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

Solution 1

Python Code
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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])

Java Code
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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';
    }
}

C++ Code
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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';
    }
};

Go Code
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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')
}

JavaScript Code
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/**
 * @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];
};

C# Code
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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';
    }
}