S = “aaacbb”
Output: bbc
Explanation: The frequencies of ‘a’, ‘b’, and ‘c’ in the given string are 3, 2, 1. Only the character ‘a’ has frequency (= 3) which is not a power of 2. Therefore, removing ‘a’ from the string S modifies it to “cbb”. Therefore, the modified string after sorting is “bbc”.Input: S = “neveropen”
Output: eeeefggkkorss
Approach: The given problem can be solved by using Hashing. Follow the steps below to solve the given problem:
- Initialize an auxiliary array of size 26, say freq[], to store the frequency of each character in the string S such that index 0 represents character ‘a’, 1 represents the character ‘b’, and so on.
- Traverse the given string S and increment the frequency of each character S[i] by 1 in the array freq[].
- Traverse the array freq[] and if the value of freq[i] is a power of 2, then print the character (‘a’ + i), freq[i] number of times.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to remove all the characters// from a string that whose frequencies// are not equal to a perfect power of 2void countFrequency(string S, int N){ // Stores the frequency of // each character in S int freq[26] = { 0 }; // Iterate over characters of string for (int i = 0; i < N; i++) { // Update frequency of current // character in the array freq[] freq[S[i] - 'a']++; } // Traverse the array freq[] for (int i = 0; i < 26; i++) { // Check if the i-th letter // is absent from string S if (freq[i] == 0) continue; // Calculate log of frequency // of the current character // in the string S int lg = log2(freq[i]); // Calculate power of 2 of lg int a = pow(2, lg); // Check if freq[i] // is a power of 2 if (a == freq[i]) { // Print letter i + 'a' // freq[i] times while (freq[i]--) cout << (char)(i + 'a'); } }}// Driver Codeint main(){ string S = "aaacbb"; int N = S.size(); countFrequency(S, N); return 0;} |
Java
// Java program for the above approachclass GFG{ // Function to remove all the characters// from a string that whose frequencies// are not equal to a perfect power of 2static void countFrequency(String S, int N){ // Stores the frequency of // each character in S int []freq = new int[26]; //Array.Clear(freq, 0, freq.Length); // Iterate over characters of string for(int i = 0; i < N; i++) { // Update frequency of current // character in the array freq[] freq[(int)S.charAt(i) - 'a'] += 1; } // Traverse the array freq[] for(int i = 0; i < 26; i++) { // Check if the i-th letter // is absent from string S if (freq[i] == 0) continue; // Calculate log of frequency // of the current character // in the string S int lg = (int)(Math.log(freq[i]) / Math.log(2)); // Calculate power of 2 of lg int a = (int)Math.pow(2, lg); // Check if freq[i] // is a power of 2 if (a == freq[i]) { // Print letter i + 'a' // freq[i] times while (freq[i] > 0) { freq[i] -= 1; System.out.print((char)(i + 'a')); } } }}// Driver Codepublic static void main(String args[]){ String S = "aaacbb"; int N = S.length(); countFrequency(S, N);}}// This code is contributed by AnkThon |
Python3
# Python3 program for the above approachfrom math import log2# Function to remove all the characters# from a that whose frequencies are not# equal to a perfect power of 2def countFrequency(S, N): # Stores the frequency of # each character in S freq = [0] * 26 # Iterate over characters of string for i in range(N): # Update frequency of current # character in the array freq[] freq[ord(S[i]) - ord('a')] += 1 # Traverse the array freq[] for i in range(26): # Check if the i-th letter # is absent from S if (freq[i] == 0): continue # Calculate log of frequency # of the current character # in the S lg = int(log2(freq[i])) # Calculate power of 2 of lg a = pow(2, lg) # Check if freq[i] # is a power of 2 if (a == freq[i]): # Print letter i + 'a' # freq[i] times while (freq[i]): print(chr(i + ord('a')), end = "") freq[i] -= 1# Driver Codeif __name__ == '__main__': S = "aaacbb" N = len(S) countFrequency(S, N)# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ // Function to remove all the characters// from a string that whose frequencies// are not equal to a perfect power of 2static void countFrequency(string S, int N){ // Stores the frequency of // each character in S int []freq = new int[26]; Array.Clear(freq, 0, freq.Length); // Iterate over characters of string for(int i = 0; i < N; i++) { // Update frequency of current // character in the array freq[] freq[(int)S[i] - 'a'] += 1; } // Traverse the array freq[] for(int i = 0; i < 26; i++) { // Check if the i-th letter // is absent from string S if (freq[i] == 0) continue; // Calculate log of frequency // of the current character // in the string S int lg = (int)Math.Log((double)freq[i], 2.0); // Calculate power of 2 of lg int a = (int)Math.Pow(2, lg); // Check if freq[i] // is a power of 2 if (a == freq[i]) { // Print letter i + 'a' // freq[i] times while (freq[i] > 0) { freq[i] -= 1; Console.Write((char)(i + 'a')); } } }}// Driver Codepublic static void Main(){ string S = "aaacbb"; int N = S.Length; countFrequency(S, N);}}// This code is contributed by SURENDRA_GANGWAR |
Javascript
<script> // JavaScript program for the above approach // Function to remove all the characters // from a string that whose frequencies // are not equal to a perfect power of 2 function countFrequency(S, N) { // Stores the frequency of // each character in S var freq = new Array(26).fill(0); // Iterate over characters of string for (var i = 0; i < N; i++) { // Update frequency of current // character in the array freq[] freq[S[i].charCodeAt(0) - "a".charCodeAt(0)]++; } // Traverse the array freq[] for (var i = 0; i < 26; i++) { // Check if the i-th letter // is absent from string S if (freq[i] === 0) continue; // Calculate log of frequency // of the current character // in the string S var lg = parseInt(Math.log2(freq[i])); // Calculate power of 2 of lg var a = Math.pow(2, lg); // Check if freq[i] // is a power of 2 if (a === freq[i]) { // Print letter i + 'a' // freq[i] times while (freq[i]--) document.write(String.fromCharCode(i + "a".charCodeAt(0))); } } } // Driver Code var S = "aaacbb"; var N = S.length; countFrequency(S, N); // This code is contributed by rdtank. </script> |
Output:
bbc
Time Complexity: O(N)
Auxiliary Space: O(1)
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