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Count of palindromic strings of size upto N consisting of first K alphabets occurring at most twice

  • Last Updated : 26 Aug, 2021

Given two integers N and K, the task is to find the number of palindromic strings of size at most N consisting of the first K lowercase alphabets such that each character in a string doesn’t appear more than twice.

Examples:

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Input: N = 3, K = 2
Output: 6
Explanation:
The possible strings are:
“a”, “b”, “aa”, “bb”, “aba”, “bab”.



Input: N = 4, K = 3
Output: 18
Explanation:
The possible strings are: 
“a”, “b”, “c”, “aa”, “bb”, “cc”, “aba”, “aca”, “bab”, “bcb”, “cac”, “cbc”,  “abba”, “acca”, “baab”, “bccb”, “caac”, “cbbc”.

Approach: The given problem can be solved based on the following observations:

  • Let’s try to build a 4 digit palindrome(N = 4) with only the first 3 English letters(K = 3). So, the idea is to create an empty string ( _ _ _ _ ) and now to get a palindrome out of it, only the two digits in one in its one half can be filled because the other two would be decided on the basis of them, i.e., if first 2 places are chosen to fill with a character of choice the last two will be same to that so that the string should be a palindrome.
  • Here in this case, if a is filled in first position and b in second then the only choice remaining for 3rd and 4th is to be filled with b and a respectively.
  • So this means, to find the number of palindromic strings with length 4 (N = 4) with only the first 3 letters (K = 3), count all the combinations possible for the first 2 digits(= N/2) i.e., 3*2=6 (3 choices for the first position and 2 choices for the second).
  • The above-explained case is for an even length string (N is even), and for an odd length string (N is odd), N/2 + 1 indexes can be filled.

Follow the steps below to solve the given problem:

  1. Fo finding the count palindromic strings of length at most N, then count palindromic strings of each length from 1 to N and then add them together.
  2. For the value of N as:
    • If N is even, then find all combinations possible till N/2 because only half of the positions can be filled.
    • If N is odd, then find all combinations possible till N/2 +1, and extra + 1 for the element as the middle element.
  3. Add all of them together and print the answer accordingly.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function of return the number of
// palindromic strings of length N with
// first K alphabets possible
int lengthNPalindrome(int N, int K)
{
    int half = N / 2;
 
    // If N is odd, half + 1 position
    // can be filled to cope with the
    // extra middle element
    if (N & 1) {
        half += 1;
    }
 
    int ans = 1;
    for (int i = 1; i <= half; i++) {
        ans *= K;
 
        // K is reduced by one, because
        // count of choices for the next
        // position is  reduced by 1 as
        // a element can only once
        K--;
    }
 
    // Return the possible count
    return ans;
}
 
// Function to find the count of palindromic
// string of first K characters according
// to the given criteria
int palindromicStrings(int N, int K)
{
    // If N=1, then only K palindromic
    // strings possible.
    if (N == 1) {
        return K;
    }
 
    // If N=2, the 2*K palindromic strings
    // possible, K for N=1 and K for N=2
    if (N == 2) {
        return 2 * K;
    }
 
    int ans = 0;
 
    // Initialize ans with the count of
    // strings possible till N = 2
    ans += (2 * K);
 
    for (int i = 3; i <= N; i++) {
        ans += lengthNPalindrome(i, K);
    }
 
    // Return the possible count
    return ans;
}
 
// Driver Code
int main()
{
    int N = 4, K = 3;
    cout << palindromicStrings(N, K);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG {
    // Function of return the number of
    // palindromic strings of length N with
    // first K alphabets possible
    static int lengthNPalindrome(int N, int K)
    {
        int half = N / 2;
 
        // If N is odd, half + 1 position
        // can be filled to cope with the
        // extra middle element
        if (N % 2 == 1) {
            half += 1;
        }
 
        int ans = 1;
        for (int i = 1; i <= half; i++) {
            ans *= K;
 
            // K is reduced by one, because
            // count of choices for the next
            // position is  reduced by 1 as
            // a element can only once
            K--;
        }
 
        // Return the possible count
        return ans;
    }
 
    // Function to find the count of palindromic
    // string of first K characters according
    // to the given criteria
    static int palindromicStrings(int N, int K)
    {
        // If N=1, then only K palindromic
        // strings possible.
        if (N == 1) {
            return K;
        }
 
        // If N=2, the 2*K palindromic strings
        // possible, K for N=1 and K for N=2
        if (N == 2) {
            return 2 * K;
        }
 
        int ans = 0;
 
        // Initialize ans with the count of
        // strings possible till N = 2
        ans += (2 * K);
 
        for (int i = 3; i <= N; i++) {
            ans += lengthNPalindrome(i, K);
        }
 
