Count of strings where adjacent characters are of difference one

iven a number n, count number of strings of length n such that every string has adjacent characters with difference between ASCII values as 1.

Examples:

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

For strings starting with character ‘A’ and length ‘i’, we consider all strings of length ‘i-1’ and starting with character ‘B’



For strings starting with character ‘G’ and length ‘i’, we consider all strings of length ‘i-1’ and starting with character ‘H’ and all strings of length ‘i-1’ and starting with ‘F’.

We take the base case for n = 1, and set result for all 26 characters as 1. This simply means when 1 character string is consider all alphabets from a-z are taken only once.

For N = 2,

For N = 3,

Conclusion : For 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++

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// CPP Program to count strings with adjacent
// characters.
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python 3

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# 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

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C#

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// 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_

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Output:

Total strings are : 98

This article is contributed by Shubham Rana. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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