400

400. Nth Digit (Medium)

Find the n^th digit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...

Note:
n is positive and will fit within the range of a 32-bit signed integer (n < 2³¹).

Example 1:

Input:
3
Output:
3

Example 2:

Input:
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.

Companies:
Google

Related Topics:
Math

Solution 1.

There are 9 one-digit numbers (1 ~ 9) which have 1 * 9 = 9 digits in total.

There are 90 two-digit numbers (10 ~ 99) which have 2 * 90 = 180 digits in total.

There are 900 three-digit numbers (100 ~ 999) which have 3 * 900 = 2700 digits in total.

There are 9000 four-digit numbers (1000 ~ 9999) which have 4 * 9000 = 36000 digits in total.

...

Assume number k contains the nth digit. To find out the value of the nth digit, we'd better know the value of k and the index of the digit within k.

Assume n = 361, then k must be a three-digit number because 361 > 9 + 180 = 189 and 361 < 9 + 180 + 2700 = 2889.

So n is the 361 - 9 - 180 - 1 = 171st digit within the three-digit numbers, and it's within the floor(171 / 3) = 57th number. So k = 100 + 57 = 157.

And n is the 171 % 3 = 0st digit within k, i.e. 1.

Now we can generalize this algorithm.

We keep subtracting cnt = 9 * len * pow(10, len - 1) from n until n is not greater than cnt.

Then n - 1 is the index of the digit within the len-digit numbers.

Then we have k = pow(10, len - 1) + (n - 1) / len and r = (n - 1) % len.

The last step is simply get the rth digit within k.

// OJ: https://leetcode.com/problems/nth-digit/
// Author: github.com/lzl124631x
// Time: O(lgN)
// Space: O(1)
class Solution {
public:
    int findNthDigit(int n) {
        long long len = 1, p = 1, cnt = 9;
        while (n > cnt) {
            n -= cnt;
            ++len;
            cnt = 9 * len * (p *= 10);
        }
        int k = p + (n - 1) / len, r = (n - 1) % len, i = len - r - 1;
        while (i--) k /= 10;
        return k % 10;
    }
};