打印字符串的字典序最小的非回文置换的最后一个字符

定的字符串str，任务是打印出给定的字符串的字典序最小的非回文置换的最后一个字符。如果不存在这样的排列，请打印“ -1” 。

例子：

Input: str = “deepqvu”
Output: v
Explanation: The string “deepquv” is the lexicographically smallest permutation which is not a palindrome.
Therefore, the last character is v.

Input: str = “zyxaaabb”
Output: z
Explanation: The string “aaabbxyz” is the lexicographically smallest permutation which is not a palindrome.
Therefore, the last character is z.

天真的方法：解决问题的最简单方法是生成给定字符串的所有可能排列，并针对每个排列检查它是否是回文。在所有获得的非回文排列中，按字典顺序打印最小的排列的最后一个字符。请按照以下步骤操作：

1. 对给定的字符串str进行排序。
2. 使用函数next_permutation()检查回文字符串的所有后续排列。
3. 如果存在任何非回文排列，则打印其最后一个字符。
4. 否则，打印“ -1”。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to check whether a string
// s is a palindrome or not
bool isPalin(string s, int N)
{
    // Traverse the string
    for (int i = 0; i < N; i++) {
 
        // If unequal character
        if (s[i] != s[N - i - 1]) {
            return false;
        }
    }
 
    return true;
}
 
// Function to find the smallest
// non-palindromic lexicographic
// permutation of string s
void lexicographicSmallestString(
    string s, int N)
{
    // Base Case
    if (N == 1) {
        cout << "-1";
    }
 
    // Sort the given string
    sort(s.begin(), s.end());
 
    int flag = 0;
 
    // If the formed string is
    // non palindromic
    if (isPalin(s, N) == false)
        flag = 1;
 
    if (!flag) {
 
        // Check for all permutations
        while (next_permutation(s.begin(),
                                s.end())) {
 
            // Check palindromic
            if (isPalin(s, N) == false) {
                flag = 1;
                break;
            }
        }
    }
 
    // If non palindromic string found
    // print its last character
    if (flag == 1) {
        int lastChar = s.size() - 1;
        cout << s[lastChar] << ' ';
    }
 
    // Otherwise, print "-1"
    else {
        cout << "-1";
    }
}
 
// Driver Code
int main()
{
    // Given string str
    string str = "deepqvu";
 
    // Length of the string
    int N = str.length();
 
    // Function Call
    lexicographicSmallestString(str, N);
 
    return 0;
}

Java

// Java program for the
// above approach
import java.util.*;
class GFG{
 
// Function to check whether
// a String s is a palindrome
// or not
static boolean isPalin(String s,
                       int N)
{
  // Traverse the String
  for (int i = 0; i < N; i++)
  {
    // If unequal character
    if (s.charAt(i) !=
        s.charAt(N - i - 1))
    {
      return false;
    }
  }
 
  return true;
}
   
static boolean next_permutation(char[] p)
{
  for (int a = p.length - 2;
           a >= 0; --a)
    if (p[a] < p[a + 1])
      for (int b = p.length - 1;; --b)
        if (p[b] > p[a])
        {
          char t = p[a];
          p[a] = p[b];
          p[b] = t;
          for (++a, b = p.length - 1;
                 a < b; ++a, --b)
          {
            t = p[a];
            p[a] = p[b];
            p[b] = t;
          }
           
          return true;
        }
   
  return false;
}
   
//Method to sort a string alphabetically
static String sortString(String inputString)
{
  // convert input string
  // to char array
  char tempArray[] =
       inputString.toCharArray();
 
  // Sort tempArray
  Arrays.sort(tempArray);
 
  // Return new sorted string
  return new String(tempArray);
} 
 
// Function to find the smallest
// non-palindromic lexicographic
// permutation of String s
static void lexicographicSmallestString(String s,
                                        int N)
{
  // Base Case
  if (N == 1)
  {
    System.out.print("-1");
  }
 
  // Sort the given String
  s =  sortString(s);
 
  int flag = 0;
 
  // If the formed String is
  // non palindromic
  if (isPalin(s, N) == false)
    flag = 1;
 
  if (flag != 0)
  {
    // Check for all permutations
    while (next_permutation(s.toCharArray()))
    {
      // Check palindromic
      if (isPalin(s, N) == false)
      {
        flag = 1;
        break;
      }
    }
  }
 
