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Problem solution · Java

Minimum Unique Word Abbreviation

Minimum Unique Word Abbreviation: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
85 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

Direct simulation

For Minimum Unique Word Abbreviation, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 85 lines of Java from the credited upstream file 411.java.
  • The implementation visibly relies on sequence storage.
  • 7 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeMinimum Unique Word Abbreviation · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public String minAbbreviation(String target, String[] dictionary) {    final int m = target.length();    List<Integer> masks = new ArrayList<>();     for (final String word : dictionary) {      if (word.length() != m)        continue;      masks.add(getMask(target, word));    }     if (masks.isEmpty())      return String.valueOf(m);     List<String> abbrs = new ArrayList<>();     final int maxCand = (int) Math.pow(2, m);    // all the candidate representation of the target    for (int i = 0; i < maxCand; ++i) {      final int cand = i;      // All the masks have at lease one bit different from the candidate.      if (masks.stream().allMatch(mask -> (cand & mask) > 0))        abbrs.add(getAbbr(target, cand));    }     String ans = target;     for (final String abbr : abbrs)      if (getAbbrLen(abbr) < getAbbrLen(ans))        ans = abbr;     return ans;  }   private int getMask(final String target, final String word) {    final int m = target.length();    // mask[i] = 0 := target[i] == word[i]    // mask[i] = 1 := target[i] != word[i]    // e.g. target = "apple"    //        word = "blade"    //        mask =  11110    int mask = 0;    for (int i = 0; i < m; ++i)      if (word.charAt(i) != target.charAt(i))        mask |= 1 << m - 1 - i;    return mask;  }   String getAbbr(final String target, int cand) {    final int m = target.length();    StringBuilder sb = new StringBuilder();    int replacedCount = 0;    for (int i = 0; i < m; ++i)      if ((cand >> m - 1 - i & 1) == 1) {        // If cand[i] = 1, `abbr` should show the original character.        if (replacedCount > 0)          sb.append(replacedCount);        sb.append(target.charAt(i));        replacedCount = 0;      } else {        // If cand[i] = 0, `abbr` can be replaced.        ++replacedCount;      }    if (replacedCount > 0)      sb.append(replacedCount);    return sb.toString();  }   int getAbbrLen(final String abbr) {    int abbrLen = 0;    int i = 0;    int j = 0;    while (i < abbr.length()) {      if (Character.isAlphabetic(abbr.charAt(j)))        ++j;      else        while (j < abbr.length() && Character.isDigit(abbr.charAt(j)))          ++j;      ++abbrLen;      i = j;    }    return abbrLen;  }}

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