Given a positive integer N, the task is to find the number of alphanumeric palindromic strings of length N. Since the count of such strings can be very large, print the answer modulo 109 + 7.
Examples:
Input: N = 2
Output: 62
Explanation: There are 26 palindromes of the form {“AA”, “BB”, …, “ZZ”}, 26 palindromes of the form {“aa”, “bb”, …, “cc”} and 10 palindromes of the form {“00”, “11”, …, “99”}. Therefore, the total number of palindromic strings = 26 + 26 + 10 = 62.Input: N = 3
Output: 3844
Naive Approach: The simplest approach is to generate all possible alphanumeric strings of length N and for each string, check if it is a palindrome or not. Since, at each position, 62 characters can be placed in total. Hence, there are 62N possible strings.
Time Complexity: O(N*62N)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to use the property of palindrome. It can be observed that if the length of the string is even, then for each index i from 0 to N/2, characters at indices i and (N – 1 – i) are the same. Therefore, for each position from 0 to N/ 2 there are 62N/2 options. Similarly, if the length is odd, 62(N+1)/2 options are there. Hence, it can be said that, for some N, there are 62ceil(N/2) possible palindromic strings.
Follow the steps below to solve the problem:
- For the given value of N, calculate 62ceil(N/2) mod 109 + 7 using Modular Exponentiation.
- Print 62ceil(N/2) mod 109 + 7 as the required answer.
Below is the implementation of the above approach:
C++14
// C++ program for the // above approach#include <bits/stdc++.h> using namespace std; // Function to calculate // (x ^ y) mod pint power(int x, int y, int p){ // Initialize result int res = 1; // Update x if it is more // than or equal to p x = x % p; if (x == 0) return 0; while (y > 0) { // If y is odd, multiply // x with result if ((y & 1) == 1) res = (res * x) % p; // y must be even now y = y >> 1; x = (x * x) % p; } // Return the final // result return res;}// Driver Codeint main() { // Given N int N = 3; int flag, k, m; // Base Case if((N == 1) || (N == 2)) cout << 62; else m = 1000000000 + 7; // Check whether n // is even or odd if(N % 2 == 0) { k = N / 2; flag = true; } else { k = (N - 1) / 2; flag = false; } if(flag != 0) { // Function Call int a = power(62, k, m); cout << a; } else { // Function Call int a = power(62, (k + 1), m); cout << a; }}// This code is contributed by sanjoy_62 |
Java
// Java program for the // above approachimport java.util.*;class GFG{// Function to calculate// (x ^ y) mod pstatic int power(int x, int y, int p) { // Initialize result int res = 1; // Update x if it is more // than or equal to p x = x % p; if (x == 0) return 0; while (y > 0) { // If y is odd, multiply // x with result if ((y & 1) == 1) res = (res * x) % p; // y must be even now y = y >> 1; x = (x * x) % p; } // Return the final // result return res;}// Driver Codepublic static void main(String[] args) { // Given N int N = 3; int flag, k, m =0; // Base Case if ((N == 1) || (N == 2)) System.out.print(62); else m = 1000000000 + 7; // Check whether n // is even or odd if (N % 2 == 0) { k = N / 2; flag = 1; } else { k = (N - 1) / 2; flag = 0; } if (flag != 0) { // Function Call int a = power(62, k, m); System.out.print(a); } else { // Function Call int a = power(62, (k + 1), m); System.out.print(a); }}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approach# Function to calculate (x ^ y) mod pdef power(x, y, p): # Initialize result res = 1 # Update x if it is more # than or equal to p x = x % p if (x == 0): return 0 while (y > 0): # If y is odd, multiply # x with result if ((y & 1) == 1): res = (res * x) % p # y must be even now y = y >> 1 x = (x * x) % p # Return the final result return res# Driver Code# Given NN = 3# Base Caseif((N == 1) or (N == 2)): print(62) else: m = (10**9)+7 # Check whether n # is even or odd if(N % 2 == 0): k = N//2 flag = True else: k = (N - 1)//2 flag = False if(flag): # Function Call a = power(62, k, m) print(a) else: # Function Call a = power(62, (k + 1), m) print(a) |
C#
// C# program for the // above approach using System;class GFG{// Function to calculate// (x ^ y) mod pstatic int power(int x, int y, int p) { // Initialize result int res = 1; // Update x if it is more // than or equal to p x = x % p; if (x == 0) return 0; while (y > 0) { // If y is odd, multiply // x with result if ((y & 1) == 1) res = (res * x) % p; // y must be even now y = y >> 1; x = (x * x) % p; } // Return the final // result return res;}// Driver Codepublic static void Main() { // Given N int N = 3; int flag, k, m = 0; // Base Case if ((N == 1) || (N == 2)) Console.Write(62); else m = 1000000000 + 7; // Check whether n // is even or odd if (N % 2 == 0) { k = N / 2; flag = 1; } else { k = (N - 1) / 2; flag = 0; } if (flag != 0) { // Function Call int a = power(62, k, m); Console.Write(a); } else { // Function Call int a = power(62, (k + 1), m); Console.Write(a); }}}// This code is contributed by code_hunt |
Javascript
<script>// JavaScript program to implement// the above approach// Function to calculate// (x ^ y) mod pfunction power(x, y, p){ // Initialize result let res = 1; // Update x if it is more // than or equal to p x = x % p; if (x == 0) return 0; while (y > 0) { // If y is odd, multiply // x with result if ((y & 1) == 1) res = (res * x) % p; // y must be even now y = y >> 1; x = (x * x) % p; } // Return the final // result return res;}// Driver Code // Given N let N = 3; let flag, k, m =0; // Base Case if ((N == 1) || (N == 2)) document.write(62); else m = 1000000000 + 7; // Check whether n // is even or odd if (N % 2 == 0) { k = N / 2; flag = 1; } else { k = (N - 1) / 2; flag = 0; } if (flag != 0) { // Function Call let a = power(62, k, m); document.write(a); } else { // Function Call let a = power(62, (k + 1), m); document.write(a); } </script> |
3844
Time Complexity: O(log N)
Auxiliary Space: O(N)
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