Given a number N, the task is to find the sum of all N-digit palindromic numbers which are divisible by 9 and the number doesn’t contain 0 in it.
Note: The sum can be very large, take modulo with 109+7.
Example:
Input: N = 2
Output: 99
Explanation:
There is only one 2-digit palindromic number divisible by 9 is 99.Input: N = 3
Output: 4995
Naive Approach: The idea is to iterate over all the N-digit numbers and check whether the numbers are palindromic and divisible by 9 and doesn’t contain zero in it. If yes then the summation of all the numbers given the required result.
Efficient Approach: Below are the some observation for this problem statement:
- If N is odd, then the number can be of the form “ab..x..ba” and if N is even then the number can be “ab..xx..ba”, where ‘a’, ‘b’ and ‘x’ can be replaced by digit from 1 to 9.
- If number “ab..x..ba” divisible by 9, (2*(a+b) + x) must be divisible by 9.
- For every possible pair of (a, b) there always exist one such ‘x’ so that the number is divisible by 9.
- Then the count of N-digit palindromic numbers which are divisible by 9 can be calculated from below formula:
Since we have 9 digits to fill at the position a and b. Therefore, total possible combinations will be: if N is Odd then, count = pow(9, N/2) if N is Even then, count = pow(9, N/2 - 1) as N/2 digits are repeated at the end to get the number palindromic.
Proof of Uniqueness of ‘x’:
- Minimum value of a and b is 1, therefore minimum sum = 2.
- Maximum value of a and b is 9, therefore minimum sum = 18.
- As the numbers are palindromic, then the sum is given by:
sum = 2*(a+b) + x;
- 2*(a+b) will be of the form 2, 4, 6, 8, …, 36. To make sum divisible by 9 the value of x can be:
sum one of the possible value of x sum + x 2 7 9 4 5 9 6 3 9 8 1 9 . . . . . . . . . 36 9 45
Below are the steps:
- After the above observations, the sum of the digit of all N-digit palindromic number which is divisible by 9 at any kth position is given by:
sum at kth position = 45*(number of combinations)/9
- Iterate a loop over [1, N] and find the number of possible combination to placed a digit at current index.
- Find the sum of all the digits at current digit using the above mentioned formula.
- The summation of all the sum calculated at each iteration given the required result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;long long int MOD = 1000000007;// Function to find a^b efficientlylong long int power(long long int a, long long int b){ // Base Case if (b == 0) return 1; // To store the value of a^b long long int temp = power(a, b / 2); temp = (temp * temp) % MOD; // Multiply temp by a until b is not 0 if (b % 2 != 0) { temp = (temp * a) % MOD; } // Return the final ans a^b return temp;}// Function to find sum of all N-digit// palindromic number divisible by 9void palindromicSum(int N){ long long int sum = 0, res, ways; // Base Case if (N == 1) { cout << "9" << endl; return; } if (N == 2) { cout << "99" << endl; return; } ways = N / 2; // If N is even, decrease ways by 1 if (N % 2 == 0) ways--; // Find the total number of ways res = power(9, ways - 1); // Iterate over [1, N] and find the // sum at each index for (int i = 0; i < N; i++) { sum = sum * 10 + 45 * res; sum %= MOD; } // Print the final Sum cout << sum << endl;}// Driver Codeint main(){ int N = 3; palindromicSum(N); return 0;} |
Java
