Count all possible N digit numbers that satisfy the given condition - GeeksforGeeks

19 May, 2021
Given an integer N, the task is to count all possible N digit numbers such that A + reverse(A) = 10N – 1 where A is an N digit number and reverse(A) is reverse of A. A shouldn’t have any leading 0s.Examples:  
Input: N = 2 Output: 9 All possible 2 digit numbers are 90, 81, 72, 63, 54, 45, 36, 27 and 18.Input: N = 4 Output: 90  
 
Approach: First we have to conclude that if N is odd then there is no number which will satisfy the given condition, let’s prove it for N = 3,  
, so and . which is impossible as it is a floating point number.  
Now Find answer for when N is even. For example, N=4,  
and now if x + y = 9 then the number of pairs which satisfy this condition are 10. (0, 9), (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0) Now, 1st and Nth digit cannot have the pair (0, 9) as there shouldn’t be any leading 0s in A but for all the remaining N/2-1 pairs there can be 10 pairs. So the answer is , As N is large so we will print 9 followed by N/2-1 number of 0s.  
Below is the implementation of the above approach: 
C++
Java
Python3
C#
PHP
Javascript
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std; // Function to return the count of required numbersstring getCount(int N){     // If N is odd then return 0    if (N % 2 == 1)        return 0;     string result = "9";     for (int i = 1; i <= N / 2 - 1; i++)        result += "0";     return result;} // Driver Codeint main(){     int N = 4;    cout << getCount(N);     return 0;}
// Java implementation of above approachclass GFG{    // Function to return the count of required numbers    static String getCount(int N)    {             // If N is odd then return 0        if (N % 2 == 1)            return "0";             String result = "9";        for (int i = 1; i <= N / 2 - 1; i++)            result += "0";        return result;    }         // Driver Code    public static void main(String []args)    {             int N = 4;        System.out.println(getCount(N));    }} // This code is contributed by ihritik
# Python3 implementation of above approach # Function to return the count of required numbersdef getCount(N):     # If N is odd then return 0    if (N % 2 == 1):        return "0"     result = "9"     for i in range (1, N // 2 ):        result = result + "0"     return result # Driver CodeN = 4print(getCount(N)) # This code is contributed by ihritik
// C# implementation of above approachusing System; class GFG{    // Function to return the count of required numbers    static string getCount(int N)    {             // If N is odd then return 0        if (N % 2 == 1)            return "0";        string result = "9";        for (int i = 1; i <= N / 2 - 1; i++)            result += "0";        return result;    }         // Driver Code    public static void Main()    {             int N = 4;        Console.WriteLine(getCount(N));    }} // This code is contributed by ihritik
<?php// PHP implementation of above approach // Function to return the count of// required numbersfunction getCount($N){     // If N is odd then return 0    if ($N % 2 == 1)        return 0;     $result = "9";     for ($i = 1; $i <= $N / 2 - 1; $i++)        $result .= "0";     return $result;} // Driver Code$N = 4;echo getCount($N); // This code is contributed by Ryuga?>
<script> // Javascript implementation of the approach     // Function to return the count of required numbers    function getCount(N)    {             // If N is odd then return 0        if (N % 2 == 1)            return "0";             let result = "9";        for (let i = 1; i <= N / 2 - 1; i++)            result += "0";        return result;    }      // Driver code     let N = 4;    document.write(getCount(N));     </script>
90
 
Time Complexity: O(N) 
ihritik
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ManasChhabra2
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