长度为3的可能的回文字符串通过使用给定的字符串的字符

给定一个由N 个字符组成的字符串S ，任务是按字典顺序打印所有长度为3 的回文字符串，这些字符串可以使用给定字符串S 的字符形成。

例子：

Input: S = “aabc”
Output:
aba
aca

Input: S = “ddadbac”
Output:
aba
aca
ada
dad
dbd
dcd
ddd

方法：给定的问题可以通过使用哈希的概念来解决，通过使用 Map 来实现它来存储字符。请按照以下步骤解决问题：

• 初始化一个 unordered_map，比如Hash ，它存储每个字符的计数。
• 将字符串的每个字符的频率存储在映射Hash 中。
• 初始化一组字符串，比如st ，以存储所有长度为3 的回文字符串。
• 使用变量i迭代字符[a, z] ，并执行以下步骤：
  • 如果Hash[i] 的值为2 ，则使用变量j迭代范围 [a, z] 上的字符。如果Hash[j]的值大于0且i不等于j ，则将字符串(i + j + i)插入到 Set st 中。
  • 否则，如果Hash[i]的值大于2则使用变量j迭代范围 [a, z] 上的字符，如果Hash[j]的值大于0 ，则插入字符串（ i + j + i)在集合st 中。
• 完成上述步骤后，遍历集合st ，打印其中存储的字符串。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to print all palindromic
// strings of length 3 that can be
// formed using characters of string S
void generatePalindrome(string S)
{
    // Stores the count of character
    unordered_map Hash;
 
    // Traverse the string S
    for (auto ch : S) {
        Hash[ch]++;
    }
 
    // Stores all palindromic strings
    set st;
 
    // Iterate over the charchaters
    // over the range ['a', 'z']
    for (char i = 'a'; i <= 'z'; i++) {
 
        // If Hash[ch] is equal to 2
        if (Hash[i] == 2) {
 
            // Iterate over the characters
            // over the range ['a', 'z']
            for (char j = 'a'; j <= 'z'; j++) {
 
                // Stores all the
                // palindromic string
                string s = "";
                if (Hash[j] && i != j) {
 
                    s += i;
                    s += j;
                    s += i;
 
                    // Push the s into
                    // the set st
                    st.insert(s);
                }
            }
        }
 
        // If Hash[i] is greater than
        // or equal to 3
        if (Hash[i] >= 3) {
 
            // Iterate over charchaters
            // over the range ['a', 'z']
            for (char j = 'a';
                 j <= 'z'; j++) {
 
                // Stores all the
                // palindromic string
                string s = "";
 
                // If Hash[j] is positive
                if (Hash[j]) {
                    s += i;
                    s += j;
                    s += i;
 
                    // Push s into
                    // the set st
                    st.insert(s);
                }
            }
        }
    }
 
    // Iterate over the set
    for (auto ans : st) {
        cout << ans << "\n";
    }
}
 
// Driver Code
int main()
{
    string S = "ddabdac";
    generatePalindrome(S);
 
    return 0;
}

Python3

# Python3 program for the above approach
 
# Function to print all palindromic
# strings of length 3 that can be
# formed using characters of string S
def generatePalindrome(S):
     
    # Stores the count of character
    Hash = {}
 
    # Traverse the string S
    for ch in S:
        Hash[ch] = Hash.get(ch,0) + 1
 
    # Stores all palindromic strings
    st = {}
 
    # Iterate over the charchaters
    # over the range ['a', 'z']
    for i in range(ord('a'), ord('z') + 1):
 
        # If Hash[ch] is equal to 2
        if (chr(i) in Hash and Hash[chr(i)] == 2):
             
            # Iterate over the characters
            # over the range ['a', 'z']
            for j in range(ord('a'), ord('z') + 1):
                 
                # Stores all the
                # palindromic string
                s = ""
                 
                if (chr(j) in Hash and i != j):
                    s += chr(i)
                    s += chr(j)
                    s += chr(i)
 
                    # Push the s into
                    # the set st
                    st[s] = 1
 
        # If Hash[i] is greater than
        # or equal to 3
        if ((chr(i) in Hash) and Hash[chr(i)] >= 3):
             
            # Iterate over charchaters
            # over the range ['a', 'z']
            for j in range(ord('a'), ord('z') + 1):
 
                # Stores all the
                # palindromic string
                s = ""
 
                # If Hash[j] is positive
                if (chr(j) in Hash):
                    s += chr(i)
                    s += chr(j)
                    s += chr(i)
 
