# Count of strings where adjacent characters are of difference one

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

**Examples**:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.

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++

`// 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;` `}` |

## 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

`<script>` `// 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));` ` ` `</script>` |

**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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.