相邻字符相差一的字符串计数

给定一个数字 n，计算长度为 n 的字符串的数量，使得每个字符串都有相邻的字符，并且 ASCII 值之间的差异为 1。

例子：

Input :  N = 1
Output : Total strings are 26
         Explanation : For N=1, strings 
         are a, b, c,, ...., x, y, z 

Input :  N = 2
Output : Total strings are 50
         Explanation : For N = 2, strings
         are ab, ba, bc, cb, .., yx, yz, zy

对于开始字符“A”和长度“我”的字符串，我们认为长的“I-1”的所有字符串，并开始用字符“B”
为了启动用字符“G”和长度“I”的字符串，我们考虑长度“I-1”的所有字符串，并具有字符“H”开始，并用“F”长度“I-1”和开始的所有字符串。
我们取基础案例对于n = 1，和设定结果为所有26字符为1。这仅仅意味着当1个是考虑从AZ采取只有一旦所有字母。
对于N = 2 ，

对于N = 3 ，

结论：对于N = n

countAdjacent(n)
    dp[i][j] finally stores count of strings
             of length i and starting with 
             character j.

    Initialize dp[n+1][27] as 0
    Initialize dp[1][j] = 1 where j = 0 to 25
    for i = 2 to n
      for j = 0 to 25
         if (j = 0)
           dp[i][j] = dp[i-1][j+1];
         else
           dp[i][j] = dp[i-1][j-1] + dp[i-1][j+1];
    Sum of n-th row from 0 to 25 is the result.

C++

// CPP Program to count strings with adjacent
// characters.
#include 
using namespace std;
 
int countStrs(int n)
{
    long int dp[n + 1][27];
 
    // Initializing arr[n+1][27] to 0
    memset(dp, 0, sizeof(dp));
 
    // Initialing 1st row all 1 from 0 to 25
    for (int i = 0; i <= 25; i++)
        dp[1][i] = 1;
 
    // Begin evaluating from i=2 since 1st row is set
    for (int i = 2; i <= n; i++) {
        for (int j = 0; j <= 25; j++)
 
            // j=0 is 'A' which can make strings
            // of length i using strings of length
            // i-1 and starting with 'B'
            if (j == 0)
                dp[i][j] = dp[i - 1][j + 1];
            else
                dp[i][j] = (dp[i - 1][j - 1] +
                            dp[i - 1][j + 1]);
    }
 
    // Our result is sum of last row.
    long int sum = 0;
    for (int i = 0; i <= 25; i++)
        sum = (sum + dp[n][i]);
    return sum;
}
 
// Driver's Code
int main()
{
    int n = 3;
    cout << "Total strings are : " << countStrs(n);
    return 0;
}

Java

// Java Program to count strings
// with adjacent characters.
class GFG
{
 
    static long countStrs(int n)
    {
        long[][] dp = new long[n + 1][27];
 
        // Initializing arr[n+1][27] to 0
        for (int i = 0; i < n + 1; i++)
        {
            for (int j = 0; j < 27; j++)
            {
                dp[i][j] = 0;
            }
        }
 
        // Initialing 1st row all 1 from 0 to 25
        for (int i = 0; i <= 25; i++)
        {
            dp[1][i] = 1;
        }
 
        // Begin evaluating from i=2
        // since 1st row is set
        for (int i = 2; i <= n; i++)
        {
             
            // j=0 is 'A' which can make strings
            for (int j = 0; j <= 25; j++) 
             
            // of length i using strings of length
            // i-1 and starting with 'B'
            {
                if (j == 0)
                {
                    dp[i][j] = dp[i - 1][j + 1];
                }
                else
                {
                    dp[i][j] = (dp[i - 1][j - 1]
                            + dp[i - 1][j + 1]);
                }
            }
        }
 
        // Our result is sum of last row.
        long sum = 0;
        for (int i = 0; i <= 25; i++)
        {
            sum = (sum + dp[n][i]);
        }
        return sum;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 3;
        System.out.println("Total strings are : " +
                                        countStrs(n));
    }
}
 
