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 : 1

Asked in: Facebook

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 keylength 
  ....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 node
struct 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 node
void 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 where 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 functions
int 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.

Python3

# Python program to find minimum breaks needed
# to break a string in dictionary words.

import sys


class TrieNode:

    def __init__(self):

        self.endOfTree = False
        self.children = [None for i in range(26)]



root = TrieNode()
minWordBreak = sys.maxsize

# If not present, inserts a key into the trie
    # If the key is the prefix of trie node, just
    # marks leaf node
def insert(key):

    global root,minWordBreak

    length = len(key)

    pcrawl = root

    for i in range(length):

        index = ord(key[i])- ord('a')

        if(pcrawl.children[index] == None):
            pcrawl.children[index] = 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.
def _minWordBreak(key):

    global minWordBreak

    minWordBreak = sys.maxsize

    minWordBreakUtil(root, key, 0, sys.maxsize, 0)

def minWordBreakUtil(node,key,start,min_Break,level):

    global minWordBreak,root

    pCrawl = node

    # base case, update minimum Break
    if (start == len(key)):
            min_Break = min(min_Break, level - 1)
            if(min_Break<minWordBreak):
                minWordBreak = min_Break

            return


    # traverse given key(pattern)
    for i in range(start,len(key)):
        index = ord(key[i]) - ord('a')
        if (pCrawl.children[index]==None):
            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
keys=["cat", "mat", "ca", "ma", "at", "c", "dog", "og", "do" ]

for i in range(len(keys)):
    insert(keys[i])

_minWordBreak("catmatat")

print(minWordBreak)

# This code is contributed by shinjanpatra

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 node
function 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 code
let 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

Time Complexity: O(n*n*L) where n is the length of the input string and L is the maximum length of word in the dictionary.
Auxiliary Space: O(ALPHABET_SIZE^L+n*L)

Approach 2: Using Dynamic Programming

• We maintain an array dp where dp[i] represents the minimum number of breaks needed to break the substring s[0…i-1] into dictionary words. 
• We initialize dp[0] = 0 since the empty string can be broken into zero words. 
• For each position i in the string, we iterate over all dictionary words and check if the substring ending at position i matches the current dictionary word. 
• If it does, we update dp[i] to be the minimum of dp[i] and dp[i – len] + 1, where len is the length of the current dictionary word. 
• Finally, we return dp[n] – 1, where n is the length of the input string, since the number of breaks needed is one less than the number of words.

C++

#include <bits/stdc++.h>
using namespace std;

const int INF = 1e9;

int minWordBreak(string s, vector<string>& dict) {
    int n = s.length();
    vector<int> dp(n + 1, INF);
    dp[0] = 0;

    for (int i = 1; i <= n; i++) {
        for (string word : dict) {
            int len = word.length();
            if (i >= len && s.substr(i - len, len) == word) {
                dp[i] = min(dp[i], dp[i - len] + 1);
            }
        }
    }

    return dp[n] - 1;
}

int main() {
    vector<string> dict = {"Cat", "Mat", "Ca", "Ma", "at", "C", "Dog", "og", "Do"};
    string s = "CatMatat";
    cout << minWordBreak(s, dict) << endl;
    return 0;
}

Java

import java.util.*;

public class WordBreak {
    static final int INF = (int)1e9;

    public static int minWordBreak(String s,
                                   List<String> dict)
    {
        int n = s.length();
        int[] dp = new int[n + 1];
        Arrays.fill(dp, INF);
        dp[0] = 0;

        for (int i = 1; i <= n; i++) {
            for (String word : dict) {
                int len = word.length();
                if (i >= len
                    && s.substring(i - len, i)
                           .equals(word)) {
                    dp[i]
                        = Math.min(dp[i], dp[i - len] + 1);
                }
            }
        }

        return dp[n] - 1;
    }

    public static void main(String[] args)
    {
        List<String> dict
            = Arrays.asList("Cat", "Mat", "Ca", "Ma", "at",
                            "C", "Dog", "og", "Do");
        String s = "CatMatat";
        System.out.println(minWordBreak(s, dict));
    }
}

Python3

def min_word_break(s, dict):
    n = len(s)
    dp = [float('inf')] * (n + 1)  # Initialize a list to store minimum word breaks needed
    dp[0] = 0  # No breaks needed for an empty string

    for i in range(1, n + 1):
        for word in dict:
            length = len(word)
            if i >= length and s[i - length:i] == word:
                # Check if the substring from i-length to i matches a word in the dictionary
                dp[i] = min(dp[i], dp[i - length] + 1)
                # Update the minimum word breaks needed at position i

    return dp[n] - 1  # Return the minimum word breaks needed for the entire string

def main():
    dict = ["Cat", "Mat", "Ca", "Ma", "at", "C", "Dog", "og", "Do"]
    s = "CatMatat"
    min_breaks = min_word_break(s, dict)
    print(f"The minimum word breaks needed: {min_breaks}")

if __name__ == "__main__":
    main()
# This code is contributed by shivamgupta0987654321

C#

using System;
using System.Collections.Generic;

public class GFG
{
    const int INF = 1000000000;
    public static int MinWordBreak(string s, List<string> dict)
    {
        int n = s.Length;
        int[] dp = new int[n + 1];
        for (int i = 0; i <= n; i++)
            dp[i] = INF;
        dp[0] = 0;
        for (int i = 1; i <= n; i++)
        {
            foreach (string word in dict)
            {
                int len = word.Length;
                if (i >= len && s.Substring(i - len, len) == word)
                {
                    dp[i] = Math.Min(dp[i], dp[i - len] + 1);
                }
            }
        }
        return dp[n] - 1;
    }
    public static void Main(string[] args)
    {
        List<string> dict = new List<string> { "Cat", "Mat", "Ca", "Ma", "at", "C", "Dog", "og", "Do" };
        string s = "CatMatat";
        int result = MinWordBreak(s, dict);
        Console.WriteLine(result);
    }
}

Javascript

// Define a constant for infinity, which will be used as an initial value in the DP array
const INF = 1e9;

// Function to find the minimum word break count
function minWordBreak(s, dict) {
    const n = s.length; // Length of the input string
    const dp = new Array(n + 1).fill(INF); // Initialize a DP array with initially infinite values
    dp[0] = 0; // The base case: No breaks needed for an empty string

    // Iterate over the characters in the input string
    for (let i = 1; i <= n; i++) {
        // Iterate over the words in the dictionary
        for (const word of dict) {
            const len = word.length; // Length of the current dictionary word

            // Check if the current substring (ending at position i) matches the dictionary word
            if (i >= len && s.substring(i - len, i) === word) {
                // If it matches, update the DP array with the minimum break count
                dp[i] = Math.min(dp[i], dp[i - len] + 1);
            }
        }
    }

    return dp[n] - 1; // Subtract 1 to get the minimum break count
}

// Driver code
const dict = ["Cat", "Mat", "Ca", "Ma", "at", "C", "Dog", "og", "Do"];
const s = "CatMatat";
console.log("Minimum Word Break Count:", minWordBreak(s, dict));

Output

2

Note that this code has a time complexity of O(n*m), where n is the length of the input string and m is the size of the dictionary. This is because we iterate over all positions in the string and for each position, we iterate over all words in the dictionary.

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