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Sum of all N digit palindrome numbers

  • Difficulty Level : Easy
  • Last Updated : 03 Aug, 2021

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 

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 its value to the answer.

Below is the implementation of the above approach:  

CPP




// C++ program for the
// above approach
#include <bits/stdc++.h>
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 palindrome
        // 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 palindrome
        // 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 palindrome
        # 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 palindrome
            // 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

Javascript




<script>
 
// JavaScript program for the
// above approach
 
// Function to check
// palindrome
function isPalindrome(s)
{
    var 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
function getSum(n)
{
 
    var start = Math.pow(10, n - 1);
    var end = Math.pow(10, n) - 1;
 
    var sum = 0;
 
    // Run a loop to check
    // all possible palindrome
    for (var i = start; i <= end; i++) {
        var s = (i.toString());
        // If palindrome
        // append sum
        if (isPalindrome(s)) {
            sum += i;
        }
    }
 
    return sum;
}
 
// Driver code
var n = 1;
var ans = getSum(n);
document.write( ans + "<br>");
 
 
</script>
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, by 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 the above approach:  

C++




// C++ program for
// the above approach
#include <bits/stdc++.h>
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

Javascript




<script>
 
 
// Javascript program for
// the above approach
 
// Function to calculate
// sum of n digit number
function getSum(n)
{
 
    var 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, parseInt((n - 1) / 2));
    }
    return sum;
}
 
// Driver code
var n = 3;
var ans = getSum(n);
document.write(ans + "<br>");
 
 
</script>
Output: 
49500

 

Time Complexity: O(1)
 


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