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Number of strings of length N with no palindromic sub string
  • Difficulty Level : Medium
  • Last Updated : 21 May, 2018

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 Program
int 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 Program
n = 2
m = 3
print (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.
?>



Output
6

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