Skip to main content

word break | Dynamic programming | geeks for geeks solution

 Given a string A and a dictionary of n words B, find out if A can be segmented into a space-separated sequence of dictionary words.

Note: From the dictionary B each word can be taken any number of times and in any order.
Example 1:

Input:
n = 12
B = { "i", "like", "sam",
"sung", "samsung", "mobile",
"ice","cream", "icecream",
"man", "go", "mango" }
A = "ilike"
Output:
1
Explanation:
The string can be segmented as "i like".


Example 2:

Input:
n = 12
B = { "i", "like", "sam",
"sung", "samsung", "mobile",
"ice","cream", "icecream", 
"man", "go", "mango" }
A = "ilikesamsung"
Output:
1
Explanation:
The string can be segmented as 
"i like samsung" or "i like sam sung".

 

Your Task:
Complete wordBreak() function which takes a string and list of strings as a parameter and returns 1 if it is possible to break words, else return 0. You don't need to read any input or print any output, it is done by driver code.


Answer : 


class Solution

{

    bool find_word(vector<string>&b , string a)

    {

        for(auto i = b.begin() ; i!= b.end() ; i++)

        {

            if(a == *i)

             return true;

        }

        return false;

    }

public:

    int wordBreak(string A, vector<string> &B) {

         

         if(A.length() == 0)

         {

             return 1;

         }

        

        for(int i =0 ; i < A.length() ; i++)

        { 

            string p = A.substr(0 , i+1);

            if(find_word(B, p))

               {

                  if(wordBreak(A.substr(i+1) , B))

                    return 1;

               }

            

        }

        return 0;

    }

};





Labels:

Comments

Post a Comment

Popular posts from this blog

leetcode 48 solution

  48 .  Rotate Image You are given an  n x n  2D  matrix  representing an image, rotate the image by  90  degrees (clockwise). You have to rotate the image  in-place , which means you have to modify the input 2D matrix directly.  DO NOT  allocate another 2D matrix and do the rotation.   Example 1: Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [[7,4,1],[8,5,2],[9,6,3]] Example 2: Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]] Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]   Constraints: n == matrix.length == matrix[i].length 1 <= n <= 20 -1000 <= matrix[i][j] <= 1000 solution: class Solution { public:     void swap(int& a , int &b)     {         int c ;         c = a;         a = b;         b = c;     }     void transpose (vector<vector<int>...
Post a Comment
Read more

2485. Find the Pivot Integer | Binary search

  Given a positive integer   n , find the   pivot integer   x   such that: The sum of all elements between  1  and  x  inclusively equals the sum of all elements between  x  and  n  inclusively. Return  the pivot integer  x . If no such integer exists, return  -1 . It is guaranteed that there will be at most one pivot index for the given input.   Example 1: Input: n = 8 Output: 6 Explanation: 6 is the pivot integer since: 1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8 = 21. Example 2: Input: n = 1 Output: 1 Explanation: 1 is the pivot integer since: 1 = 1. Example 3: Input: n = 4 Output: -1 Explanation: It can be proved that no such integer exist.   Constraints: 1 <= n <= 1000 Solution : class Solution { publ ic:     int pivotInteger( int n ) {         int sum = (( n )*( n + 1 ))/ 2 ;         int i = 1 ;         int j =...
Post a Comment
Read more

Regular Expression Matching Leetcode Solution

Regular Expression Matching Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where: '.' Matches any single character.​​​​ '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). Example 1: Input: s = "aa", p = "a"  Output: false  Explanation: "a" does not match the entire string "aa". Example 2: Input: s = "aa", p = "a*"  Output: true  Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa". Example 3: Input: s = "ab", p = ".*"  Output: true  Explanation: ".*" means "zero or more (*) of any character (.)". Constraints: 1 <= s.length <= 20 1 <= p.length <= 20 s contains only lowercase English letters. p contains only lowercase Englis...
Post a Comment
Read more