Last updated on October 7th, 2024 at 02:09 am

Here, We see Nth Digit LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Nth Digit LeetCode Solution

Problem Statement

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:
Input: n = 3
Output: 3

Example 2:
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.

1. Nth Digit Leetcode Solution C++

class Solution {
public:
    int findNthDigit(int n) {
        int a = 9, d = 1;
        while (n - (long)a * d > 0)
        {
            n -= a * d;
            a *= 10;
            d += 1;
        }
        int num = pow(10, d - 1) + (n - 1) / d;
        cout<<num;
        return to_string(num)[(n - 1) % d] - '0';
    }
};

2. Nth Digit Leetcode Solution Java

class Solution {
    public int findNthDigit(int n) {
        int digit = 1;
        int num_digits_in_interval = 9;
        while (n - num_digits_in_interval > 0) {
            n -= num_digits_in_interval; 
            digit += 1;
            num_digits_in_interval = 9 * ((int) Math.pow(10, digit - 1))*digit;
            if (num_digits_in_interval < 0) break;
        }
        int base = ((int) Math.pow(10, digit - 1));
        int number =  base + (n - 1) / digit;
        char digit_in_number = String.valueOf(number).charAt((n - 1) % digit);
        return digit_in_number - '0';
    }
}

3. Nth Digit Solution JavaScript

var findNthDigit = function(n) {
    var len = 1;
    var range = 9;
    var base = 1;
    while(n>len*range)
    {
        n -= len *range;
        range *= 10;
        base *= 10;
        len++;
    }
    var num = base + Math.floor((n-1)/len);
    var s = num.toString();
    return parseInt(s[(n-1)%len]);
};

4. Nth Digit Solution Python

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