N-th character in the string made by concatenating natural numbers

Given an integer N, the task is to find the N-th character in the string made by concatenating natural numbers (Integers beginning from 1). The starting string will be “12345678910111213..”. 

Examples: 

Input: N = 3 
Output: 3
3rd character in the string "12345678910111213.." is 3.

Input: N = 11
Output: 0
11th character in the string "12345678910111213..." is 0 

The idea is to generate the required string until string length does exceed N. 

• Initialize a Null string and c=1
• Add the c to the string by type-casting it to a character
• If c is a single-digit number, append it to the string
• if c is greater than 9, then store it in a temporary string and reverse it and append to the original string
• If at any moment the string length exceeds N, then return s[n-1].

Below is the implementation of the above approach:  

C++

// C++ program to find the N-th character
// in the string "1234567891011.."
#include <bits/stdc++.h>
using namespace std;

// Function that returns the N-th character
char NthCharacter(int n)
{
    // initially null string
    string s = "";

    // starting integer
    int c = 1;

    // add integers in string
    for (int i = 1;; i++) {

        // one digit numbers added
        if (c < 10)
            s += char(48 + c);

        // more than 1 digit number, generate
        // equivalent number in a string s1
        // and concatenate s1 into s.
        else
        {
            string s1 = "";
            int dup = c;

            // add the number in string
            while (dup) {
                s1 += char((dup % 10) + 48);
                dup /= 10;
            }

            // reverse the string
            reverse(s1.begin(), s1.end());

            // attach the number
            s += s1;
        }
        c++;

        // if the length exceeds N
        if (s.length() >= n) {
            return s[n - 1];
        }
    }
}

// Driver Code
int main()
{
    int n = 11;

    cout << NthCharacter(n);

    return 0;
}

Java

// Java program to find the N-th character
// in the string "1234567891011.."


class GFG
{
    // Function that returns the N-th character
    static char NthCharacter(int n)
    {
        // initially null string
        String s = "";

        // starting integer
        int c = 1;

        // add integers in string
        for (int i = 1;; i++) {

            // one digit numbers added
            if (c < 10)
                s += Integer.toString(c);

            // more than 1 digit number, generate
            // equivalent number in a string s1
            // and concatenate s1 into s.
            else
            {
                String s1 = "";
                int dup = c;

                // add the number in string
                while (dup >0) {
                    s1 += Integer.toString(dup % 10);
                    dup /= 10;
                }

                // reverse the string
                StringBuilder temp = new StringBuilder();
                temp.append(s1);
                temp = temp.reverse();

                // attach the number
                s += temp;
            }
            c++;

            // if the length exceeds N
            if (s.length() >= n) {
                return s.charAt(n - 1);
            }
        }
    }

    // Driver Code
    public static void main(String []args)
    {
        int n = 11;

        System.out.println( NthCharacter(n));


    }

}

// This article is contributed by ihritik

Python3

# Python 3 program to find the N-th character
# in the string "1234567891011.."

# Function that returns the N-th character
def NthCharacter(n):

    # initially null string
    s = ""

    # starting integer
    c = 1

    # add integers in string
    while(True) :

        # one digit numbers added
        if (c < 10):
            s += chr(48 + c)

        # more than 1 digit number, generate
        # equivalent number in a string s1
        # and concatenate s1 into s.
        else:
            s1 = ""
            dup = c

            # add the number in string
            while (dup > 0):
                s1 += chr((dup % 10) + 48)
                dup //= 10

            # reverse the string
            s1 = "".join(reversed(s1))

            # attach the number
            s += s1
        c += 1

        # if the length exceeds N
        if (len(s) >= n):
            return s[n - 1]

# Driver Code
if __name__ == "__main__":

    n = 11
    print(NthCharacter(n))

# This code is contributed by ita_c

C#

// C# program to find the N-th character
// in the string "1234567891011.."
using System;

class GFG
{
    // Function that returns the N-th character
    static char NthCharacter(int n)
    {
        // initially null string
        String s = "";

        // starting integer
        int c = 1;

        // add integers in string
        for (int i = 1;; i++)
        {

            // one digit numbers added
            if (c < 10)
                s += c.ToString();

            // more than 1 digit number, generate
            // equivalent number in a string s1
            // and concatenate s1 into s.
            else
            {
                String s1 = "";
                int dup = c;

                // add the number in string
                while (dup > 0)
                {
                    s1 += (dup % 10).ToString();
                    dup /= 10;
                }

                // reverse the string
                String temp = reverse(s1);

                // attach the number
                s += temp;
            }
            c++;

            // if the length exceeds N
            if (s.Length >= n)
            {
                return s[n - 1];
            }
        }
    }

    static String reverse(String input)
    {
        char[] a = input.ToCharArray();
        int l, r = 0;
        r = a.Length - 1;

        for (l = 0; l < r; l++, r--)
        {

            // Swap values of l and r
            char temp = a[l];
            a[l] = a[r];
            a[r] = temp;
        }
        return String.Join("", a);
    }

    // Driver Code
    public static void Main(String []args)
    {
        int n = 11;

        Console.WriteLine( NthCharacter(n));
    }
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript program to find the N-th character
// in the string "1234567891011.."

// Function that returns the N-th character
function NthCharacter(n)
{
    // initially null string
        let s = "";

        // starting integer
        let c = 1;

        // add integers in string
        for (let i = 1;; i++) {

            // one digit numbers added
            if (c < 10)
                s += c.toString();

            // more than 1 digit number, generate
            // equivalent number in a string s1
            // and concatenate s1 into s.
            else
            {
                let s1 = "";
                let dup = c;

                // add the number in string
                while (dup >0) {
                    s1 += (dup % 10).toString();
                    dup = Math.floor(dup/10);
                }

                // reverse the string
                s1=s1.split("").reverse().join("");

                // attach the number
                s += s1;
            }
            c++;

            // if the length exceeds N
            if (s.length >= n) {
                return s[n-1];
            }
        }
}

// Driver Code

let n = 11;
document.write( NthCharacter(n));



// This code is contributed by avanitrachhadiya2155

</script>

Output

0

Complexity Analysis:

• Time Complexity: O(N²), as we are traversing N times using a loop and in each traversal we are using reverse function which will cost O (N), so in the worst case the program will effectively cost O (N*N).
• Auxiliary Space: O(N), as we are using extra space for the string.