Leetcode: Minimum Unique Word Abbreviation

A string such as "word" contains the following abbreviations:

["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]
Given a target string and a set of strings in a dictionary, find an abbreviation of this target string with the smallest possible length such that it does not conflict with abbreviations of the strings in the dictionary.

Each number or letter in the abbreviation is considered length = 1. For example, the abbreviation "a32bc" has length = 4.

Note:
In the case of multiple answers as shown in the second example below, you may return any one of them.
Assume length of target string = m, and dictionary size = n. You may assume that m ≤ 21, n ≤ 1000, and log2(n) + m ≤ 20.
Examples:
"apple", ["blade"] -> "a4" (because "5" or "4e" conflicts with "blade")

"apple", ["plain", "amber", "blade"] -> "1p3" (other valid answers include "ap3", "a3e", "2p2", "3le", "3l1").

Refer to https://discuss.leetcode.com/topic/61799/java-bit-mask-dfs-with-pruning

bit mask refer to http://bookshadow.com/weblog/2016/10/02/leetcode-minimum-unique-word-abbreviation/

 

1. The key idea of my solution is to preprocess the dictionary to transfer all the words to bit sequences (int):

比如apple 和 amper 字母相同设1,不同设0,所以得到10100

又比如target='apple'，word='amble'，则word的bitmask为10011

在这个过程中ignore与target长度不一样的，得到每个字典里面的bitmask，下面称呼它为mask，所以mask里为1的位表示与target字母相同

a p p l e
a m b l e
1 0 0 1 1

2. 开始缩写target单词，只不过我们不直接缩写成十进制数，而是先缩写成二进制，1表示保留字母，0表示替换为数字。下面称呼这个当前的缩写结果为curResult, 因为curResult由target缩写而来，所以curResult里为1的位与target字母相同

例如单词manipulation的缩写m2ip6n可以转化为100110000001

m a n i p u l a t i o n
m  2  i p      6      n
1 0 0 1 1 0 0 0 0 0 0 1

3. mask里为1的位表示与target字母相同， 同时curResult里为1的位也与target字母相同， 如果

 curResult & mask == curResult,

则说明curResult里为1的位子上，mask也一定是1，那么curResult也一定是mask所对应的那个string的一个缩写，所以这里conflict出现了，所以这个curResult要被skip掉

 

4. 在所有没有conflict的curResult里面，找到长度最短的一个，如何找到长度最短，可以在recursion里面维护一个当前curResult长度大小的变量，同时有一个变量保存最小长度用以更新

 

5. 最后将长度最短那个curResult(它目前只是一个二进制缩写)，复原成十进制缩写

 

这是我做过最难的Bitmask的题了

public class Solution {
    private int minLen;
    private int result;
    
    public String minAbbreviation(String target, String[] dictionary) {
        // only keep words whose length == target in new dict, then compute their bit masks
        Set<Integer> maskSet = new HashSet<>();
        for(String s: dictionary){
            if(target.length() == s.length()){
                maskSet.add(getBitMask(s,target));
            }
        }
    
        // dfs with pruning
        minLen = target.length()+1;
        result = -1;
    
        dfs(target,maskSet,0,0,0);
    
        if(minLen > target.length()){
            return "";
        }
    
        // convert result to word
        int zeroCnt = 0;
        String res = "";
        for (int i = target.length()-1; i>=0; i--) {
            //遇到0要累加连续零个数,遇到1填原char
            int digit = (result & 1);
            if(digit == 0){
                ++zeroCnt;
            } else {
                if(zeroCnt > 0){
                    res = zeroCnt + res;
                    zeroCnt =0;
                }
                res = target.charAt(i) + res;
            }
            result >>= 1;
        }
        if(zeroCnt > 0) res = zeroCnt + res;
        return res;
    }
    
    /**
     *
     * @param target
     * @param maskSet masks of words in dict
     * @param start idx at target
     * @param curLen current abbr's length
     */
    private void dfs(String target,Set<Integer> maskSet,int start,int curLen,int curResult){
        // pruning, no need to continue, already not min length
        if(curLen >= minLen) return;
    
        if(start == target.length()){
            // check whether curResult mask conflicts with words in dict
            for(int mask:maskSet){
                /**
                 * 单词manipulation的缩写m2ip6n可以转化为100110000001
                 *  m a n i p u l a t i o n
                    m  2  i p      6      n
                    1 0 0 1 1 0 0 0 0 0 0 1
                 * 0代表随意不care,如果这个mask和dict中某个mask的所有1重合代表在意的位置完全相同,
                 * 说明这个mask和dict中那个词冲突
                 * 我们要找的是不冲突的mask
                 */
                if((curResult & mask) == curResult){
                    return; // conflict
                }
            }
            // no conflict happens, can use
            if(minLen > curLen){
                minLen = curLen;
                result = curResult;
            }
            return;
        }
    
        // case 1: replace chars from start in target with number
        for (int i = start; i < target.length(); i++) {
            //被replace掉的char位置由0代替所以是curResult<<(i+1-start),没replace掉的这里不管,我们只管到i,之后的由backtrack内决定
            //注意:不允许word => w11d这种用数字代替但含义不同
            if(curLen == 0 || (curResult &1) == 1){
                //后者即上一次是保留了字母
                dfs(target,maskSet,i+1,curLen+1,curResult<<(i+1-start));
            }
        }
    
        // case 2: no replace from start (curResult << 1)+1代表新的这位保留了char,所以是加一
        dfs(target,maskSet,start+1,curLen+1,(curResult << 1)+1);
    }
    
    // 比如apple 和 amper 字母相同设1,不同设0,所以得到10100
    private int getBitMask(String s1,String s2){
        int mask = 0;
        for (int i = 0; i < s1.length(); i++) {
            mask <<= 1;
            if(s1.charAt(i) == s2.charAt(i)){
                mask += 1;
            }
        }
        return mask;
    }
}