# Sum of all N digit palindrome numbers

Given a number N. The task is to find the sum of all N digit palindromes.

Examples:

```Input: N = 2
Output: 495
Explanation:
11 + 22 + 33 + 44 + 55 +
66 + 77 + 88 + 99
= 495

Input: N = 7
Output: 49500000000
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach:

Run a loop from 10^(n-1) to 10^(n) – 1 and check when the current number is palindrome or not. If it adds it’s value to answer.

Below is the implementation of above approach:

## CPP

 `// C++ program for the ` `// above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to check ` `// palindrome ` `bool` `isPalindrome(string& s) ` `{ ` `    ``int` `left = 0, right = s.size() - 1; ` `    ``while` `(left <= right) { ` `        ``if` `(s[left] != s[right]) { ` `            ``return` `false``; ` `        ``} ` `        ``left++; ` `        ``right--; ` `    ``} ` `    ``return` `true``; ` `} ` ` `  `// Function to calculate ` `// the sum of n-digit ` `// palindrome ` `long` `long` `getSum(``int` `n) ` `{ ` ` `  `    ``int` `start = ``pow``(10, n - 1); ` `    ``int` `end = ``pow``(10, n) - 1; ` ` `  `    ``long` `long` `sum = 0; ` ` `  `    ``// Run a loop to check ` `    ``// all possible palindrome ` `    ``for` `(``int` `i = start; i <= end; i++) { ` `        ``string s = to_string(i); ` `        ``// If palndrome ` `        ``// append sum ` `        ``if` `(isPalindrome(s)) { ` `            ``sum += i; ` `        ``} ` `    ``} ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `n = 1; ` `    ``long` `long` `ans = getSum(n); ` `    ``cout << ans << ``'\n'``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program for the ` `// above approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to check ` `// palindrome ` `static` `boolean` `isPalindrome(String s) ` `{ ` `    ``int` `left = ``0``, right = s.length() - ``1``; ` `    ``while` `(left <= right) ` `    ``{ ` `        ``if` `(s.charAt(left) != s.charAt(right)) ` `        ``{ ` `            ``return` `false``; ` `        ``} ` `        ``left++; ` `        ``right--; ` `    ``} ` `    ``return` `true``; ` `} ` ` `  `// Function to calculate ` `// the sum of n-digit ` `// palindrome ` `static` `long` `getSum(``int` `n) ` `{ ` ` `  `    ``int` `start = (``int``) Math.pow(``10``, n - ``1``); ` `    ``int` `end = (``int``) (Math.pow(``10``, n) - ``1``); ` ` `  `    ``long` `sum = ``0``; ` ` `  `    ``// Run a loop to check ` `    ``// all possible palindrome ` `    ``for` `(``int` `i = start; i <= end; i++) ` `    ``{ ` `        ``String s = String.valueOf(i); ` `         `  `        ``// If palndrome ` `        ``// append sum ` `        ``if` `(isPalindrome(s)) ` `        ``{ ` `            ``sum += i; ` `        ``} ` `    ``} ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``1``; ` `    ``long` `ans = getSum(n); ` `    ``System.out.print(ans); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python

 `# Python program for the above approach  ` `import` `math ` ` `  `# Function to check  ` `# palindrome  ` `def` `isPalindrome(s): ` `    ``left ``=` `0` `    ``right ``=` `len``(s) ``-` `1` `    ``while` `(left <``=` `right): ` `        ``if` `(s[left] !``=` `s[right]): ` `            ``return` `False` `         `  `        ``left ``=` `left ``+` `1` `        ``right ``=` `right ``-` `1` ` `  `    ``return` `True` ` `  `# Function to calculate  ` `# the sum of n-digit  ` `# palindrome  ` `def` `getSum(n): ` `    ``start ``=` `int``(math.``pow``(``10``, n ``-` `1``))  ` `    ``end ``=` `int``(math.``pow``(``10``, n)) ``-` `1` ` `  `    ``sum` `=` `0` ` `  `    ``# Run a loop to check  ` `    ``# all possible palindrome  ` `    ``for` `i ``in` `range``(start, end ``+` `1``): ` `        ``s ``=` `str``(i)  ` `         `  `        ``# If palndrome  ` `        ``# append sum  ` `        ``if` `(isPalindrome(s)): ` `            ``sum` `=` `sum` `+` `i ` ` `  `    ``return` `sum` ` `  `# Driver code  ` ` `  `n ``=` `1` `ans ``=` `getSum(n) ` `print``(ans) ` ` `  `# This code is contributed by Sanjit_Prasad `

## C#

 `// C# program for the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to check ` `    ``// palindrome ` `    ``static` `bool` `isPalindrome(``string` `s) ` `    ``{ ` `        ``int` `left = 0, right = s.Length - 1; ` `        ``while` `(left <= right) ` `        ``{ ` `            ``if` `(s[left] != s[right]) ` `            ``{ ` `                ``return` `false``; ` `            ``} ` `            ``left++; ` `            ``right--; ` `        ``} ` `        ``return` `true``; ` `    ``} ` `     `  `    ``// Function to calculate ` `    ``// the sum of n-digit ` `    ``// palindrome ` `    ``static` `long` `getSum(``int` `n) ` `    ``{ ` `     `  `        ``int` `start = (``int``) Math.Pow(10, n - 1); ` `        ``int` `end = (``int``) (Math.Pow(10, n) - 1); ` `     `  `        ``long` `sum = 0; ` `     `  `        ``// Run a loop to check ` `        ``// all possible palindrome ` `        ``for` `(``int` `i = start; i <= end; i++) ` `        ``{ ` `            ``string` `s = i.ToString();; ` `             `  `            ``// If palndrome ` `            ``// append sum ` `            ``if` `(isPalindrome(s)) ` `            ``{ ` `                ``sum += i; ` `            ``} ` `        ``} ` `     `  `        ``return` `sum; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 1; ` `        ``long` `ans = getSum(n); ` `        ``Console.Write(ans); ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```45
```

