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++
// 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
// 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
# 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
Output:
0
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