                    # Push s into
                    # the set st
                    st[s] = 1
 
    # Iterate over the set
    for ans in st:
        print(ans)
 
# Driver Code
if __name__ == '__main__':
     
    S = "ddabdac"
     
    generatePalindrome(S)
 
# 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 print all palindromic
// strings of length 3 that can be
// formed using characters of string S
static void generatePalindrome(string S)
{
     
    // Stores the count of character
    Dictionary Hash = new Dictionary();
 
    // Traverse the string S
    foreach (char ch in S)
    {
        if (Hash.ContainsKey(ch))
            Hash[ch]++;
        else
            Hash.Add(ch, 1);
    }
 
    // Stores all palindromic strings
    HashSet st = new HashSet();
 
    // Iterate over the charchaters
    // over the range ['a', 'z']
    for(char i = 'a'; i <= 'z'; i++)
    {
         
        // If Hash[ch] is equal to 2
        if (Hash.ContainsKey(i) && Hash[i] == 2)
        {
             
            // Iterate over the characters
            // over the range ['a', 'z']
            for(char j = 'a'; j <= 'z'; j++)
            {
                 
                // Stores all the
                // palindromic string
                string s = "";
                 
                if (Hash.ContainsKey(j) && i != j)
                {
                    s += i;
                    s += j;
                    s += i;
 
                    // Push the s into
                    // the set st
                    st.Add(s);
                }
            }
        }
 
        // If Hash[i] is greater than
        // or equal to 3
        if (Hash.ContainsKey(i) && Hash[i] >= 3)
        {
             
            // Iterate over charchaters
            // over the range ['a', 'z']
            for(char j = 'a'; j <= 'z'; j++)
            {
                 
                // Stores all the
                // palindromic string
                string s = "";
 
                // If Hash[j] is positive
                if (Hash.ContainsKey(j))
                {
                    s += i;
                    s += j;
                    s += i;
 
                    // Push s into
                    // the set st
                    st.Add(s);
                }
            }
        }
    }
 
    // Iterate over the set
    foreach(string ans in st)
    {
        Console.WriteLine(ans);
    }
}
 
// Driver Code
public static void Main()
{
    string S = "ddabdac";
     
    generatePalindrome(S);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript

// Javascript program for the above approach
 
// Function to print all palindromic
// strings of length 3 that can be
// formed using characters of string S
function generatePalindrome(S)
{
    // Stores the count of character
    let Hash=new Map();
  
    // Traverse the string S
    for (let ch=0;ch< S.length;ch++) {
        if(!Hash.has(S[ch]))
            Hash.set(S[ch],1);
        else
        {
            Hash.set(S[ch],Hash.get(S[ch])+1)
        }
         
    }
  
    // Stores all palindromic strings
    let st=new Set();
  
    // Iterate over the charchaters
    // over the range ['a', 'z']
    for (let i = 'a'.charCodeAt(0); i <= 'z'.charCodeAt(0); i++) {
  
        // If Hash[ch] is equal to 2
        if (Hash.get(String.fromCharCode(i)) == 2) {
  
            // Iterate over the characters
            // over the range ['a', 'z']
            for (let j = 'a'.charCodeAt(0); j <= 'z'.charCodeAt(0); j++) {
  
                // Stores all the
                // palindromic string
                let s = "";
                if (Hash.get(String.fromCharCode(j)) && i != j) {
  
                    s += String.fromCharCode(i);
                    s += String.fromCharCode(j);
                    s += String.fromCharCode(i);
  
                    // Push the s into
                    // the set st
                    st.add(s);
                }
            }
        }
  
        // If Hash[i] is greater than
        // or equal to 3
        if (Hash.get(String.fromCharCode(i)) >= 3) {
  
            // Iterate over charchaters
            // over the range ['a', 'z']
            for (let j = 'a'.charCodeAt(0);
                 j <= 'z'.charCodeAt(0); j++) {
  
                // Stores all the
                // palindromic string
                let s = "";
  
                // If Hash[j] is positive
                if (Hash.get(String.fromCharCode(j))) {
                    s += String.fromCharCode(i);
                    s += String.fromCharCode(j);
                    s += String.fromCharCode(i);
  
                    // Push s into
                    // the set st
                    st.add(s);
                }
            }
        }
    }
  
    // Iterate over the set
    for (let item of st.values()) {
        document.write(item+"<br>")
    }
}
 
// Driver Code
let S = "ddabdac";
generatePalindrome(S);
 
// This code is contributed by unknown2108

输出：

aba
aca
ada
dad
dbd
dcd
ddd

时间复杂度： O(26*26)
辅助空间： O(26)

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