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Problem solution · Java

Smallest Divisible Digit Product II

Smallest Divisible Digit Product II: a Java solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
140 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Hash-based lookup

For Smallest Divisible Digit Product II, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 140 lines of Java from the credited upstream file 3348.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 9 loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeSmallest Divisible Digit Product II · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public String smallestNumber(String num, long t) {    Pair<Map<Integer, Integer>, Boolean> primeCountResult = getPrimeCount(t);    Map<Integer, Integer> primeCount = primeCountResult.getKey();    boolean isDivisible = primeCountResult.getValue();    if (!isDivisible)      return "-1";     Map<Integer, Integer> factorCount = getFactorCount(primeCount);    if (sumValues(factorCount) > num.length())      return construct(factorCount);     Map<Integer, Integer> primeCountPrefix = getPrimeCount(num);    int firstZeroIndex = num.indexOf('0');    if (firstZeroIndex == -1) {      firstZeroIndex = num.length();      if (isSubset(primeCount, primeCountPrefix))        return num;    }     for (int i = num.length() - 1; i >= 0; --i) {      final int d = num.charAt(i) - '0';      // Remove the current digit's factors from primeCountPrefix.      primeCountPrefix = subtract(primeCountPrefix, FACTOR_COUNTS.get(d));      final int spaceAfterThisDigit = num.length() - 1 - i;      if (i > firstZeroIndex)        continue;      for (int biggerDigit = d + 1; biggerDigit < 10; ++biggerDigit) {        // Compute the required factors after replacing with a larger digit.        Map<Integer, Integer> factorsAfterReplacement = getFactorCount(            subtract(subtract(primeCount, primeCountPrefix), FACTOR_COUNTS.get(biggerDigit)));        // Check if the replacement is possible within the available space.        if (sumValues(factorsAfterReplacement) <= spaceAfterThisDigit) {          // Fill extra space with '1', if any, and construct the result.          final int fillOnes = spaceAfterThisDigit - sumValues(factorsAfterReplacement);          return num.substring(0, i) + // Keep the prefix unchanged.              biggerDigit +            // Replace the current digit.              "1".repeat(fillOnes) + // Fill remaining space with '1'.              construct(factorsAfterReplacement);        }      }    }     // No solution of the same length exists, so we need to extend the number    // by prepending '1's and adding the required factors.    Map<Integer, Integer> factorsAfterExtension = getFactorCount(primeCount);    return "1".repeat(num.length() + 1 - sumValues(factorsAfterExtension)) +        construct(factorsAfterExtension);  }   private static final Map<Integer, Map<Integer, Integer>> FACTOR_COUNTS = Map.of(      0, Map.of(), 1, Map.of(), 2, Map.of(2, 1), 3, Map.of(3, 1), 4, Map.of(2, 2), 5, Map.of(5, 1),      6, Map.of(2, 1, 3, 1), 7, Map.of(7, 1), 8, Map.of(2, 3), 9, Map.of(3, 2));   // Returns the prime count of t and if t is divisible by 2, 3, 5, 7.  private Pair<Map<Integer, Integer>, Boolean> getPrimeCount(long t) {    Map<Integer, Integer> count = new HashMap<>(Map.of(2, 0, 3, 0, 5, 0, 7, 0));    for (int prime : new int[] {2, 3, 5, 7}) {      while (t % prime == 0) {        t /= prime;        count.put(prime, count.get(prime) + 1);      }    }    return new Pair<>(count, t == 1);  }   // Returns the prime count of `num`.  private Map<Integer, Integer> getPrimeCount(String num) {    Map<Integer, Integer> count = new HashMap<>(Map.of(2, 0, 3, 0, 5, 0, 7, 0));    for (final char c : num.toCharArray()) {      Map<Integer, Integer> digitFactors = FACTOR_COUNTS.get(c - '0');      for (Map.Entry<Integer, Integer> entry : digitFactors.entrySet()) {        final int prime = entry.getKey();        final int freq = entry.getValue();        count.merge(prime, freq, Integer::sum);      }    }    return count;  }   private Map<Integer, Integer> getFactorCount(Map<Integer, Integer> count) {    // 2^3 = 8    final int count8 = count.get(2) / 3;    final int remaining2 = count.get(2) % 3;    // 3^2 = 9    final int count9 = count.get(3) / 2;    int count3 = count.get(3) % 2;    // 2^2 = 4    int count4 = remaining2 / 2;    int count2 = remaining2 % 2;    // Combine 2 and 3 to 6 if both are present    int count6 = 0;    if (count2 == 1 && count3 == 1) {      count2 = 0;      count3 = 0;      count6 = 1;    }    // Combine 3 and 4 to 2 and 6 if both are present    if (count3 == 1 && count4 == 1) {      count2 = 1;      count6 = 1;      count3 = 0;      count4 = 0;    }    return Map.of(2, count2, 3, count3, 4, count4, 5, count.get(5), 6, count6, 7, count.get(7), 8,                  count8, 9, count9);  }   private String construct(Map<Integer, Integer> factors) {    StringBuilder sb = new StringBuilder();    for (int digit = 2; digit < 10; ++digit)      sb.append(String.valueOf(digit).repeat(factors.get(digit)));    return sb.toString();  }   // Returns true if a is a subset of b.  private boolean isSubset(Map<Integer, Integer> a, Map<Integer, Integer> b) {    for (Map.Entry<Integer, Integer> entry : a.entrySet())      if (b.get(entry.getKey()) < entry.getValue())        return false;    return true;  }   // Returns a - b.  private Map<Integer, Integer> subtract(Map<Integer, Integer> a, Map<Integer, Integer> b) {    Map<Integer, Integer> res = new HashMap<>(a);    for (Map.Entry<Integer, Integer> entry : b.entrySet()) {      final int key = entry.getKey();      final int value = entry.getValue();      res.put(key, Math.max(0, res.get(key) - value));    }    return res;  }   // Returns the sum of the values in `count`.  private int sumValues(Map<Integer, Integer> count) {    return count.values().stream().mapToInt(Integer::intValue).sum();  }}

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