Shortest Impossible Sequence of Rolls

This approach uses dynamic programming to count the number of distinct subsequences for each possible length. We iterate through lengths L = 1, 2, 3, ... and for each length, we calculate how many unique subsequences of that length can be formed from the rolls array. The first length L for which this count is less than the total number of possible sequences of that length (k^L) is the answer.

1. We are looking for the smallest length L such that the number of distinct subsequences of rolls of length L is less than k^L.
2. We can iterate L from 1 upwards.
3. For each L, we calculate the total number of distinct subsequences of length L that can be formed from the rolls array.
4. Let dp[i][j] be the number of distinct subsequences of length j in the prefix rolls[0...i-1].
5. The recurrence relation is: dp[i][j] = dp[i-1][j] + (dp[i-1][j-1] - dp[p][j-1]), where p is the index of the previous occurrence of the element rolls[i-1]. This formula counts subsequences not using rolls[i-1] (dp[i-1][j]) and those formed by appending rolls[i-1] to shorter subsequences, avoiding double counting.
6. We can optimize space by noticing that computing the column for length L only requires the column for length L-1. So we can use two arrays, one for the previous length and one for the current length.
7. For each L, we compute dp[n][L]. If dp[n][L] < k^L, we have found our answer, and we return L.
8. We must handle potential overflow when calculating k^L.

Let dp[i][j] be the number of distinct subsequences of length j that can be formed using the prefix of the rolls array of length i (i.e., rolls[0...i-1]). Our goal is to find the smallest j such that dp[n][j] < k^j.

The recurrence relation for dp[i][j] is derived by considering the element rolls[i-1]:

• Subsequences of length j in rolls[0...i-1] that do not include rolls[i-1] are the same as subsequences of length j in rolls[0...i-2]. Their count is dp[i-1][j].
• Subsequences of length j in rolls[0...i-1] that do include rolls[i-1] as their last element are formed by taking any distinct subsequence of length j-1 from rolls[0...i-2] and appending rolls[i-1]. The number of such new subsequences is dp[i-1][j-1]. However, this might lead to double counting if rolls[i-1] has appeared before. If the previous occurrence of rolls[i-1] was at index p-1, then any subsequence of length j-1 from rolls[0...p-2] extended with rolls[i-1] has already been counted. We must subtract these, which amounts to dp[p-1][j-1].

So, the full recurrence is: dp[i][j] = dp[i-1][j] + dp[i-1][j-1] - dp[p-1][j-1], where p is the 1-based index of the previous occurrence of rolls[i-1].

We can optimize the space from O(n*ans) to O(n) by using only two arrays to store the DP results for the current and previous lengths.

import java.util.Arrays;

class Solution {
    public int shortestImpossibleSequenceOfRolls(int[] rolls, int k) {
        int n = rolls.length;
        // dp for length len-1. Initially for len=0, there's 1 empty subsequence.
        long[] prevDp = new long[n + 1];
        Arrays.fill(prevDp, 1);

        for (int len = 1; len <= n + 1; len++) {
            long kPowLen = 1;
            // Calculate k^len with overflow check
            for (int i = 0; i < len; i++) {
                if (kPowLen > Long.MAX_VALUE / k) {
                    kPowLen = Long.MAX_VALUE; // Represents a very large number
                    break;
                }
                kPowLen *= k;
            }

            long[] currDp = new long[n + 1]; // dp for length len
            int[] last = new int[k + 1]; // Stores last seen 1-based index of a roll value

            for (int i = 1; i <= n; i++) {
                // Number of subsequences of length 'len' in rolls[0...i-2]
                currDp[i] = currDp[i - 1];
                
                // New subsequences formed by appending rolls[i-1]
                long toAdd = prevDp[i - 1];
                int rollValue = rolls[i - 1];
                if (last[rollValue] > 0) {
                    // Subtract subsequences that would be duplicates
                    toAdd -= prevDp[last[rollValue] - 1];
                }
                currDp[i] += toAdd;
            }

            if (currDp[n] < kPowLen) {
                return len;
            }
            prevDp = currDp;
        }
        
        return n + 1; // Should not be reached given problem constraints
    }
}