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++

`// 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_ ` |

**Output:**

Total strings are : 98

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