第 i 个字母是给定单词的第 (i-1)-th、i-th 或 (i+1)-th 个字母的单词计数

给定一个字符串str 。任务是计算与 str 长度相同的单词，并且第 i 个位置的每个字母是 str 的第 (i-1)-th、i-th 或 (i+1)-th 位置字母。
注意：对于第一个字母，考虑 W 的第 i 个和第 (i+1) 个位置字母。对于最后一个字母，考虑 str 的第 (i-1) 个和第 i 个位置字母。
例子：

Input : str[] = "ab"
Output : 4
Words that can be formed: aa, ab, ba, bb.

Input : str[] = "x"
Output : 1

对于索引 i 处的任何字母，除了第一个和最后一个字母外，存在三个可能的字母，即给定单词的第 (i-1) 个、第 i 个或第 (i+1) 个字母。因此，如果其中三个不同，我们就有 3 种可能性。如果其中两个相同，我们有两种可能性。如果一切都一样，我们只有一种可能性。
因此，遍历给定的单词并找到每个字母的可能性并将它们相乘。
同样，对于第一个字母，检查第一个和第二个位置的不同字母。对于最后一个位置，检查最后一个和倒数第二个位置的不同字母。
下面是这种方法的实现：

C++

// C++ program to count words  whose ith letter
// is either (i-1)th, ith, or (i+1)th letter
// of given word.
#include
using namespace std;
 
// Return the count of words.
int countWords(char str[], int len)
{
    int count = 1;
 
    // If word contain single letter, return 1.
    if (len == 1)
        return count;
 
    // Checking for first letter.
    if (str[0] == str[1])
        count *= 1;
    else
        count *= 2;
 
    // Traversing the string and multiplying
    // for combinations.
    for (int j=1; j

---

Java

// Java program to count words  whose ith letter
// is either (i-1)th, ith, or (i+1)th letter
// of given word.
public class GFG {
 
    // Return the count of words.
    static int countWords(String str, int len)
    {
        int count = 1;
      
        // If word contain single letter, return 1.
        if (len == 1)
            return count;
      
        // Checking for first letter.
        if (str.charAt(0) == str.charAt(1))
            count *= 1;
        else
            count *= 2;
      
        // Traversing the string and multiplying
        // for combinations.
        for (int j = 1; j < len - 1; j++)
        {
            // If all three letters are same.
            if (str.charAt(j) == str.charAt(j - 1) &&
                    str.charAt(j) == str.charAt(j + 1))
                count *= 1;
      
            // If two letter are distinct.
            else if (str.charAt(j) == str.charAt(j - 1)||
                    str.charAt(j) == str.charAt(j + 1) ||
                   str.charAt(j - 1) == str.charAt(j + 1))
                count *= 2;
      
            // If all three letter are distinct.
            else
                count *= 3;
        }
      
        // Checking for last letter.
        if (str.charAt(len - 1) == str.charAt(len - 2))
            count *= 1;
        else
            count *= 2;
      
        return count;
    }
      
    // Driven Program
    public static void main(String args[])
    {
        String str = "abc";
        int len = str.length();
      
        System.out.println(countWords(str, len));
    }
}
// This code is contributed by Sumit Ghosh

Python 3

# Python 3 program to count words  whose ith letter
# is either (i-1)th, ith, or (i+1)th letter
# of given word.
  
# Return the count of words.
def countWords( str,  l):
 
    count = 1;
  
    # If word contain single letter, return 1.
    if (l == 1):
        return count
  
    # Checking for first letter.
    if (str[0] == str[1]):
        count *= 1
    else:
        count *= 2
  
    # Traversing the string and multiplying
    # for combinations.
    for j in range(1,l-1):
        # If all three letters are same.
        if (str[j] == str[j-1] and str[j] == str[j+1]):
            count *= 1
  
        # If two letter are distinct.
        else if (str[j] == str[j-1] or
                 str[j] == str[j+1] or
                 str[j-1] == str[j+1]):
            count *= 2
  
        # If all three letter are distinct.
        else:
            count *= 3
  
    # Checking for last letter.
    if (str[l - 1] == str[l - 2]):
        count *= 1
    else:
        count *= 2
  
    return count
  
# Driven Program
if __name__ == "__main__":
     
    str = "abc"
    l = len(str)
  
    print(countWords(str, l))

C#

// C# program to count words whose
// ith letter is either (i-1)th,
// ith, or (i+1)th letter of the
// given word
using System;
 
public class GFG {
 
    // Return the count of words.
    static int countWords(string str, int len)
    {
        int count = 1;
 
        // If word contain single letter,
        // return 1.
        if (len == 1)
            return count;
 
        // Checking for first letter.
        if (str[0] == str[1])
            count *= 1;
        else
            count *= 2;
 
        // Traversing the string and
        // multiplying for combinations.
        for (int j = 1; j < len - 1; j++) {
 
            // If all three letters are same.
            if (str[j] == str[j - 1] &&
                  str[j] == str[j + 1])
                count *= 1;
 
            // If two letter are distinct.
            else if (str[j] == str[j - 1] ||
                     str[j] == str[j + 1] ||
                    str[j - 1] == str[j + 1])
                count *= 2;
 
            // If all three letter are distinct.
            else
                count *= 3;
        }
 
        // Checking for last letter.
        if (str[len - 1] == str[len - 2])
            count *= 1;
        else
            count *= 2;
 
        return count;
    }
 
    // Driver Program
    public static void Main()
    {
        string str = "abc";
        int len = str.Length;
 
        Console.WriteLine(countWords(str, len));
    }
}
 
// This code is contributed by Anant Agarwal

Javascript

// Javascript program to count words  whose ith letter
// is either (i-1)th, ith, or (i+1)th letter
// of given word.
     
    // Return the count of words.
    function countWords(str,len)
    {
        let count = 1;
        
        // If word contain single letter, return 1.
        if (len == 1)
            return count;
        
        // Checking for first letter.
        if (str[0] == str[1])
            count *= 1;
        else
            count *= 2;
        
        // Traversing the string and multiplying
        // for combinations.
        for (let j = 1; j < len - 1; j++)
        {
            // If all three letters are same.
            if (str[j] == str[j-1] &&
                    str[j] == str[j+1])
                count *= 1;
        
            // If two letter are distinct.
            else if (str[j] == str[j-1]||
                    str[j] == str[j+1] ||
                   str[j-1] == str[j+1])
                count *= 2;
        
            // If all three letter are distinct.
            else
                count *= 3;
        }
        
        // Checking for last letter.
        if (str[len-1] == str[len-2])
            count *= 1;
        else
            count *= 2;
        
        return count;
    }
     // Driven Program
    let str = "abc";
    let len= str.length;
    document.write(countWords(str, len));
     
    // This code is contributed by avanitrachhadiya2155

输出：


12

时间复杂度： O（字符串长度）。