Minimum Word Break
Given a string s, break s such that every substring of the partition can be found in the dictionary. Return the minimum break needed.
Examples:
Given a dictionary ["Cat", "Mat", "Ca",
"tM", "at", "C", "Dog", "og", "Do"]
Input : Pattern "CatMat"
Output : 1
Explanation: we can break the sentences
in three ways, as follows:
CatMat = [ Cat Mat ] break 1
CatMat = [ Ca tM at ] break 2
CatMat = [ C at Mat ] break 2 so the
output is: 1
Input : Dogcat
Output : 1Asked in: Facebook
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Solution of this problem is based on the WordBreak Trie solution and level ordered graph. We start traversing given pattern and start finding a character of pattern in a trie. If we reach a node(leaf) of a trie from where we can traverse a new word of a trie(dictionary), we increment level by one and call search function for rest of the pattern character in a trie. In the end, we return minimum Break.
MinBreak(Trie, key, level, start = 0 )
.... If start == key.length()
...update min_break
for i = start to keylenght
....If we found a leaf node in trie
MinBreak( Trie, key, level+1, i )Below is the implementation of above idea
C++
// C++ program to find minimum breaks needed// to break a string in dictionary words.#include <bits/stdc++.h>using namespace std;const int ALPHABET_SIZE = 26;// trie nodestruct TrieNode { struct TrieNode* children[ALPHABET_SIZE]; // isEndOfWord is true if the node // represents end of a word bool isEndOfWord;};// Returns new trie node (initialized to NULLs)struct TrieNode* getNode(void){ struct TrieNode* pNode = new TrieNode; pNode->isEndOfWord = false; for (int i = 0; i < ALPHABET_SIZE; i++) pNode->children[i] = NULL; return pNode;}// If not present, inserts the key into the trie// If the key is the prefix of trie node, just// marks leaf nodevoid insert(struct TrieNode* root, string key){ struct TrieNode* pCrawl = root; for (int i = 0; i < key.length(); i++) { int index = key[i] - 'a'; if (!pCrawl->children[index]) pCrawl->children[index] = getNode(); pCrawl = pCrawl->children[index]; } // mark last node as leaf pCrawl->isEndOfWord = true;}// function break the string into minimum cut// such the every substring after breaking// in the dictionary.void minWordBreak(struct TrieNode* root, string key, int start, int* min_Break, int level = 0){ struct TrieNode* pCrawl = root; // base case, update minimum Break if (start == key.length()) { *min_Break = min(*min_Break, level - 1); return; } // traverse given key(pattern) int minBreak = 0; for (int i = start; i < key.length(); i++) { int index = key[i] - 'a'; if (!pCrawl->children[index]) return; // if we find a condition were we can // move to the next word in a trie // dictionary if (pCrawl->children[index]->isEndOfWord) minWordBreak(root, key, i + 1, min_Break, level + 1); pCrawl = pCrawl->children[index]; }}// Driver program to test above functionsint main(){ string dictionary[] = { "Cat", "Mat", "Ca", "Ma", "at", "C", "Dog", "og", "Do" }; int n = sizeof(dictionary) / sizeof(dictionary[0]); struct TrieNode* root = getNode(); // Construct trie for (int i = 0; i < n; i++) insert(root, dictionary[i]); int min_Break = INT_MAX; minWordBreak(root, "CatMatat", 0, &min_Break, 0); cout << min_Break << endl; return 0;} |
Java
// Java program to find minimum breaks needed// to break a string in dictionary words.public class Trie {TrieNode root = new TrieNode();int minWordBreak = Integer.MAX_VALUE; // Trie node class TrieNode { boolean endOfTree; TrieNode children[] = new TrieNode[26]; TrieNode(){ endOfTree = false; for(int i=0;i<26;i++){ children[i]=null; } } } // If not present, inserts a key into the trie // If the key is the prefix of trie node, just // marks leaf node void insert(String key){ int length = key.length(); int index; TrieNode pcrawl = root; for(int i = 0; i < length; i++) { index = key.charAt(i)- 'a'; if(pcrawl.children[index] == null) pcrawl.children[index] = new TrieNode(); pcrawl = pcrawl.children[index]; } // mark last node as leaf pcrawl.endOfTree = true; } // function break the string into minimum cut // such the every substring after breaking // in the dictionary. void minWordBreak(String key) { minWordBreak = Integer.MAX_VALUE; minWordBreakUtil(root, key, 0, Integer.MAX_VALUE, 0); } void minWordBreakUtil(TrieNode node, String key, int start, int min_Break, int level) { TrieNode pCrawl = node; // base case, update minimum Break if (start == key.length()) { min_Break = Math.min(min_Break, level - 1); if(min_Break<minWordBreak){ minWordBreak = min_Break; } return; } // traverse