Sum of all N digit palindrome numbers
Last Updated :
24 Feb, 2022
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
#include <bits/stdc++.h>
using namespace std;
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;
}
long long getSum(int n)
{
int start = pow(10, n - 1);
int end = pow(10, n) - 1;
long long sum = 0;
for (int i = start; i <= end; i++) {
string s = to_string(i);
if (isPalindrome(s)) {
sum += i;
}
}
return sum;
}
int main()
{
int n = 1;
long long ans = getSum(n);
cout << ans << '\n';
return 0;
}
|
Java
import java.util.*;
class GFG
{
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;
}
static long getSum(int n)
{
int start = (int) Math.pow(10, n - 1);
int end = (int) (Math.pow(10, n) - 1);
long sum = 0;
for (int i = start; i <= end; i++)
{
String s = String.valueOf(i);
if (isPalindrome(s))
{
sum += i;
}
}
return sum;
}
public static void main(String[] args)
{
int n = 1;
long ans = getSum(n);
System.out.print(ans);
}
}
|
Python
import math
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
def getSum(n):
start = int(math.pow(10, n - 1))
end = int(math.pow(10, n)) - 1
sum = 0
for i in range(start, end + 1):
s = str(i)
if (isPalindrome(s)):
sum = sum + i
return sum
n = 1
ans = getSum(n)
print(ans)
|
C#
using System;
class GFG
{
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;
}
static long getSum(int n)
{
int start = (int) Math.Pow(10, n - 1);
int end = (int) (Math.Pow(10, n) - 1);
long sum = 0;
for (int i = start; i <= end; i++)
{
string s = i.ToString();;
if (isPalindrome(s))
{
sum += i;
}
}
return sum;
}
public static void Main()
{
int n = 1;
long ans = getSum(n);
Console.Write(ans);
}
}
|
Javascript
<script>
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 getSum(n)
{
var start = Math.pow(10, n - 1);
var end = Math.pow(10, n) - 1;
var sum = 0;
for (var i = start; i <= end; i++) {
var s = (i.toString());
if (isPalindrome(s)) {
sum += i;
}
}
return sum;
}
var n = 1;
var ans = getSum(n);
document.write( ans + "<br>");
</script>
|
Time Complexity: O(n*log(n))
Auxiliary Space: O(1)
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++
#include <bits/stdc++.h>
using namespace std;
long double getSum(int n)
{
long double sum = 0;
if (n == 1) {
sum = 45.0;
}
else {
sum = (99.0 / 2.0) * pow(10, n - 1)
* pow(10, (n - 1) / 2);
}
return sum;
}
int main()
{
int n = 3;
long double ans = getSum(n);
cout << setprecision(12) << ans << '\n';
return 0;
}
|
Java
class GFG
{
static double getSum(int n)
{
double sum = 0;
if (n == 1)
{
sum = 45.0;
}
else
{
sum = (99.0 / 2.0) *
Math.pow(10, n - 1) *
Math.pow(10, (n - 1) / 2);
}
return sum;
}
public static void main(String[] args)
{
int n = 3;
double ans = getSum(n);
System.out.print(ans);
}
}
|
Python3
def getSum(n):
sum = 0;
if (n == 1):
sum = 45.0;
else:
sum = (99.0 / 2.0) * pow(10, n - 1)\
* pow(10, (n - 1) / 2);
return sum;
if __name__ == '__main__':
n = 3;
ans = int(getSum(n));
print(ans);
|
C#
using System;
class GFG
{
static double getSum(int n)
{
double sum = 0;
if (n == 1)
{
sum = 45.0;
}
else
{
sum = (99.0 / 2.0) *
Math.Pow(10, n - 1) *
Math.Pow(10, (n - 1) / 2);
}
return sum;
}
public static void Main(String[] args)
{
int n = 3;
double ans = getSum(n);
Console.Write(ans);
}
}
|
Javascript
<script>
function getSum(n)
{
var sum = 0;
if (n == 1) {
sum = 45.0;
}
else {
sum = (99.0 / 2.0) * Math.pow(10, n - 1)
* Math.pow(10, parseInt((n - 1) / 2));
}
return sum;
}
var n = 3;
var ans = getSum(n);
document.write(ans + "<br>");
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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