# Beats 89.41%. Java Solution Explained In Detail

• ``````/**
*
* Explanation:
* Lets consider the range of numbers based of digits.
* For example the number of digits for numbers ranging 1-9 is 1
* 			   the number of digits for numbers ranging 10-99 is 2
*  		   the number of digits for numbers ranging 100-999 is 3 and so on
*
* Now, Lets see the total count of digits present for these ranges
* For example the total number of digits for numbers ranging 1-9 is 9
* 			   the total number of digits for numbers ranging 10-99 is 180
* 			   the total number of digits for number ranging 100-999 is 2700
* The pattern observed in the total number of digits is as follows:
* 9, 180, 2700,36000 => (3^2)*(10^0)*1, (3^2)*(10^1)*2, (3^2)*(10^2)*3, (3^2)*(10^3)*4
* Therefore, the formula is Math.pow(3, 2) * Math.pow(10, i) * (i+1)
*
* Using this formula, we can calculate the digit range for every number range.
* For example, the digit range for 10-99(99=10^2-1) is 10-189(189=180+9)
* 				the digit range for 100-999(999=10^3-1) is 190-2889(2889=2700+189) and so on
*
* Now that we know the digit ranges that correspond to the number ranges, its easy to find the Nth digit.
* For a given n, find the range the digit falls under.
* For example, 21st digit falls under 10-189 range.
* To identify the 21st digit, we need to subtract 21 from 10. This gives us the number of digits we need to move.
* Since we also know that for 10-99 range, moving 2 digits is equivalent to 1 number,
* we can simply compute quotient and remainder for the number of digits we need to move, which is 11(21-10)
*
* Therefore, the quotient is 5 and the remainder is 1.We add 5 to minimum number in the range which is 10.
* Therefore the number is 10+5=15. And since remainder is 1, its number at 1st index.
*
*/
public class NthDigit {

public int findNthDigit(int n) {
if(n < 10) return n;

int i=1;
int minDigit = 10;
int maxDigit = (int) (Math.pow(3, 2) * Math.pow(10, i) * (i+1) + minDigit - 1);

while(n > maxDigit) {
minDigit = maxDigit + 1;
i++;
maxDigit = (int) (Math.pow(3, 2) * Math.pow(10, i) * (i+1) + minDigit - 1);
}

int minN = (int) Math.pow(10, i);
int moves = n - minDigit;
int add = 0, modulo = 0;
if(moves >= i+1) {
modulo = moves%(i+1);
}else {
modulo = moves;
}
int finalN = minN + add;
char[] chars = ("" + finalN).toCharArray();
return Character.getNumericValue(chars[modulo]);
}
}
``````

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