Smallest Divisible Digit Product II

• Define a recursive function, for example, solve(index, p2, p3, p5, p7, isTight).
• index: The current digit position being filled (from left to right).
• p2, p3, p5, p7: The remaining counts of prime factors (2, 3, 5, 7) needed.
• isTight: A boolean flag indicating if we are restricted to digits >= num[index]. If false, we can use any digit from 1-9.
• The function explores placing digits d from a valid range at the current index.
• For each d, it makes a recursive call for index + 1 with updated parameters.
• The base case is when index reaches the end of the number. If all factor requirements are met (p_i <= 0), it signifies a valid number.
• Memoization is used to store results for states (index, p2, p3, p5, p7, isTight) to avoid recomputation.

The core of this method is a recursive function that constructs the result string. The function tries to determine the best digit to place at each position, starting from the most significant digit.

The state of the recursion needs to track:

1. index: The current position in the number string we are building.
2. p2, p3, p5, p7: The count of prime factors (2, 3, 5, 7) that still need to be covered by the product of the remaining digits.
3. isTight: A boolean flag. If true, the digit at the current index can be from num[index] to '9'. If we choose a digit d > num[index], the isTight constraint is lifted for subsequent positions (isTight becomes false). If isTight is already false, we can choose any digit from '1' to '9'.

The function's goal is to return the smallest possible suffix that can be formed from index onwards, given the factor requirements. If no such suffix exists, it returns a special value indicating failure.

To avoid re-calculating the same state, a multi-dimensional array memo[index][p2][p3][p5][p7][isTight] is used. However, with num.length up to 2 * 10^5, the size of this memoization table becomes impractically large, making the approach infeasible.

1// Conceptual code for the recursive approach2class Solution {3    String num;4    int[] tFactors;5    String[][][][][] memo;6 7    public String solve(int index, int p2, int p3, int p5, int p7, boolean isTight) {8        if (p2 <= 0 && p3 <= 0 && p5 <= 0 && p7 <= 0) {9            // Requirement met, fill rest with '1's10            StringBuilder ones = new StringBuilder();11            for (int i = 0; i < num.length() - index; i++) {12                ones.append('1');13            }14            return ones.toString();15        }16        if (index == num.length()) {17            return null; // Failed to meet requirement18        }19        if (memo[index][p2][p3][p5][p7][isTight ? 1 : 0] != null) {20            return memo[index][p2][p3][p5][p7][isTight ? 1 : 0];21        }22 23        String ans = null;24        int lowerBound = isTight ? (num.charAt(index) - '0') : 1;25 26        for (int d = lowerBound; d <= 9; d++) {27            if (d == 0) continue;28 29            int[] digitFactors = getFactors(d);30            boolean newIsTight = isTight && (d == lowerBound);31 32            String res = solve(index + 1, p2 - digitFactors[0], p3 - digitFactors[1], 33                               p5 - digitFactors[2], p7 - digitFactors[3], newIsTight);34 35            if (res != null) {36                ans = d + res;37                break; // Found the smallest digit `d`, so we can stop38            }39        }40        return memo[index][p2][p3][p5][p7][isTight ? 1 : 0] = ans;41    }42    // Helper methods getFactors, etc.43}

• Prime Factorization: First, find the prime factorization of t. If t has any prime factor greater than 7, no solution exists. Let the required counts of factors 2, 3, 5, 7 be (r2, r3, r5, r7).
• DP Precomputation: Create a DP table dp[p2][p3][p5][p7] that stores the optimal (shortest, then lexicographically smallest) string of digits whose product yields exactly p2 factors of 2, p3 of 3, etc. This table can be populated using Dijkstra's algorithm on the state space of factor counts.
• Handle Two Cases: The problem is split into finding the smallest solution of length n (same as num) and finding the smallest solution of length > n.
• Case 1 (length n):
  • Handle any '0's in num by creating a new, larger, zero-free startNum.
  • Iterate from the rightmost digit of startNum (i = n-1 to 0).
  • At each position i, try to use a digit d greater than startNum[i].
  • For the prefix startNum[0...i-1]d, calculate the remaining factors needed.
  • Look up the optimal suffix for these factors in the precomputed dp table.
  • If the suffix fits in the remaining length, construct the number. The first one found is the smallest of length n.
• Case 2 (length > n):
  • Find the optimal digit string s for the full requirements (r2, r3, r5, r7) from the dp table.
  • The minimum length of a solution longer than n is m = max(n + 1, s.length()).
  • Construct the smallest number of length m using digits from s and padding with '1's.
• Final Result: Compare the results from both cases and return the overall smallest number.

