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

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++

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// 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;
}

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Java














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// 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 



















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Python 3














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# 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 



















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    Article Tags : 
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Practice Tags : 
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