        // Return the possible count
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 4, K = 3;
 
        System.out.println(palindromicStrings(N, K));
    }
}
// This code is contributed by Potta Lokesh

Python3




# Python3 program for the above approach
 
 
# Function of return the number of
# palindromic strings of length N with
# first K alphabets possible
def lengthNPalindrome(N, K) :
     
    half = N // 2;
 
    # If N is odd, half + 1 position
    # can be filled to cope with the
    # extra middle element
    if (N & 1) :
        half += 1;
 
    ans = 1;
    for i in range(1, half + 1) :
        ans *= K;
 
        # K is reduced by one, because
        # count of choices for the next
        # position is  reduced by 1 as
        # a element can only once
        K -= 1;
 
    # Return the possible count
    return ans;
 
# Function to find the count of palindromic
# string of first K characters according
# to the given criteria
def palindromicStrings(N, K) :
 
    # If N=1, then only K palindromic
    # strings possible.
    if (N == 1) :
        return K;
 
    # If N=2, the 2*K palindromic strings
    # possible, K for N=1 and K for N=2
    if (N == 2) :
        return 2 * K;
 
    ans = 0;
 
    # Initialize ans with the count of
    # strings possible till N = 2
    ans += (2 * K);
 
    for i in range(3, N + 1) :
        ans += lengthNPalindrome(i, K);
 
    # Return the possible count
    return ans;
 
# Driver Code
if __name__ == "__main__" :
 
    N = 4; K = 3;
    print(palindromicStrings(N, K));
 
    # This code is contributed by AnkThon

C#




// C# program for the above approach
using System;
 
class GFG
{
   
    // Function of return the number of
    // palindromic strings of length N with
    // first K alphabets possible
    static int lengthNPalindrome(int N, int K)
    {
        int half = N / 2;
 
        // If N is odd, half + 1 position
        // can be filled to cope with the
        // extra middle element
        if (N % 2 == 1) {
            half += 1;
        }
 
        int ans = 1;
        for (int i = 1; i <= half; i++) {
            ans *= K;
 
            // K is reduced by one, because
            // count of choices for the next
            // position is  reduced by 1 as
            // a element can only once
            K--;
        }
 
        // Return the possible count
        return ans;
    }
 
    // Function to find the count of palindromic
    // string of first K characters according
    // to the given criteria
    static int palindromicStrings(int N, int K)
    {
        // If N=1, then only K palindromic
        // strings possible.
        if (N == 1) {
            return K;
        }
 
        // If N=2, the 2*K palindromic strings
        // possible, K for N=1 and K for N=2
        if (N == 2) {
            return 2 * K;
        }
 
        int ans = 0;
 
        // Initialize ans with the count of
        // strings possible till N = 2
        ans += (2 * K);
 
        for (int i = 3; i <= N; i++) {
            ans += lengthNPalindrome(i, K);
        }
 
        // Return the possible count
        return ans;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int N = 4, K = 3;
 
        Console.Write(palindromicStrings(N, K));
    }
}
 
// This code is contributed by shivanisinghss2110

Javascript




<script>
 
// Javascript program for the above approach
 
// Function of return the number of
// palindromic strings of length N with
// first K alphabets possible
function lengthNPalindrome(N,K)
{
    var half = N / 2;
 
    // If N is odd, half + 1 position
    // can be filled to cope with the
    // extra middle element
    if (N & 1) {
        half += 1;
    }
 
    var ans = 1;
    var i;
    for(i = 1; i <= half; i++) {
        ans *= K;
 
        // K is reduced by one, because
        // count of choices for the next
        // position is  reduced by 1 as
        // a element can only once
        K--;
    }
 
    // Return the possible count
    return ans;
}
 
// Function to find the count of palindromic
// string of first K characters according
// to the given criteria
function palindromicStrings(N, K)
{
    // If N=1, then only K palindromic
    // strings possible.
    if (N == 1) {
        return K;
    }
 
    // If N=2, the 2*K palindromic strings
    // possible, K for N=1 and K for N=2
    if (N == 2) {
        return 2 * K;
    }
 
    ans = 0;
 
    // Initialize ans with the count of
    // strings possible till N = 2
    ans += (2 * K);
 
    for (i = 3; i <= N; i++) {
        ans += lengthNPalindrome(i, K);
    }
 
    // Return the possible count
    return ans;
}
 
// Driver Code
    var N = 4, K = 3;
    document.write(palindromicStrings(N, K));
 
// This code is contributed by ipg2016107.
</script>

 
 

Output: 
18

 

 

Time Complexity: O(N2)
Auxiliary Space: O(1)

 




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