// Java program for the above approachclass GFG{static int MOD = 1000000007;// Function to find a^b efficientlystatic int power(int a, int b){ // Base Case if (b == 0) return 1; // To store the value of a^b int temp = power(a, b / 2); temp = (temp * temp) % MOD; // Multiply temp by a until b is not 0 if (b % 2 != 0) { temp = (temp * a) % MOD; } // Return the final ans a^b return temp;}// Function to find sum of all N-digit// palindromic number divisible by 9static void palindromicSum(int N){ int sum = 0, res, ways; // Base Case if (N == 1) { System.out.print("9" + "\n"); return; } if (N == 2) { System.out.print("99" + "\n"); return; } ways = N / 2; // If N is even, decrease ways by 1 if (N % 2 == 0) ways--; // Find the total number of ways res = power(9, ways - 1); // Iterate over [1, N] and find the // sum at each index for (int i = 0; i < N; i++) { sum = sum * 10 + 45 * res; sum %= MOD; } // Print the final Sum System.out.print(sum + "\n");}// Driver Codepublic static void main(String[] args){ int N = 3; palindromicSum(N);}}// This code is contributed by Amit Katiyar |
Python3
# Python 3 program for the above approachMOD = 1000000007# Function to find a^b efficientlydef power(a, b): # Base Case if (b == 0): return 1 # To store the value of a^b temp = power(a, b // 2) temp = (temp * temp) % MOD # Multiply temp by a until b is not 0 if (b % 2 != 0): temp = (temp * a) % MOD # Return the final ans a^b return temp# Function to find sum of all N-digit# palindromic number divisible by 9def palindromicSum(N): sum = 0 # Base Case if (N == 1): print("9") return if (N == 2): print("99") return ways = N // 2 # If N is even, decrease ways by 1 if (N % 2 == 0): ways -= 1 # Find the total number of ways res = power(9, ways - 1) # Iterate over [1, N] and find the # sum at each index for i in range(N): sum = sum * 10 + 45 * res sum %= MOD # Print the final Sum print(sum)# Driver Codeif __name__ == "__main__": N = 3 palindromicSum(N)# This code is contributed by chitranayal |
C#
// C# program for the above approachusing System;class GFG{static int MOD = 1000000007;// Function to find a^b efficientlystatic int power(int a, int b){ // Base Case if (b == 0) return 1; // To store the value of a^b int temp = power(a, b / 2); temp = (temp * temp) % MOD; // Multiply temp by a until b is not 0 if (b % 2 != 0) { temp = (temp * a) % MOD; } // Return the readonly ans a^b return temp;}// Function to find sum of all N-digit// palindromic number divisible by 9static void palindromicSum(int N){ int sum = 0, res, ways; // Base Case if (N == 1) { Console.Write("9" + "\n"); return; } if (N == 2) { Console.Write("99" + "\n"); return; } ways = N / 2; // If N is even, decrease ways by 1 if (N % 2 == 0) ways--; // Find the total number of ways res = power(9, ways - 1); // Iterate over [1, N] and find the // sum at each index for(int i = 0; i < N; i++) { sum = sum * 10 + 45 * res; sum %= MOD; } // Print the readonly sum Console.Write(sum + "\n");}// Driver Codepublic static void Main(String[] args){ int N = 3; palindromicSum(N);}}// This code is contributed by AbhiThakur |
Javascript
<script>// Javascript program for the above approachlet MOD = 1000000007; // Function to find a^b efficientlyfunction power(a, b){ // Base Case if (b == 0) return 1; // To store the value of a^b let temp = power(a, b / 2); temp = (temp * temp) % MOD; // Multiply temp by a until b is not 0 if (b % 2 != 0) { temp = (temp * a) % MOD; } // Return the final ans a^b return temp;} // Function to find sum of all N-digit// palindromic number divisible by 9function palindromicSum(N){ let sum = 0, res, ways; // Base Case if (N == 1) { document.write("9" + "\n"); return; } if (N == 2) { document.write("99" + "\n"); return; } ways = Math.floor(N / 2); // If N is even, decrease ways by 1 if (N % 2 == 0) ways--; // Find the total number of ways res = power(9, ways - 1); // Iterate over [1, N] and find the // sum at each index for (let i = 0; i < N; i++) { sum = sum * 10 + 45 * res; sum %= MOD; } // Print the final Sum document.write(sum + "<br/>");}// Driver Code let N = 3; palindromicSum(N); </script> |
Output:
4995
Time Complexity: O(N), where N is the given number.
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
