FilesExpand file tree

 master

/

WordAbbreviation527.java

Copy path

Blame

More file actions

Blame

More file actions

Latest commit

 

History

History

History

186 lines (162 loc) · 5.96 KB

 master

/

WordAbbreviation527.java

Copy path

Top

File metadata and controls

• Code
• Blame

186 lines (162 loc) · 5.96 KB

Raw

Copy raw file

Download raw file

Open symbols panel

Edit and raw actions

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

121

122

123

124

125

126

127

128

129

130

131

132

133

134

135

136

137

138

139

140

141

142

143

144

145

146

147

148

149

150

151

152

153

154

155

156

157

158

159

160

161

162

163

164

165

166

167

168

169

170

171

172

173

174

175

176

177

178

179

180

181

182

183

184

185

186

/**

* Given an array of n distinct non-empty strings, you need to generate minimal

* possible abbreviations for every word following rules below.

*

* Begin with the first character and then the number of characters abbreviated,

* which followed by the last character.

*

* If there are any conflict, that is more than one words share the same

* abbreviation, a longer prefix is used instead of only the first character

* until making the map from word to abbreviation become unique. In other

* words, a final abbreviation cannot map to more than one original words.

* If the abbreviation doesn't make the word shorter, then keep it as original.

*

* Example:

* Input: ["like", "god", "internal", "me", "internet", "interval", "intension", "face", "intrusion"]

* Output: ["l2e","god","internal","me","i6t","interval","inte4n","f2e","intr4n"]

*

* Note:

* Both n and the length of each word will not exceed 400.

* The length of each word is greater than 1.

* The words consist of lowercase English letters only.

* The return answers should be in the same order as the original array.

*/

public class WordAbbreviation527 {

public List<String> wordsAbbreviation(List<String> dict) {

int N = dict.size();

Abbr[] abbrs = new Abbr[N];

for (int i=0; i<N; i++) {

String word = dict.get(i);

abbrs[i] = new Abbr(word, encode(word, 0), 0);

}

for (int i=0; i<N; i++) {

while (true) {

Set<Abbr> duplicates = new HashSet<>();

for (int j=i+1; j<N; j++) {

if (abbrs[j].abbr.equals(abbrs[i].abbr)) {

duplicates.add(abbrs[j]);

}

}

if (duplicates.isEmpty()) break;

duplicates.add(abbrs[i]);

for (Abbr abbr: duplicates) {

abbr.offset++;

abbr.abbr = encode(abbr.word, abbr.offset);

}

}

}

List<String> res = new ArrayList<>();

for (Abbr abbr: abbrs) {

res.add(abbr.abbr);

}

return res;

}

private String encode(String word, int offset) {

if (offset + 2 >= word.length()) return word;

String encoded = word.substring(0, offset+1) + Integer.toString(word.length()-offset-2) + word.charAt(word.length()-1);

return encoded.length() >= word.length() ? word : encoded;

}

class Abbr {

String word;

String abbr;

int offset;

Abbr(String word, String abbr, int offset) {

this.word = word;

this.abbr = abbr;

this.offset = offset;

}

}

/**

* https://leetcode.com/problems/word-abbreviation/solution/s

*/

public List<String> wordsAbbreviation2(List<String> words) {

Map<String, List<IndexedWord>> groups = new HashMap();

String[] ans = new String[words.size()];

int index = 0;

for (String word: words) {

String ab = abbrev(word, 0);

if (!groups.containsKey(ab))

groups.put(ab, new ArrayList());

groups.get(ab).add(new IndexedWord(word, index));

index++;

}

for (List<IndexedWord> group: groups.values()) {

Collections.sort(group, (a, b) -> a.word.compareTo(b.word));

int[] lcp = new int[group.size()];

for (int i = 1; i < group.size(); ++i) {

int p = longestCommonPrefix(group.get(i-1).word, group.get(i).word);

lcp[i] = p;

lcp[i-1] = Math.max(lcp[i-1], p);

}

for (int i = 0; i < group.size(); ++i)

ans[group.get(i).index] = abbrev(group.get(i).word, lcp[i]);

}

return Arrays.asList(ans);

}

public String abbrev(String word, int i) {

int N = word.length();

if (N - i <= 3) return word;

return word.substring(0, i+1) + (N - i - 2) + word.charAt(N-1);

}

public int longestCommonPrefix(String word1, String word2) {

int i = 0;

while (i < word1.length() && i < word2.length()

&& word1.charAt(i) == word2.charAt(i))

i++;

return i;

}

class IndexedWord {

String word;

int index;

IndexedWord(String w, int i) {

word = w;

index = i;

}

}

/**

* https://leetcode.com/problems/word-abbreviation/solution/s

*/

public List<String> wordsAbbreviation3(List<String> words) {

Map<String, List<IndexedWord>> groups = new HashMap();

String[] ans = new String[words.size()];

int index = 0;

for (String word: words) {

String ab = abbrev(word, 0);

if (!groups.containsKey(ab))

groups.put(ab, new ArrayList());

groups.get(ab).add(new IndexedWord(word, index));

index++;

}

for (List<IndexedWord> group: groups.values()) {

TrieNode trie = new TrieNode();

for (IndexedWord iw: group) {

TrieNode cur = trie;

for (char letter: iw.word.substring(1).toCharArray()) {

if (cur.children[letter - 'a'] == null)

cur.children[letter - 'a'] = new TrieNode();

cur.count++;

cur = cur.children[letter - 'a'];

}

}

for (IndexedWord iw: group) {

TrieNode cur = trie;

int i = 1;

for (char letter: iw.word.substring(1).toCharArray()) {

if (cur.count == 1) break;

cur = cur.children[letter - 'a'];

i++;

}

ans[iw.index] = abbrev(iw.word, i-1);

}

}

return Arrays.asList(ans);

}

class TrieNode {

TrieNode[] children;

int count;

TrieNode() {

children = new TrieNode[26];

count = 0;

}

}

}