通过将每个字符向右循环移动相应频率来修改字符串

给定一个由小写英文字母组成的字符串S ，任务是将给定字符串S 的每个字符循环右移其频率。

Circular shifting of characters refers to shifting character ‘z’ to ‘a’, as its next character.

例子：

Input: S = “geeksforgeeks”
Output: iiimugpsiiimu 
Explanation:
Following changes are made on the string S:

1. Frequency of ‘g’ is 2. Therefore, shifting the character ‘g’ by 2 becomes ‘i’.
2. Frequency of ‘e’ is 4. Therefore, shifting the character ‘e’ by 4 becomes ‘i’.
3. Frequency of ‘k’ is 2. Therefore, shifting the character ‘k’ by 2 becomes ‘m’.
4. Frequency of ‘s’ is 2. Therefore, shifting the character ‘s’ by 2 becomes ‘u’.
5. Frequency of ‘f’ is 1. Therefore, shifting the character ‘f’ by 1 becomes ‘g’.
6. Frequency of ‘o’ is 1. Therefore, shifting the character ‘o’ by 1 becomes ‘p’.
7. Frequency of ‘r’ is 1. Therefore, shifting the character ‘r’ by 1 becomes ‘s’.

After the above shifting of characters, the string modifies to “iiimugpsiiimu”.

Input: S = “aabcadb”
Output: ddddded

办法：解决这个问题的想法是遍历字符串，找到字符串中的每个字符出现的频率，然后通过它的频率增加每个字符。请按照以下步骤解决问题：

• 初始化一个数组，比如frequency[] ，它存储字符串S中每个字符的出现次数。
• 遍历给定的字符串S并执行以下步骤：
  • 查找当前字符S[i]的频率。
  • 按频率增加当前字符并将S[i]的值更新为其更新后的字符。
• 完成上述步骤后，将字符串S打印为结果修改后的字符串。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to replace the characters
// by its frequency of character in it
void addFrequencyToCharacter(string S)
{
    // Stores frequencies of characters
    // in the string S
    int frequency[26] = { 0 };
 
    int N = S.length();
 
    // Traverse the string S
    for (int i = 0; i < N; i++) {
 
        // Increment the frequency of
        // each character by 1
        frequency[S[i] - 'a'] += 1;
    }
 
    // Traverse the string S
    for (int i = 0; i < N; i++) {
 
        // Find the frequency of
        // the current character
        int add = frequency[S[i] - 'a'] % 26;
 
        // Update the character
        if (int(S[i]) + add
            <= int('z'))
            S[i] = char(int(S[i])
                        + add);
 
        else {
            add = (int(S[i]) + add)
                  - (int('z'));
            S[i] = char(int('a')
                        + add - 1);
        }
    }
 
    // Print the resultant string
    cout << S;
}
 
// Driver Code
int main()
{
    string S = "geeksforgeeks";
    addFrequencyToCharacter(S);
 
    return 0;
}

Java

// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG {
 
    // Function to replace the characters
    // by its frequency of character in it
    static void addFrequencyToCharacter(String Str)
    {
        // Stores frequencies of characters
        // in the string S
        int frequency[] = new int[26];
 
        int N = Str.length();
        char S[] = Str.toCharArray();
 
        // Traverse the string S
        for (int i = 0; i < N; i++) {
 
            // Increment the frequency of
            // each character by 1
            frequency[S[i] - 'a'] += 1;
        }
 
        // Traverse the string S
        for (int i = 0; i < N; i++) {
 
            // Find the frequency of
            // the current character
            int add = frequency[S[i] - 'a'] % 26;
 
            // Update the character
            if ((int)(S[i]) + add <= (int)('z'))
                S[i] = (char)((int)(S[i]) + add);
 
            else {
                add = ((int)(S[i]) + add) - ((int)('z'));
                S[i] = (char)((int)('a') + add - 1);
            }
        }
 
        // Print the resultant string
        System.out.println(new String(S));
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String S = "geeksforgeeks";
        addFrequencyToCharacter(S);
    }
}
 
// This code is contributed by Kingash.

