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Problem solution · Java

Maximize the Number of Partitions After Operations

Maximize the Number of Partitions After Operations: 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
39 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 Maximize the Number of Partitions After Operations, 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

  • 39 lines of Java from the credited upstream file 3003.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 codeMaximize the Number of Partitions After Operations · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int maxPartitionsAfterOperations(String s, int k) {    Map<Long, Integer> mem = new HashMap<>();    return maxPartitionsAfterOperations(s, 0, true, 0, k, mem) + 1;  }   // Returns the maximum number of partitions of s[i..n), where `canChange` is  // true if we can still change a letter, and `mask` is the bitmask of the  // letters we've seen.  private int maxPartitionsAfterOperations(final String s, int i, boolean canChange, int mask,                                           int k, Map<Long, Integer> mem) {    if (i == s.length())      return 0;     Long key = (long) i << 27 | (canChange ? 1 : 0) << 26 | mask;    if (mem.containsKey(key))      return mem.get(key);     // Initialize the result based on the current letter.    int res = getRes(s, i, canChange, mask, 1 << (s.charAt(i) - 'a'), k, mem);     // If allowed, explore the option to change the current letter.    if (canChange)      for (int j = 0; j < 26; ++j)        res = Math.max(res, getRes(s, i, false, mask, 1 << j, k, mem));     mem.put(key, res);    return res;  }   private int getRes(final String s, int i, boolean nextCanChange, int mask, int newBit, int k,                     Map<Long, Integer> mem) {    final int newMask = mask | newBit;    if (Integer.bitCount(newMask) > k) // fresh start      return 1 + maxPartitionsAfterOperations(s, i + 1, nextCanChange, newBit, k, mem);    return maxPartitionsAfterOperations(s, i + 1, nextCanChange, newMask, k, mem);  }}

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