  // If non palindromic String found
  // print its last character
  if (flag == 1)
  {
    int lastChar = s.length() - 1;
    System.out.print(s.charAt(lastChar) + " ");
  }
 
  // Otherwise, print "-1"
  else
  {
    System.out.print("-1");
  }
}
 
// Driver Code
public static void main(String[] args)
{
  // Given String str
  String str = "deepqvu";
 
  // Length of the String
  int N = str.length();
 
  // Function Call
  lexicographicSmallestString(str, N);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the
# above approach
 
# Function to check whether
# a String s is a palindrome
# or not
def isPalin(s, N):
   
    # Traverse the String
    for i in range(N):
       
        # If unequal character
        if (s[i] != s[N - i - 1]):
            return False;         
    return True;
 
def next_permutation(p):
   
    for a in range(len(p) - 2, 0, -1):
        if (p[a] < p[a + 1]):
            for b in range(len(p) - 1, -1):
                if (p[b] > p[a]):
                    t = p[a];
                    p[a] = p[b];
                    p[b] = t;
                     
                    for a in range(a + 1,
                                   a < b,):
                        b = len(p) - 1;
                        if(a < b):
                            t = p[a];
                            p[a] = p[b];
                            p[b] = t;
                        b -= 1;
 
                    return True;
                   
    return False;
 
# Method to sort a string
# alphabetically
def sortString(inputString):
   
    # convert input string
    # to char array
    # Sort tempArray
    tempArray = ''.join(sorted(inputString));
 
    # Return new sorted string
    return tempArray;
 
# Function to find the smallest
# non-palindromic lexicographic
# permutation of String s
def lexicographicSmallestString(s, N):
   
    # Base Case
    if (N == 1):
        print("-1");
 
    # Sort the given String
    s = sortString(s);
 
    flag = 0;
 
    # If the formed String is
    # non palindromic
    if (isPalin(s, N) == False):
        flag = 1;
 
    if (flag != 0):
       
        # Check for all permutations
        while (next_permutation(s)):
           
            # Check palindromic
            if (isPalin(s, N) == False):
                flag = 1;
                break;
 
    # If non palindromic String
    # found print last character
    if (flag == 1):
        lastChar = len(s) - 1;
        print(s[lastChar],
              end = "");
 
    # Otherwise, pr"-1"
    else:
        print("-1");
 
# Driver Code
if __name__ == '__main__':
   
    # Given String str
    str = "deepqvu";
 
    # Length of the String
    N = len(str);
 
    # Function Call
    lexicographicSmallestString(str, N);
 
# This code is contributed by shikhasingrajput

C#

// C# program for the
// above approach
using System;
class GFG{
 
// Function to check whether
// a String s is a palindrome
// or not
static bool isPalin(String s,
                    int N)
{
  // Traverse the String
  for (int i = 0; i < N; i++)
  {
    // If unequal character
    if (s[i] !=
        s[N - i - 1])
    {
      return false;
    }
  }
 
  return true;
}
   
static bool next_permutation(char[] p)
{
  for (int a = p.Length - 2;
           a >= 0; --a)
    if (p[a] < p[a + 1])
      for (int b = p.Length - 1;; --b)
        if (p[b] > p[a])
        {
          char t = p[a];
          p[a] = p[b];
          p[b] = t;
          for (++a, b = p.Length - 1;
                 a < b; ++a, --b)
          {
            t = p[a];
            p[a] = p[b];
            p[b] = t;
          }
           
          return true;
        }
   
  return false;
}
   
//Method to sort a string alphabetically
static String sortString(String inputString)
{
  // convert input string
  // to char array
  char []tempArray =
       inputString.ToCharArray();
 
  // Sort tempArray
  Array.Sort(tempArray);
 
  // Return new sorted string
  return new String(tempArray);
} 
 
// Function to find the smallest
// non-palindromic lexicographic
// permutation of String s
static void lexicographicSmallestString(String s,
                                        int N)
{
  // Base Case
  if (N == 1)
  {
    Console.Write("-1");
  }
 
  // Sort the given String
  s =  sortString(s);
 
  int flag = 0;
 
  // If the formed String is
  // non palindromic
  if (isPalin(s, N) == false)
    flag = 1;
 
  if (flag != 0)
  {
    // Check for all permutations
    while (next_permutation(s.ToCharArray()))
    {
      // Check palindromic
      if (isPalin(s, N) == false)
      {
        flag = 1;
        break;
      }
    }
  }
 