Time Complexity: O(n*log(n))

Efficient approach:

On carefully observing the sum of n digit palindrome a series is formed i.e 45, 495, 49500, 495000, 49500000, 495000000. Therefore, on deducing this to a mathematical formula we get for n = 1 sum = 45 and for n > 1 put sum = (99/2)*10^n-1*10^(n-1)/2

Below is the implementation of above approach:

## C++

 `// C++ program for ` `// the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to calculate ` `// sum of n digit number ` `long` `double` `getSum(``int` `n) ` `{ ` ` `  `    ``long` `double` `sum = 0; ` `    ``// Corner case ` `    ``if` `(n == 1) { ` `        ``sum = 45.0; ` `    ``} ` `    ``// Using above approach ` `    ``else` `{ ` `        ``sum = (99.0 / 2.0) * ``pow``(10, n - 1) ` `              ``* ``pow``(10, (n - 1) / 2); ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `n = 3; ` `    ``long` `double` `ans = getSum(n); ` `    ``cout << setprecision(12) << ans << ``'\n'``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program for ` `// the above approach ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to calculate ` `    ``// sum of n digit number ` `    ``static` `double` `getSum(``int` `n) ` `    ``{ ` ` `  `        ``double` `sum = ``0``; ` `         `  `        ``// Corner case ` `        ``if` `(n == ``1``)  ` `        ``{ ` `            ``sum = ``45.0``; ` `        ``} ` `         `  `        ``// Using above approach ` `        ``else`  `        ``{ ` `            ``sum = (``99.0` `/ ``2.0``) *  ` `                    ``Math.pow(``10``, n - ``1``) *  ` `                    ``Math.pow(``10``, (n - ``1``) / ``2``); ` `        ``} ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``3``; ` `        ``double` `ans = getSum(n); ` `        ``System.out.print(ans); ` `    ``} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python program for ` `# the above approach ` ` `  `# Function to calculate ` `# sum of n digit number ` `def` `getSum(n): ` ` `  `    ``sum` `=` `0``; ` ` `  `    ``# Corner case ` `    ``if` `(n ``=``=` `1``): ` `        ``sum` `=` `45.0``; ` `     `  `    ``# Using above approach ` `    ``else``: ` `        ``sum` `=` `(``99.0` `/` `2.0``) ``*` `pow``(``10``, n ``-` `1``)\ ` `        ``*` `pow``(``10``, (n ``-` `1``) ``/` `2``); ` `     `  `    ``return` `sum``; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `3``; ` `    ``ans ``=` `int``(getSum(n)); ` `    ``print``(ans); ` ` `  `# This code is contributed by 29AjayKumar `

## C#

 `// C# program for ` `// the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to calculate ` `    ``// sum of n digit number ` `    ``static` `double` `getSum(``int` `n) ` `    ``{ ` `        ``double` `sum = 0; ` `         `  `        ``// Corner case ` `        ``if` `(n == 1)  ` `        ``{ ` `            ``sum = 45.0; ` `        ``} ` `         `  `        ``// Using above approach ` `        ``else` `        ``{ ` `            ``sum = (99.0 / 2.0) *  ` `                    ``Math.Pow(10, n - 1) *  ` `                    ``Math.Pow(10, (n - 1) / 2); ` `        ``} ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int` `n = 3; ` `        ``double` `ans = getSum(n); ` `        ``Console.Write(ans); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```49500
```

Time Complexity: O(1)

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