检查每个单词的字符是否可以重新排列以形成算术级数(AP)

定的字符串str，任务是检查是否有可能重新排列字符串，使得定字符串的每个字的字符是算术级数。

例子：

Input: str = “ace yzx fbd”
Output: true
Explanation: Rearrange the given string to “ace xyz bdf”.
All the characters of the word “ace” are in AP with a common difference of 2.
All the characters of the word “xyz” are in AP with a common difference of 1
All the characters of the word “bdf” are in AP with a common difference of 2.
Therefore, the required output is true.

Input: str = “geeks for geeks”
Output: false

方法：想法是对给定字符串的每个单词进行排序，并检查所有单词中相邻字符之间的差异是否相等。如果发现是真的，则打印“是” 。否则，打印No。请按照以下步骤解决问题。

1. 遍历字符串str ，并用空格定界符分割str的每个单词。
2. 以升序对给定字符串的每个单词进行排序。
3. 检查单词的所有相邻字符之间的差异是否相等。
4. 如果发现是真的，则打印“是” 。否则，打印No。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to check str can be
// rearranged such that characters
// of each word forms an AP
 
int checkWordAP(string str)
{
    // Stores the string
    // in stringstream
    stringstream ss(str);
 
    // Stores each word of
    // the given string
    string temp;
    while (getline(ss, temp, ' ')) {
        // Sort the current word
        sort(temp.begin(), temp.end());
 
        // Check if the current word
        // of the given string is in AP
        for (int i = 2; i < temp.length();
             i++) {
            // Store the value of difference
            // between adjacent characters
            int diff = temp[1] - temp[0];
 
            // Check if difference between all
            // adjacent characters are equal
            if (diff != temp[i] - temp[i - 1]) {
                return false;
            }
        }
    }
 
    // If all words are in AP.
    return true;
}
 
// Driver Code
int main()
{
    string str = "ace yzx fbd";
 
    // If all words of the given
    // string are in AP
    if (checkWordAP(str)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG{
 
// Function to check str can be
// rearranged such that characters
// of each word forms an AP
static boolean checkWordAP(String s)
{
  // Stores the String
  // in Stringstream
  String str[] = s.split(" ");
 
  // Stores each word of
  // the given String
 
  for (String temp : str )
  {
    // Sort the current word
    temp = sort(temp);
 
    // Check if the current word
    // of the given String is in AP
    for (int i = 2; i < temp.length(); i++)
    {
      // Store the value of difference
      // between adjacent characters
      int diff = temp.charAt(1) - temp.charAt(0);
 
      // Check if difference between all
      // adjacent characters are equal
      if (diff != temp.charAt(i) - temp.charAt(i - 1))
      {
        return false;
      }
    }
  }
 
  // If all words are in AP.
  return true;
}
   
static String sort(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);
}
 
// Driver Code
public static void main(String[] args)
{
  String str = "ace yzx fbd";
 
  // If all words of the given
  // String are in AP
  if (checkWordAP(str))
  {
    System.out.print("Yes");
  }
  else
  {
    System.out.print("No");
  }
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 program to implement
# the above approach
 
# Function to check st can be
# rearranged such that characters
# of each word forms an AP
def checkWordAP(st):
 
    # Stores each word of
    # the given sting
    st = st.split(" ")
     
    for temp in st:
         
        # Sort the current word
        temp = sorted(temp)
 
        # Check if the current word
        # of the given is in AP
        for i in range(2, len(temp)):
             
            # Store the value of difference
            # between adjacent characters
            diff = ord(temp[1]) - ord(temp[0])
 
            # Check if difference between all
            # adjacent characters are equal
            if (diff != ord(temp[i]) -
                        ord(temp[i - 1])):
                return False
 
    # If all words are in AP.
    return True
 
# Driver Code
if __name__ == '__main__':
     
    st = "ace yzx fbd"
 
    # If all words of the given
    # are in AP
    if (checkWordAP(st)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to check str can be
// rearranged such that characters
// of each word forms an AP
static bool checkWordAP(String s)
{
  // Stores the String
  // in Stringstream
  String []str = s.Split(' ');
 
  // Stores each word of
  // the given String
  String temp = "";
  foreach (String temp1 in str )
  {
    // Sort the current word
    temp = sort(temp1);
 
    // Check if the current word
    // of the given String is in AP
    for (int i = 2; i < temp.Length; i++)
    {
      // Store the value of difference
      // between adjacent characters
      int diff = temp[1] - temp[0];
 
      // Check if difference between all
      // adjacent characters are equal
      if (diff != temp[i] - temp[i - 1])
      {
        return false;
      }
    }
  }
 
  // If all words are in AP.
  return true;
}
   
static String sort(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);
}
 
// Driver Code
public static void Main(String[] args)
{
  String str = "ace yzx fbd";
 
  // If all words of the given
  // String are in AP
  if (checkWordAP(str))
  {
    Console.Write("Yes");
  }
  else
  {
    Console.Write("No");
  }
}
}
 
// This code is contributed by Princi Singh

输出

Yes

时间复杂度： O(N log ₂ N)
辅助空间： O(N)