通过去除频率不等于2的幂的字符后对字符进行排序来修改字符串

给定一个字符串S选自N小写字母，该任务是从其频率不是2的幂，并且然后进行排序以升序字符串的字符串中删除的字符。

例子：

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 = “geeksforgeeks”
Output: eeeefggkkorss

方法：给定的问题可以通过使用哈希来解决。请按照以下步骤解决给定的问题：

• 初始化一个大小为26的辅助数组，比如freq[] ，以存储字符串S中每个字符的频率，使得索引0表示字符‘a’ ， 1表示字符‘b’ ，依此类推。
• 遍历给定的字符串S并将数组freq[]中每个字符S[i]的频率增加1 。
• 遍历数组freq[] ，如果freq[i]的值是 2 的幂，则打印字符(‘a’ + i), freq[i]次。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to remove all the characters
// from a string that whose frequencies
// are not equal to a perfect power of 2
void 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 Code
int main()
{
    string S = "aaacbb";
    int N = S.size();
    countFrequency(S, N);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
  
// Function to remove all the characters
// from a string that whose frequencies
// are not equal to a perfect power of 2
static 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 Code
public 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 approach
from math import log2
 
# Function to remove all the characters
# from a that whose frequencies are not
# equal to a perfect power of 2
def 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 Code
if __name__ == '__main__':
     
    S = "aaacbb"
    N = len(S)
     
    countFrequency(S, N)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using 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 2
static 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 Code
public static void Main()
{
    string S = "aaacbb";
    int N = S.Length;
     
    countFrequency(S, N);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript

      // 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.
    

输出：

bbc

时间复杂度： O(N)
辅助空间： O(N)

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