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:
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.
Implementation:
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)); // Initializing 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 Codeint main(){ int n = 3; cout << "Total strings are : " << countStrs(n); return 0;} |
Java
// Java Program to count strings// with adjacent characters.import java.io.*;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; } } // Initializing 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)] # Initializing 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; // Initializing 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; } } // Initializing 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
Time Complexity: O(26*n)
Auxiliary Space: O(26*n)
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