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Problem solution · Java

Count Beautiful Numbers

Count Beautiful Numbers: 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
46 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 Count Beautiful Numbers, 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

  • 46 lines of Java from the credited upstream file 3490.java.
  • The implementation visibly relies on hash lookup, ordered lookup.
  • 1 loop block 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 codeCount Beautiful Numbers · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int beautifulNumbers(int l, int r) {    return count(String.valueOf(r), 0, /*tight=*/true, /*isLeadingZero=*/true,                 /*hasZero=*/false, /*sum=*/0, /*prod=*/1, new HashMap<>()) -        count(String.valueOf(l - 1), 0, /*tight=*/true, /*isLeadingZero=*/true,              /*hasZero=*/false, /*sum=*/0, /*prod=*/1, new HashMap<>());  }   private int count(final String s, int i, boolean tight, boolean isLeadingZero, boolean hasZero,                    int sum, int prod, Map<String, Integer> mem) {    if (i == s.length()) {      if (isLeadingZero)        return 0;      return (hasZero || prod % sum == 0) ? 1 : 0;    }    final String key = hash(i, tight, isLeadingZero, hasZero, sum, prod);    if (!isLeadingZero && hasZero && !tight) {      final int val = (int) Math.pow(10, s.length() - i);      mem.put(key, val);      return val;    }    if (mem.containsKey(key))      return mem.get(key);     int res = 0;    final int maxDigit = tight ? s.charAt(i) - '0' : 9;     for (int d = 0; d <= maxDigit; ++d) {      final boolean nextTight = tight && (d == maxDigit);      final boolean nextIsLeadingZero = isLeadingZero && d == 0;      final boolean nextHasZero = !nextIsLeadingZero && d == 0;      final int nextProd = nextIsLeadingZero ? 1 : prod * d;      res += count(s, i + 1, nextTight, nextIsLeadingZero, nextHasZero, sum + d, nextProd, mem);    }     mem.put(key, res);    return res;  }   private String hash(int i, boolean tight, boolean isLeadingZero, boolean hasZero, int sum,                      int prod) {    return i + "_" + (tight ? "1" : "0") + "_" + (isLeadingZero ? "1" : "0") + "_" +        (hasZero ? "1" : "0") + "_" + sum + "_" + prod;  }}

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