Given two positive integers N, M. The task is to find the number of strings of length N under the alphabet set of size M such that no substrings of size greater than 1 is palindromic.
Examples:
Input : N = 2, M = 3
Output : 6
In this case, set of alphabet are 3, say {A, B, C}
All possible string of length 2, using 3 letters are:
{AA, AB, AC, BA, BB, BC, CA, CB, CC}
Out of these {AA, BB, CC} contain palindromic substring,
so our answer will be
8 - 2 = 6.
Input : N = 2, M = 2
Output : 2
Out of {AA, BB, AB, BA}, only {AB, BA} contain
non-palindromic substrings.
First, observe, a string does not contain any palindromic substring if the string doesn’t have any palindromic substring of the length 2 and 3, because all the palindromic string of the greater lengths contains at least one palindromic substring of the length of 2 or 3, basically in the center.
So, the following is true:
- There are M ways to choose the first symbol of the string.
- Then there are (M – 1) ways to choose the second symbol of the string. Basically, it should not be equal to first one.
- Then there are (M – 2) ways to choose any next symbol. Basically, it should not coincide with the previous symbols, that aren’t equal.
Knowing this, we can evaluate the answer in the following ways:
- If N = 1, then the answer will be M.
- If N = 2, then the answer is M*(M – 1).
- If N >= 3, then M * (M – 1) * (M – 2)N-2.
Below is the implementation of above idea :
C++
// CPP program to count number of strings of// size m such that no substring is palindrome.#include <bits/stdc++.h>using namespace std;// Return the count of strings with// no palindromic substring.int numofstring(int n, int m){ if (n == 1) return m; if (n == 2) return m * (m - 1); return m * (m - 1) * pow(m - 2, n - 2);}// Driven Programint main(){ int n = 2, m = 3; cout << numofstring(n, m) << endl; return 0;} |
Java
// Java program to count number of strings of// size m such that no substring is palindrome.import java.io.*;class GFG { // Return the count of strings with // no palindromic substring. static int numofstring(int n, int m) { if (n == 1) return m; if (n == 2) return m * (m - 1); return m * (m - 1) * (int)Math.pow(m - 2, n - 2); } // Driven Program public static void main (String[] args) { int n = 2, m = 3; System.out.println(numofstring(n, m)); }}// This code is contributed by ajit. |
Python3
# Python3 program to count number of strings of# size m such that no substring is palindrome# Return the count of strings with# no palindromic substring.def numofstring(n, m): if n == 1: return m if n == 2: return m * (m - 1) return m * (m - 1) * pow(m - 2, n - 2)# Driven Programn = 2m = 3print (numofstring(n, m))# This code is contributed# by Shreyanshi Arun. |
C#
// C# program to count number of strings of// size m such that no substring is palindrome.using System;class GFG { // Return the count of strings with // no palindromic substring. static int numofstring(int n, int m) { if (n == 1) return m; if (n == 2) return m * (m - 1); return m * (m - 1) * (int)Math.Pow(m - 2, n - 2); } // Driver Code public static void Main () { int n = 2, m = 3; Console.Write(numofstring(n, m)); }}// This code is contributed by Nitin Mittal. |
PHP
<?php// PHP program to count number// of strings of size m such // that no substring is palindrome.// Return the count of strings with// no palindromic substring.function numofstring($n, $m){ if ($n == 1) return $m; if ($n == 2) return $m * ($m - 1); return $m * ($m - 1) * pow($m - 2, $n - 2);}// Driver Code{ $n = 2; $m = 3; echo numofstring($n, $m) ; return 0;}// This code is contributed by nitin mittal.?> |
Javascript
<script>// JavaScript program to count number of strings of// size m such that no substring is palindrome. // Return the count of strings with // no palindromic substring. function numofstring(n, m) { if (n == 1) return m; if (n == 2) return m * (m - 1); return m * (m - 1) * Math.pow(m - 2, n - 2); }// Driver Code let n = 2, m = 3; document.write(numofstring(n, m));// This code is contributed by code_hunt.</script> |
Output
6
Time Complexity: O(log n), for using of pow function where n is the given input.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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