440. K-th Smallest in Lexicographical Order

Problem

Given two integers n and k, return the kth lexicographically smallest integer in the range [1, n].

  Example 1:

Input: n = 13, k = 2
Output: 10
Explanation: The lexicographical order is [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9], so the second smallest number is 10.

Example 2:

Input: n = 1, k = 1
Output: 1

  Constraints:

• 1 <= k <= n <= 10^9

Solution (Java)

public class Solution {
    public int findKthNumber(int n, int k) {
        int curr = 1;
        k = k - 1;
        while (k > 0) {
            int steps = calSteps(n, curr, curr + 1L);
            if (steps <= k) {
                curr += 1;
                k -= steps;
            } else {
                curr *= 10;
                k -= 1;
            }
        }
        return curr;
    }

    // use long in case of overflow
    private int calSteps(int n, long n1, long n2) {
        long steps = 0;
        while (n1 <= n) {
            steps += Math.min(n + 1L, n2) - n1;
            n1 *= 10;
            n2 *= 10;
        }
        return (int) steps;
    }
}

Solution (Javascript)

/**
 * @param {number} n
 * @param {number} k
 * @return {number}
 */
var findKthNumber = function(n, k) {
    let i, prefix = '';
    while (k !== 0) {
        for (i = 0; i <= 9; i++) {
            const count = countForPrefix(n, prefix + i);
            if (count < k)
                k -= count;
            else
                break;
        }
        prefix = prefix + i;
        k--; // number equal to prefix
    }

    return parseInt(prefix, 10);
};

// Calculates the amount of
// numbers <= n that starts with prefix.

function countForPrefix (n, prefix) {
    let a = parseInt(prefix);
    let b = a + 1;
    if (a > n || a === 0)
        return 0;

    let res = 1;
    a *= 10; b *= 10;
    while (a <= n) {
        res += Math.min(n + 1, b) - a;
        a *= 10; b *= 10;
    }

    return res;
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).