given key(pattern) for (int i = start; i < key.length(); i++) { int index = key.charAt(i) - 'a'; if (pCrawl.children[index]==null) return; // if we find a condition were we can // move to the next word in a trie // dictionary if (pCrawl.children[index].endOfTree) { minWordBreakUtil(root, key, i + 1, min_Break, level + 1); } pCrawl = pCrawl.children[index]; } } // Driver code public static void main(String[] args) { String keys[] = {"cat", "mat", "ca", "ma", "at", "c", "dog", "og", "do" }; Trie trie = new Trie(); // Construct trie int i; for (i = 0; i < keys.length ; i++) trie.insert(keys[i]); trie.minWordBreak("catmatat"); System.out.println(trie.minWordBreak); }}// This code is contributed by Pavan Koli. |
C#
// C# program to find minimum breaks needed// to break a string in dictionary words.using System;class Trie{ TrieNode root = new TrieNode(); int minWordBreak = int.MaxValue ; // Trie node public class TrieNode { public bool endOfTree; public TrieNode []children = new TrieNode[26]; public TrieNode() { endOfTree = false; for(int i = 0; i < 26; i++) { children[i] = null; } } } // If not present, inserts a key // into the trie If the key is the // prefix of trie node, just marks leaf node void insert(String key) { int length = key.Length; int index; TrieNode pcrawl = root; for(int i = 0; i < length; i++) { index = key[i]- 'a'; if(pcrawl.children[index] == null) pcrawl.children[index] = new TrieNode(); pcrawl = pcrawl.children[index]; } // mark last node as leaf pcrawl.endOfTree = true; } // function break the string into minimum cut // such the every substring after breaking // in the dictionary. void minWordBreaks(String key) { minWordBreak = int.MaxValue; minWordBreakUtil(root, key, 0, int.MaxValue, 0); } void minWordBreakUtil(TrieNode node, String key, int start, int min_Break, int level) { TrieNode pCrawl = node; // base case, update minimum Break if (start == key.Length) { min_Break = Math.Min(min_Break, level - 1); if(min_Break < minWordBreak) { minWordBreak = min_Break; } return; } // traverse given key(pattern) for (int i = start; i < key.Length; i++) { int index = key[i] - 'a'; if (pCrawl.children[index]==null) return; // if we find a condition were we can // move to the next word in a trie // dictionary if (pCrawl.children[index].endOfTree) { minWordBreakUtil(root, key, i + 1, min_Break, level + 1); } pCrawl = pCrawl.children[index]; } } // Driver code public static void Main(String[] args) { String []keys = {"cat", "mat", "ca", "ma", "at", "c", "dog", "og", "do" }; Trie trie = new Trie(); // Construct trie int i; for (i = 0; i < keys.Length ; i++) trie.insert(keys[i]); trie.minWordBreaks("catmatat"); Console.WriteLine(trie.minWordBreak); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript program to find minimum breaks needed// to break a string in dictionary words.class TrieNode{ constructor() { this.endOfTree=false; this.children=new Array(26); for(let i=0;i<26;i++) this.children[i]=null; }}let root = new TrieNode();let minWordBreak = Number.MAX_VALUE;// If not present, inserts a key into the trie // If the key is the prefix of trie node, just // marks leaf nodefunction insert(key){ let length = key.length; let index; let pcrawl = root; for(let i = 0; i < length; i++) { index = key[i].charAt(0)- 'a'.charAt(0); if(pcrawl.children[index] == null) pcrawl.children[index] = new TrieNode(); pcrawl = pcrawl.children[index]; } // mark last node as leaf pcrawl.endOfTree = true;}// function break the string into minimum cut // such the every substring after breaking // in the dictionary.function _minWordBreak(key){ minWordBreak = Number.MAX_VALUE; minWordBreakUtil(root, key, 0, Number.MAX_VALUE, 0);}function minWordBreakUtil(node,key,start,min_Break,level){ let pCrawl = node; // base case, update minimum Break if (start == key.length) { min_Break = Math.min(min_Break, level - 1); if(min_Break<minWordBreak){ minWordBreak = min_Break; } return; } // traverse given key(pattern) for (let i = start; i < key.length; i++) { let index = key[i].charAt(0) - 'a'.charAt(0); if (pCrawl.children[index]==null) return; // if we find a condition were we can // move to the next word in a trie // dictionary if (pCrawl.children[index].endOfTree) { minWordBreakUtil(root, key, i + 1, min_Break, level + 1); } pCrawl = pCrawl.children[index]; }}// Driver codelet keys=["cat", "mat", "ca", "ma", "at", "c", "dog", "og", "do" ];let i;for (i = 0; i < keys.length ; i++) insert(keys[i]);_minWordBreak("catmatat");document.write(minWordBreak);// This code is contributed by rag2127</script> |
Output:
2