  // If non palindromic String found
  // print its last character
  if (flag == 1)
  {
    int lastChar = s.Length - 1;
    Console.Write(s[lastChar] + " ");
  }
 
  // Otherwise, print "-1"
  else
  {
    Console.Write("-1");
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given String str
  String str = "deepqvu";
 
  // Length of the String
  int N = str.Length;
 
  // Function Call
  lexicographicSmallestString(str, N);
}
}
 
// This code is contributed by Amit Katiyar

输出

v 

时间复杂度： O(N * N！)
辅助空间： O(1)

高效方法：为了优化上述方法，其思想是存储给定字符串str的每个字符的频率。如果所有字符都相同，则打印“ -1” 。否则，打印给定字符串str的最大字符。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the smallest non
// palindromic lexicographic string
void lexicographicSmallestString(
    string s, int N)
{
    // Stores the frequency of each
    // character of the string s
    map M;
 
    // Traverse the string
    for (char ch : s) {
        M[ch]++;
    }
 
    // If there is only one element
    if (M.size() == 1) {
        cout << "-1";
    }
 
    // Otherwise
    else {
        auto it = M.rbegin();
        cout << it->first;
    }
}
 
// Driver Code
int main()
{
    // Given string str
    string str = "deepqvu";
 
    // Length of the string
    int N = str.length();
 
    // Function Call
    lexicographicSmallestString(str, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
   
// Function to find the smallest non
// palindromic lexicographic string
static void lexicographicSmallestString(String s,
                                        int N)
{
     
    // Stores the frequency of each
    // character of the string s
    SortedMap M = new TreeMap();
   
    // Converting given string to char array
    char[] str = s.toCharArray(); 
 
    // Traverse the string
    for(char c : str)
    {
        if (M.containsKey(c))
        {
             
            // If char is present in M
            M.put(c, M.get(c) + 1);
        }
        else
        {
             
            // If char is not present in M
            M.put(c, 1);
        }
    }
 
    // If there is only one element
    if (M.size() == 1)
    {
        System.out.print("-1");
    }
 
    // Otherwise
    else
    {
        System.out.print( M.lastKey() );
    }
}
 
// Driver Code
public static void main (String[] args)
{
     
    // Given string str
    String str = "deepqvu";
 
    // Length of the string
    int N = str.length();
 
    // Function Call
    lexicographicSmallestString(str, N);
}
}
 
// This code is contributed by math_lover

Python3

# Python3 program for the above approach
 
# Function to find the smallest non
# palindromic lexicographic string
def lexicographicSmallestString(s, N):
     
    # Stores the frequency of each
    # character of the s
    M = {}
 
    # Traverse the string
    for ch in s:
        M[ch] = M.get(ch, 0) + 1
 
    # If there is only one element
    if len(M) == 1:
        print("-1")
         
    # Otherwise
    else:
        x = list(M.keys())[-2]
        print(x)
         
# Driver Code
if __name__ == '__main__':
     
    # Given str
    str = "deepqvu"
 
    # Length of the string
    N = len(str)
 
    # Function call
    lexicographicSmallestString(str, 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 find the smallest non
// palindromic lexicographic string
static void lexicographicSmallestString(String s,
                                        int N)
{   
  // Stores the frequency of each
  // character of the string s
  SortedDictionary M = new SortedDictionary();
 
  // Converting given string
  // to char array
  char[] str = s.ToCharArray(); 
 
  // Traverse the string
  foreach(char c in str)
  {
    if (M.ContainsKey(c))
    {
      // If char is present
      // in M
      M = M + 1;
    }
    else
    {
      // If char is not present
      // in M
      M.Add(c, 1);
    }
  }
 
  // If there is only
  // one element
  if (M.Count == 1)
  {
    Console.Write("-1");
  }
 
  // Otherwise
  else        
  {
    int count = 0;
    foreach(KeyValuePair m in M)
    {
      count++;
      if(count == M.Count)
        Console.Write(m.Key);
    }
  }
}
 
// Driver Code
public static void Main(String[] args)
{   
  // Given string str
  String str = "deepqvu";
 
  // Length of the string
  int N = str.Length;
 
  // Function Call
  lexicographicSmallestString(str, N);
}
}
 
// This code is contributed by gauravrajput1

输出

v

时间复杂度： O(N)
辅助空间： O(26)