leetcode527 - Word Abbreviation - hard

class Solution {
    public List<String> wordsAbbreviation(List<String> dict) {
        if (dict == null || dict.size() == 0) {
            return new ArrayList<String>();
        }
        
        int[] prefix = new int[dict.size()];
        String[] result = new String[dict.size()];
        Map<String, Integer> count = new HashMap<>();
        
        // initialization
        for (int i = 0; i < dict.size(); i++) {
            prefix[i]++;
            String abbr = abbreviate(dict.get(i), prefix[i]);
            result[i] = abbr;
            count.put(abbr, count.getOrDefault(abbr, 0) + 1);
        }
        
        boolean hasDup = true;
        while (hasDup) {
            hasDup = false;
            for (int i = 0; i < dict.size(); i++) {
                if (count.get(result[i]) > 1) {
                    hasDup = true;
                    prefix[i]++;
                    String newAbbr = abbreviate(dict.get(i), prefix[i]);
                    result[i] = newAbbr;
                    count.put(newAbbr, count.getOrDefault(newAbbr, 0) + 1);
                }
            }
        }
        
        // P3: 生成List
        List<String> resultList = new ArrayList<>();
        Collections.addAll(resultList, result);
        return resultList;
    }
    
    private String abbreviate(String s, int prefix) {
        if (s.length() - prefix - 1 <= 1) {
            return s;
        } else {
            // P2: 注意题意不是每次均匀往中间缩两格，只是prefix缩一格进去。
            return s.substring(0, prefix) + (s.length() - prefix - 1) + s.substring(s.length() - 1, s.length());
        }
    }
}

class Solution {
    public List<String> wordsAbbreviation(List<String> dict) {
        if (dict == null || dict.size() == 0) {
            return new ArrayList<String>();
        }
        
        Map<String, List<Integer>> strIdxes = new HashMap<>();
        int prefix = 1;
        
        // initialization
        for (int i = 0; i < dict.size(); i++) {
            String abbreved = abbreviate(dict.get(i), prefix);
            strIdxes.putIfAbsent(abbreved, new ArrayList<>());
            strIdxes.get(abbreved).add(i);
        }
        
        while (strIdxes.size() < dict.size()) {
            prefix++;
            System.out.println(prefix);
            // P1: 每次要开新map的原因: 你不能在遍历keySet()的时候又给keySet()插入删除key值，这是逻辑不对而且编译会报错ConcurrentModificationException。
            Map<String, List<Integer>> newMap = new HashMap<>();
            for (String s : strIdxes.keySet()) {
                List<Integer> indexes = strIdxes.get(s);
                if (indexes.size() == 1) {
                    newMap.put(s, indexes);
                    continue;
                }
                for (int index : indexes) {
                    String newAbbreved = abbreviate(dict.get(index), prefix);
                    newMap.putIfAbsent(newAbbreved, new ArrayList<>());
                    newMap.get(newAbbreved).add(index);
                }
            }
            strIdxes = newMap;
        }
        
        // P3: 生成List
        String[] resultArr = new String[dict.size()];
        for (String s : strIdxes.keySet()) {
            resultArr[strIdxes.get(s).get(0)] = s;
        }
        List<String> result = new ArrayList<>();
        Collections.addAll(result, resultArr);
        return result;
    }
    
    private String abbreviate(String s, int prefix) {
        if (s.length() - prefix - 1 <= 1) {
            return s;
        } else {
            // P2: 注意题意不是每次均匀往中间缩两格，只是prefix缩一格进去。
            return s.substring(0, prefix) + (s.length() - prefix - 1) + s.substring(s.length() - 1, s.length());
        }
    }
}