This method breaks the problem down into a manageable precomputation step and a final construction step.

1. Precomputation using DP on Factors

First, we observe that the product of digits only involves prime factors 2, 3, 5, and 7. If t has other prime factors, we return "-1". We find the required powers of these primes in t, let's say (r2, r3, r5, r7).

The subproblem is: what is the best set of digits to produce a certain combination of prime factors? We define "best" as having the minimum number of digits, and for a tie, being lexicographically smallest when sorted. We can solve this using Dijkstra's algorithm on the state space of factor counts (p2, p3, p5, p7).

• State: A tuple (p2, p3, p5, p7) representing factor counts.
• Distance: A pair (length, string), where length is the number of digits and string is the sorted string of digits.
• Priority Queue: Stores (distance, state), ordered by distance.

We initialize distance to (0, "") for state (0,0,0,0) and infinity for all others. The algorithm explores states, adding digits 2-9, and updates distances if a shorter or lexicographically smaller digit string is found. The result is a table, dp[p2][p3][p5][p7], storing the best digit string for each factor requirement.

2. Constructing the Solution

We find the answer by considering two possibilities for the result's length.

Case A: Solution has the same length as num We want the smallest zero-free number x >= num of length n.

1. If num contains '0's, we must find a larger number. The smallest zero-free number >= num can be found by finding the first '0' at index k, then finding the rightmost non-'9' digit at j < k, incrementing it, and setting digits from j+1 to n-1 to '1'. This becomes our new base num. If no such j exists, no solution of length n is possible.
2. We check if this new num is a solution. If not, we iterate from i = n-1 down to 0. For each i, we try digits d from num[i] + 1 to 9. This forms a prefix num[0...i-1]d.
3. We calculate the prime factors provided by this prefix and determine the remaining factors needed.
4. We query our dp table for the best suffix for these remaining factors. To do this, we check all dp states that satisfy the requirement and take the best one.
5. If the length of this optimal suffix is less than or equal to the remaining space (n-1-i), we've found a candidate. We pad with '1's, sort the suffix digits, and form the number. The first such number found is the answer for this case.

Case B: Solution is longer than num The smallest number longer than num must have the minimum possible length m > n. The minimum number of digits required to satisfy t is given by the length of the optimal string for (r2, r3, r5, r7), let's call it s. So, m = max(n + 1, s.length()). The result is formed by sorting the digits of s along with m - s.length() ones.

Finally, we compare the results from Case A and B and return the smaller one.

1// Conceptual code for the efficient approach2class Solution {3    // DP table and factor arrays4    private String[][][][] dp; 5    private int[] tFactors;6 7    public String smallestDivisibleDigitProductII(String num, long t) {8        // 1. Factorize t and check for invalid primes9        // ...10 11        // 2. Precompute DP table using Dijkstra12        precomputeDpTable();13 14        // 3. Find solution for length n15        String solN = findSolutionOfLengthN(num);16 17        // 4. Find solution for length > n18        String solNPlus = findSolutionOfGreaterLength(num.length());19 20        // 5. Compare and return the best result21        if (solN == null) return solNPlus;22        if (solNPlus == null) return solN;23        if (solN.length() < solNPlus.length()) return solN;24        if (solN.length() > solNPlus.length()) return solNPlus;25        return solN.compareTo(solNPlus) < 0 ? solN : solNPlus;26    }27 28    private void precomputeDpTable() { /* ... Dijkstra implementation ... */ }29 30    private String findSolutionOfLengthN(String num) { /* ... Case A logic ... */ return null; }31 32    private String findSolutionOfGreaterLength(int n) { /* ... Case B logic ... */ return null; }33}