Kth Smallest Instructions

Bob is standing at cell (0, 0), and he wants to reach destination: (row, column). He can only travel right and down. You are going to help Bob by providing instructions for him to reach destination.

The instructions are represented as a string, where each character is either:

'H', meaning move horizontally (go right), or
'V', meaning move vertically (go down).

Multiple instructions will lead Bob to destination. For example, if destination is (2, 3), both "HHHVV" and "HVHVH" are valid instructions.

However, Bob is very picky. Bob has a lucky number k, and he wants the kth lexicographically smallest instructions that will lead him to destination. k is 1-indexed.

Given an integer array destination and an integer k, return the kth lexicographically smallest instructions that will take Bob to destination.

Example 1:

Input: destination = [2,3], k = 1
Output: "HHHVV"
Explanation: All the instructions that reach (2, 3) in lexicographic order are as follows:
["HHHVV", "HHVHV", "HHVVH", "HVHHV", "HVHVH", "HVVHH", "VHHHV", "VHHVH", "VHVHH", "VVHHH"].

Example 2:

Input: destination = [2,3], k = 2
Output: "HHVHV"

Example 3:

Input: destination = [2,3], k = 3
Output: "HHVVH"

Constraints:

destination.length == 2
1 <= row, column <= 15
1 <= k <= nCr(row + column, row), where nCr(a, b) denotes a choose b.

Solution:

https://stackoverflow.com/questions/34230266/how-to-calculate-the-lexicographical-rank-of-a-given-permutation/34235272#34235272

class Solution {
    public String kthSmallestPath(int[] destination, int k) {
        int v = destination[0], h = destination[1];
        StringBuilder sb = new StringBuilder();
        long[][] combs = new long[31][31];
        for (int i = 0; i <= 30; i ++) {
            for (int j = 0; j <= i; j ++) {
                if (j == 0 || i == j) {
                    combs[i][j] = 1;
                } else {
                    // if A is picked, the rest possibilty is N-1 C j - 1
                    // if A is not picked, the rest is N - 1 C j
                    combs[i][j] = (combs[i - 1][j] + combs[i - 1][j - 1]);
                }
            }
        }
        nthP(sb, (long) k, h, v, combs);
        return sb.toString();
    }
    
    private void nthP(StringBuilder sb, long k, int h, int v, long[][] combs) {
        int t = h + v;
        for (int i = t; i > 0; i --) {
            if (i > v) {
                // assume we pick H, c is the max rank we can get
                // for example, for HHVVV, i = 5, c = 4C2 = 6
                // meaning if we pick H as first, there are ["HHHVV", "HHVHV", "HHVVH", "HVHHV", "HVHVH", "HVVHH"]
                // and if k is 7, then we have to pick V, and remove 6 from k
                long c = combs[i - 1][v];
                // we have to pick V
                if (k > c) {
                    k -= c;
                    sb.append('V');
                    v --;
                } else {
                    sb.append('H');
                }
            } else {
                sb.append('V');
            }
        }
    }
}