# Nth Digit Problem

## Description

LeetCode Problem 400.

Given an integer n, return the n^th digit of the infinite integer sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, …].

Example 1:

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Input: n = 3
Output: 3
``````

Example 2:

``````1
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Input: n = 11
Output: 0
Explanation: The 11^th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10.
``````

Constraints:

• 1 <= n <= 2^31 - 1

## Sample C++ Code

First, we calculate how many digits the target number has. For example, if the input is 302, we need to find out the 302th digit belongs to a 3-digit number. This is because for 1-digit numbers, there are 9 of them and they give us 9 digits. For 2-digit numbers, there are 90 of them and they give us 180 digits. For n = 302, 302 - 9 - 90 * 2 = 113, so it belongs to a 3-digit number.

Second, we calculate which number it belongs to. When n = 300, we know it belongs to a 3-digit number, number = 100 + 113 / 3 = 100 + 37 = 137.

Third, we find out which digit in the number it is. When n = 300, we know the digit is in the number 137, and the digit index is 113 % 3 = 2. So the digit is the second one in 137, which is 3.

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class Solution {
public:
int findNthDigit(int n) {
int base = 9;
long digits = 1;
while (n - base * digits > 0) {
n -= base * digits;
base *= 10;
digits++;
}

int index = (n - 1) % digits;
int offset = (n - 1) / digits;
long start = pow(10, digits - 1);
}
};
``````