Nth character in Concatenated Decimal String

If all decimal numbers are concatenated in a string then we will get a string which looks like string P as shown below. We need to tell Nth character of this string.
P = “12345678910111213141516171819202122232425262728293031….”
Examples:

N = 10    10th character is 1
N = 11    11th character is 0
N = 50    50th character is 3
N = 190    190th character is 1

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

We can solve this problem by break-ing the string length-wise. We know that in decimal 9 numbers are of length 1, 90 numbers are of length 2, 900 numbers are of length 3 and so on, so we can skip these numbers according to the given N and can get the desired character.

Processing for N = 190 is explained below,
P[184..195] = “979899100101”
First getting length of number at N,
190 – 9 = 181        number length is more than 1
181 – 90*2 = 1       number length is more than 2
1 – 900*3 < 0        number length is 3
Now getting actual character at N,
1 character after maximum 2 length number(99) is,  1

Processing for N = 251 is explained below,
P[250..255] = “120121”
First getting length of number at N,
251 - 9 = 242             number length is more than 1
242 – 90*2 = 62           number length is more than 2
62 – 900*3 < 0            number length is 3
Now getting actual character at N,
62 characters after maximum 2 length number(99) is,
Ceil(62/3) = 21,  99 + 21 = 120
120 is the number at N, now getting actual digit,
62%3 = 2,
2nd digit of 120 is 2, so our answer will be 2 only.

Please see below code for better understanding,

C/C++

 // C++ program to get Nth character in // concatenated Decimal String #include using namespace std;    //  Utility method to get dth digit of number N char getDigit(int N, int d) {     string str;     stringstream ss;     ss << N;     ss >> str;     return str[d - 1]; }    // Method to return Nth character in concatenated // decimal string char getNthChar(int N) {     //  sum will store character escaped till now     int sum = 0, nine = 9;        //  dist will store numbers escaped till now     int dist = 0, len;        //  loop for number lengths     for (len = 1; ; len++)     {         // nine*len will be incremented characters         // and nine will be incremented numbers         sum += nine*len;         dist += nine;            if (sum >= N)         {             //  restore variables to previous correct state             sum -= nine*len;             dist -= nine;             N -= sum;             break;         }         nine *= 10;     }        // get distance from last one digit less maximum     // number     int diff = ceil((double)N / len);        // d will store dth digit of current number     int d = N % len;     if (d == 0)         d = len;        // method will return dth numbered digit     // of (dist + diff) number     return getDigit(dist + diff, d); }    //  Driver code to test above methods int main() {     int N = 251;     cout << getNthChar(N) << endl;     return 0; }

Java

 // Java program to get Nth character in  // concatenated Decimal String     class GFG {        // Utility method to get dth digit of number N  static char getDigit(int N, int d)  {      String str=Integer.toString(N);     return str.charAt(d - 1);  }     // Method to return Nth character in concatenated  // decimal string  static char getNthChar(int N)  {      // sum will store character escaped till now      int sum = 0, nine = 9;         // dist will store numbers escaped till now      int dist = 0, len;         // loop for number lengths      for (len = 1; ; len++)      {          // nine*len will be incremented characters          // and nine will be incremented numbers          sum += nine * len;          dist += nine;             if (sum >= N)          {              // restore variables to previous correct state              sum -= nine * len;              dist -= nine;              N -= sum;              break;          }          nine *= 10;      }         // get distance from last one digit      // less maximum number      int diff = (int)(Math.ceil((double)(N) / (double)(len)));         // d will store dth digit of current number      int d = N % len;      if (d == 0)          d = len;         // method will return dth numbered digit      // of (dist + diff) number      return getDigit(dist + diff, d);  }     // Driver code public static void main (String[] args) {     int N = 251;      System.out.println(getNthChar(N));  }  }    // This code is contributed by mits

Python

 # Python program to get Nth character in # concatenated Decimal String     # Method to get dth digit of number N def getDigit(N, d):     string = str(N)     return string[d-1];     # Method to return Nth character in concatenated # decimal string def getNthChar(N):        #  sum will store character escaped till now     sum = 0     nine = 9         #  dist will store numbers escaped till now     dist = 0         #  loop for number lengths     for len in range(1,N):                # nine*len will be incremented characters         # and nine will be incremented numbers         sum += nine*len         dist += nine         if (sum >= N):                        #  restore variables to previous correct state             sum -= nine*len             dist -= nine             N -= sum             break                        nine *= 10         # get distance from last one digit less maximum     # number     diff = (N / len) + 1         # d will store dth digit of current number     d = N % len     if (d == 0):         d = len         # method will return dth numbered digit     # of (dist + diff) number     return getDigit(dist + diff, d);     #  Driver code to test above methods N = 251 print getNthChar(N)    # Contributed by Afzal_Saan

C#

 // C# program to get Nth character in  // concatenated Decimal String  using System;    class GFG {        // Utility method to get dth digit of number N  static char getDigit(int N, int d)  {      string str = Convert.ToString(N);     return str[d - 1];  }     // Method to return Nth character in  // concatenated decimal string  static char getNthChar(int N)  {      // sum will store character     // escaped till now      int sum = 0, nine = 9;         // dist will store numbers      // escaped till now      int dist = 0, len;         // loop for number lengths      for (len = 1; ; len++)      {          // nine*len will be incremented characters          // and nine will be incremented numbers          sum += nine * len;          dist += nine;             if (sum >= N)          {              // restore variables to previous              // correct state              sum -= nine * len;              dist -= nine;              N -= sum;              break;          }          nine *= 10;      }         // get distance from last one digit      // less maximum number      int diff = (int)(Math.Ceiling((double)(N) /                                   (double)(len)));         // d will store dth digit of      // current number      int d = N % len;      if (d == 0)          d = len;         // method will return dth numbered      // digit of (dist + diff) number      return getDigit(dist + diff, d);  }     // Driver code static void Main()  {     int N = 251;      Console.WriteLine(getNthChar(N));  }  }    // This code is contributed by mits

PHP

 = \$N)         {             // restore variables to             // previous correct state             \$sum -= \$nine * \$len;             \$dist -= \$nine;             \$N -= \$sum;             break;         }          \$nine *= 10;     }            // get distance from last one      // digit less maximum number     \$diff = (\$N / \$len) + 1;        // d will store dth digit      // of current number     \$d = \$N % \$len;     if (\$d == 0)         \$d = \$len;        // method will return dth numbered      // digit of (dist + diff) number     return getDigit(\$dist + \$diff, \$d); }    // Driver code \$N = 251; echo getNthChar(\$N);    // This code is contributed by mits ?>

Output:

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