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 sting 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

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

The idea is to generate the required string until string length does exceeds 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 

Python 3

# 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 

Output:

My Personal Notes

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

C++

Java

Python 3

C#

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