# 400. Nth Digit

## Description

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

Example 1:

Input: n = 3
Output: 3


Example 2:

Input: n = 11
Output: 0
Explanation: The 11th 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 <= 231 - 1

## Solutions

• class Solution {
public int findNthDigit(int n) {
int k = 1, cnt = 9;
while ((long) k * cnt < n) {
n -= k * cnt;
++k;
cnt *= 10;
}
int num = (int) Math.pow(10, k - 1) + (n - 1) / k;
int idx = (n - 1) % k;
return String.valueOf(num).charAt(idx) - '0';
}
}

• class Solution {
public:
int findNthDigit(int n) {
int k = 1, cnt = 9;
while (1ll * k * cnt < n) {
n -= k * cnt;
++k;
cnt *= 10;
}
int num = pow(10, k - 1) + (n - 1) / k;
int idx = (n - 1) % k;
return to_string(num)[idx] - '0';
}
};

• class Solution:
def findNthDigit(self, n: int) -> int:
k, cnt = 1, 9
while k * cnt < n:
n -= k * cnt
k += 1
cnt *= 10
num = 10 ** (k - 1) + (n - 1) // k
idx = (n - 1) % k
return int(str(num)[idx])


• func findNthDigit(n int) int {
k, cnt := 1, 9
for k*cnt < n {
n -= k * cnt
k++
cnt *= 10
}
num := int(math.Pow10(k-1)) + (n-1)/k
idx := (n - 1) % k
return int(strconv.Itoa(num)[idx] - '0')
}

• /**
* @param {number} n
* @return {number}
*/
var findNthDigit = function (n) {
let k = 1,
cnt = 9;
while (k * cnt < n) {
n -= k * cnt;
++k;
cnt *= 10;
}
const num = Math.pow(10, k - 1) + (n - 1) / k;
const idx = (n - 1) % k;
return num.toString()[idx];
};


• public class Solution {
public int FindNthDigit(int n) {
int k = 1, cnt = 9;
while ((long) k * cnt < n) {
n -= k * cnt;
++k;
cnt *= 10;
}
int num = (int) Math.Pow(10, k - 1) + (n - 1) / k;
int idx = (n - 1) % k;
return num.ToString()[idx] - '0';
}
}