Word break problem from gfg

#1

I’m attaching a link to the article

Can you tell me inside wordBreakUtil when we are caliing the function recursively what actually we are doing?

#2

Hey @div_yanshu07
Say dictionary is [ab,cd]
Now u had string abcdab

U found ab now in next call we are sending remaining string cdab
along with its length n-i i.e 4
along with formed string so far i.e ab
so wordBreakUtil(“cdab”,4,"ab ")

#3

Why are we sending the string formed so far? So, that we can concatenate??
If yes please give example!

#4

Yup since there can be multiple outputs like see this one
Consider the following dictionary
{ i, sam, sung, samsung}
Input: “isamsung”
Output: i sam sung
i samsung

Do a dry run on this and you will understand everything properly

#5

https://practice.geeksforgeeks.org/problems/word-break-part-2/0#

Yes, now my code is generating all the possible options but it doesn’t match with the output.
How am I supposed to generate those brackets

#6

Hey @div_yanshu07

        // cout<<"(";removed
        for(auto it: v)
        {
            cout<<"("<<it<<")";//updated
        }
        // cout<<")";removed
#7

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