// This code is contributed by 29AjayKumar

Python 3

# Python3 Program to count strings with
# adjacent characters.
def countStrs(n):
 
    # Initializing arr[n+1][27] to 0
    dp = [[0 for j in range(27)]
             for i in range(n + 1)]
 
    # Initialing 1st row all 1 from 0 to 25
    for i in range(0, 26):
        dp[1][i] = 1
 
    # Begin evaluating from i=2 since
    # 1st row is set    
    for i in range(2, n + 1):
        for j in range(0, 26):
 
            # j=0 is 'A' which can make strings
            # of length i using strings of length
            # i-1 and starting with 'B'
            if(j == 0):
                dp[i][j] = dp[i - 1][j + 1];
            else:
                dp[i][j] = (dp[i - 1][j - 1] +
                            dp[i - 1][j + 1])
 
    # Our result is sum of last row.        
    sum = 0
    for i in range(0, 26):
        sum = sum + dp[n][i]
 
    return sum
     
# Driver's Code
if __name__ == "__main__":
    n = 3
    print("Total strings are : ", countStrs(n))
     
# This code is contributed by Sairahul Jella

C#

// C# Program to count strings with 
// adjacent characters.
using System;
 
class GFG
{
    static long countStrs(int n)
    {
        long[,] dp = new long[n + 1, 27];
     
        // Initializing arr[n+1][27] to 0
        for(int i = 0; i < n + 1; i++)
            for(int j = 0; j < 27; j++)
                dp[i, j] = 0;
     
        // Initialing 1st row all 1 from 0 to 25
        for (int i = 0; i <= 25; i++)
            dp[1, i] = 1;
     
        // Begin evaluating from i=2 since 1st row is set
        for (int i = 2; i <= n; i++)
        {
            for (int j = 0; j <= 25; j++)
     
                // j=0 is 'A' which can make strings
                // of length i using strings of length
                // i-1 and starting with 'B'
                if (j == 0)
                    dp[i, j] = dp[i - 1, j + 1];
                else
                    dp[i, j] = (dp[i - 1, j - 1] +
                                dp[i - 1, j + 1]);
        }
     
        // Our result is sum of last row.
        long sum = 0;
        for (int i = 0; i <= 25; i++)
            sum = (sum + dp[n, i]);
        return sum;
    }
     
    // Driver Code
    static void Main()
    {
        int n = 3;
        Console.Write("Total strings are : " + countStrs(n));
    }
}
 
// This code is contributed by DrRoot_

Javascript

// JavaScript Program to count strings
// with adjacent characters.
 
    function countStrs(n)
    {
        let dp = new Array(n + 1);
        // Loop to create 2D array using 1D array
        for (var i = 0; i < dp.length; i++) {
            dp[i] = new Array(2);
        }
 
        // Initializing arr[n+1][27] to 0
        for (let i = 0; i < n + 1; i++)
        {
            for (let j = 0; j < 27; j++)
            {
                dp[i][j] = 0;
            }
        }
 
        // Initialing 1st row all 1 from 0 to 25
        for (let i = 0; i <= 25; i++)
        {
            dp[1][i] = 1;
        }
 
        // Begin evaluating from i=2
        // since 1st row is set
        for (let i = 2; i <= n; i++)
        {
             
            // j=0 is 'A' which can make strings
            for (let j = 0; j <= 25; j++) 
             
            // of length i using strings of length
            // i-1 and starting with 'B'
            {
                if (j == 0)
                {
                    dp[i][j] = dp[i - 1][j + 1];
                }
                else
                {
                    dp[i][j] = (dp[i - 1][j - 1]
                            + dp[i - 1][j + 1]);
                }
            }
        }
 
        // Our result is sum of last row.
        let sum = 0;
        for (let i = 0; i <= 25; i++)
        {
            sum = (sum + dp[n][i]);
        }
        return sum;
    }
  
// Driver Code
 
    let n = 3;
    document.write("Total strings are : " +
                                        countStrs(n));
         

输出：

Total strings are : 98

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。