Python3

# Python3 program for the above approach
 
# Function to replace the characters
# by its frequency of character in it
def addFrequencyToCharacter(S):
 
    # Stores frequencies of characters
    # in the string S
    frequency = [0 for i in range(26)]
 
    N = len(S)
     
    S = list(S)
    # Traverse the string S
    for i in range(N):
 
        # Increment the frequency of
        # each character by 1
        frequency[ord(S[i]) - ord('a')] += 1
     
    # Traverse the string S
    for i in range(N):
 
        # Find the frequency of
        # the current character
        add = frequency[ord(S[i]) - ord('a')] % 26
 
        # Update the character
        if ord(S[i]) + add <= ord('z'):
            S[i] = chr(ord(S[i]) + add)
 
        else:
            add = ord(S[i]) + add - ord('z')
            S[i] = chr(ord('a') + add - 1)
     
    # Print the resultant string
    s = ""
    print(s.join(S))
 
# Driver Code
if __name__ == '__main__':
     
    S = "geeksforgeeks"
     
    addFrequencyToCharacter(S)
     
# This code is contributed by jana_sayantan

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to replace the characters
// by its frequency of character in it
static void addFrequencyToCharacter(string Str)
{
     
    // Stores frequencies of characters
    // in the string S
    int[] frequency = new int[26];
 
    int N = Str.Length;
    char[] S = Str.ToCharArray();
 
    // Traverse the string S
    for(int i = 0; i < N; i++)
    {
         
        // Increment the frequency of
        // each character by 1
        frequency[S[i] - 'a'] += 1;
    }
 
    // Traverse the string S
    for(int i = 0; i < N; i++)
    {
         
        // Find the frequency of
        // the current character
        int add = frequency[S[i] - 'a'] % 26;
 
        // Update the character
        if ((int)(S[i]) + add <= (int)('z'))
            S[i] = (char)((int)(S[i]) + add);
 
        else
        {
            add = ((int)(S[i]) + add) - ((int)('z'));
            S[i] = (char)((int)('a') + add - 1);
        }
    }
 
    // Print the resultant string
    Console.Write(new string(S));
}
 
// Driver Code
public static void Main(string[] args)
{
    string S = "geeksforgeeks";
    addFrequencyToCharacter(S);
}
}
 
// This code is contributed by ukasp

Javascript

 
// Javascript program for the above approach
 
 
// Function to replace the characters
// by its frequency of character in it
function addFrequencyToCharacter(Str)
{
    // Stores frequencies of characters
    // in the string S
    var frequency = Array.from({length: 26}, (_, i) => 0);
 
    var N = Str.length;
    var S = Str.split('');
 
    // Traverse the string S
    for (var i = 0; i < N; i++) {
 
        // Increment the frequency of
        // each character by 1
        frequency[S[i].charCodeAt(0) -
        'a'.charCodeAt(0)] += 1;
    }
 
    // Traverse the string S
    for (var i = 0; i < N; i++) {
 
        // Find the frequency of
        // the current character
        var add = frequency[S[i].charCodeAt(0) -
        'a'.charCodeAt(0)] % 26;
 
        // Update the character
        if ((S[i].charCodeAt(0)) +
                               add <= ('z').charCodeAt(0))
            S[i] = String.fromCharCode((S[i].charCodeAt(0))
            + add);
 
        else {
            add = ((S[i].charCodeAt(0)) + add) -
            (('z').charCodeAt(0));
            S[i] = String.fromCharCode(('a'.charCodeAt(0)) +
            add - 1);
        }
    }
 
    // Print the resultant string
    document.write(S.join(''));
}
 
// Driver Code
var S = "geeksforgeeks";
addFrequencyToCharacter(S);
 
// This code contributed by shikhasingrajput
 

输出：

iiimugpsiiimu

时间复杂度： O(N)
辅助空间： O(26)

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