Shortest unique prefix to represent word in an array | shawnlincoding
Use the shorest unique prefix to represent each word in the array
input: ["zebra", "dog", "duck",”dot”]
output: {zebra: z, dog: do, duck: du}
[zebra, dog, duck, dove]
{zebra:z, dog: dog, duck: du, dove: dov}
[bearcat, bear]
{bearcat: bearc, bear: ""}
http://www.geeksforgeeks.org/find-all-shortest-unique-prefixes-to-represent-each-word-in-a-given-list/
看到prefix肯定想到的是trie咯，在trie node里面多加一个field count，表示
这个字符出现多少次
1，insert word into the trie
2, search the word, 找到第一个count为1的node，返回
因为说明这个node下面没有分支了，他就应该是唯一的
Time Complexity: O(NL), l is the max length of the word.
If we can't use extrac space:
排序，然后计算相邻字符串的最短前缀，这个时间复杂度比较高，应该是O(nLog(n))
Read full article from Shortest unique prefix to represent word in an array | shawnlincoding
Use the shorest unique prefix to represent each word in the array
input: ["zebra", "dog", "duck",”dot”]
output: {zebra: z, dog: do, duck: du}
[zebra, dog, duck, dove]
{zebra:z, dog: dog, duck: du, dove: dov}
[bearcat, bear]
{bearcat: bearc, bear: ""}
http://www.geeksforgeeks.org/find-all-shortest-unique-prefixes-to-represent-each-word-in-a-given-list/
A Simple Solution is to consider every prefix of every word (starting from the shortest to largest), and if a prefix is not prefix of any other string, then print it.
An Efficient Solution is to use Trie. The idea is to maintain a count in every node. Below are steps.
1) Construct a Trie of all words. Also maintain frequency of every node (Here frequency is number of times node is visited during insertion). Time complexity of this step is O(N) where N is total number of characters in all words.
2) Now, for every word, we find the character nearest to the root with frequency as 1. The prefix of the word is path from root to this character. To do this, we can traverse Trie starting from root. For every node being traversed, we check its frequency. If frequency is one, we print all characters from root to this node and don’t traverse down this node.
Time complexity if this step also is O(N) where N is total number of characters in all words.
root
/ \
(d, 3)/ \(z, 1)
/ \
Node1 Node2
/ \ \
(o,2)/ \(u,1) \(e,1)
/ \ \
Node1.1 Node1.2 Node2.1
/ \ \ \
(g,1)/ \ (t,1) \(c,1) \(b,1)
/ \ \ \
Leaf Leaf Node1.2.1 Node2.1.1
(dog) (dot) \ \
\(k, 1) \(r, 1)
\ \
Leaf Node2.1.1.1
(duck) \
\(a,1)
\
Leaf
(zebra)
struct trieNode{ struct trieNode *child[MAX]; int freq; // To store frequency};// Function to create a new trie node.struct trieNode *newTrieNode(void){ struct trieNode *newNode = new trieNode; newNode->freq = 1; for (int i = 0; i<MAX; i++) newNode->child[i] = NULL; return newNode;}// Method to insert a new string into Trievoid insert(struct trieNode *root, string str){ // Length of the URL int len = str.length(); struct trieNode *pCrawl = root; // Traversing over the length of given str. for (int level = 0; level<len; level++) { // Get index of child node from current character // in str. int index = str[level]; // Create a new child if not exist already if (!pCrawl->child[index]) pCrawl->child[index] = newTrieNode(); else (pCrawl->child[index]->freq)++; // Move to the child pCrawl = pCrawl->child[index]; }}// This function prints unique prefix for every word stored// in Trie. Prefixes one by one are stored in prefix[].// 'ind' is current index of prefix[]void findPrefixesUtil(struct trieNode *root, char prefix[], int ind){ // Corner case if (root == NULL) return; // Base case if (root->freq == 1) { prefix[ind] = '\0'; cout << prefix << " "; return; } for (int i=0; i<MAX; i++) { if (root->child[i] != NULL) { prefix[ind] = i; findPrefixesUtil(root->child[i], prefix, ind+1); } }}// Function to print all prefixes that uniquely// represent all words in arr[0..n-1]void findPrefixes(string arr[], int n){ // Construct a Trie of all words struct trieNode *root = newTrieNode(); root->freq = 0; for (int i = 0; i<n; i++) insert(root, arr[i]); // Create an array to store all prefixes char prefix[MAX_WORD_LEN]; // Print all prefixes using Trie Traversal findPrefixesUtil(root, prefix, 0);}看到prefix肯定想到的是trie咯，在trie node里面多加一个field count，表示
这个字符出现多少次
1，insert word into the trie
2, search the word, 找到第一个count为1的node，返回
因为说明这个node下面没有分支了，他就应该是唯一的
Time Complexity: O(NL), l is the max length of the word.
public static final int R = 256; private Node root; private class Node{ private int count; private boolean isEnd; private Node next[] = new Node[R]; public Node(){ count = 0; isEnd = false; } public Node(int count, boolean isEnd){ this.count = count; this.isEnd = isEnd; } } public void insert(String str){ if(root == null) root = new Node(); Node curr = root; for(int i = 0; i < str.length(); i++){ char c = str.charAt(i); if(curr.next[c] == null){ curr.next[c] = new Node(1, false); }else{ curr.next[c].count++; } curr = curr.next[c]; } curr.isEnd = true; }If we can't use extrac space:
排序，然后计算相邻字符串的最短前缀，这个时间复杂度比较高，应该是O(nLog(n))
Read full article from Shortest unique prefix to represent word in an array | shawnlincoding
