Given an input string and a dictionary of words, find out if the input string can be segmented into a space-separated sequence of dictionary words. See following examples for more details.
This is a famous Google interview question, also being asked by many other companies now a days.
Consider the following dictionary
{ i, like, go, hired, gohired, site, sweet,fruit, man, go, mango}
Input: ilike
Output: Yes
The string can be segmented as "i like".
Input: ilikemango
Output: Yes
The string can be segmented as "i like mango" or "i like man go".
Recursive implementation:
Very easy method to keep dictionary in Hash.
Recursively traverse each substring and find whether dictionary contains it or not
example :
SubString : i , restString :likemango
for likemango
SubString : l, restString :ikemango
i-likemango
l-ikemango
li-kemango
lik-emango
like-mango
m-ango
ma-ango
man-go
g-o
go-
Finally as for I , Like and Mango , it returns true, it can will finally result true.
Code :
#include <iostream>
using namespace std;
int dictionaryContains(string word)
{
string dictionary[] = {"i", "like", "go", "hired", "gohired", "site", "sweet","fruit", "man", "go", "mango"};
int size = sizeof(dictionary)/sizeof(dictionary[0]);
for (int i = 0; i < size; i++)
if (dictionary[i].compare(word) == 0)
return true;
return false;
}
bool wordBreak(string str)
{
int size = str.size();
if (size == 0) return true;
for (int i=1; i<=size; i++)
{
cout << str.substr(0, i) <<"-" <<str.substr(i, size-i)<<"\n";
if (dictionaryContains( str.substr(0, i) ) &&
wordBreak( str.substr(i, size-i) ))
return true;
}
return false;
}
// Driver program to test above functions
int main()
{
wordBreak("ilikemango")? cout <<"Yes\n": cout << "No\n";
return 0;
}
Working Code : http://ideone.com/rGnFmp
As this Approach can be done with Recursive way, IT can be done with Dynamic Programming..
W'll Put Dynamic Programming solution later..
Now lets see when We input String ilikemngo What can be output trace?
Hear we have found that "ilike" can be created in sentence, then once "mngo" iteration completes, we again check for likemngo, and ilikemango .... as we can see in output trace.
So in order to reduce recursion and re-computation we can store required output.
that can be done with the help of Dynamic Programming.
Now if What we can store and how to proceed
We need to store information in order to reduce computation of substring can be broken into sentence or not... so
we can store like
i - TRUE
il - False
ili - False
ilik - False
ilike - TRUE
.
.
. and so on.
So we can store such values in a Array Arr[i], which will store the value 1 if string [0 .. i] can form a sentence.
#include <iostream>
#include <string.h>
using namespace std;
int dictionaryContains(string word)
{
string dictionary[] = {"i", "like", "go", "hired", "gohired", "site", "sweet","fruit", "man", "go", "mango"};
int size = sizeof(dictionary)/sizeof(dictionary[0]);
for (int i = 0; i < size; i++)
if (dictionary[i].compare(word) == 0)
return true;
return false;
}
bool wordBreak(string str)
{
int size = str.size();
if (size == 0) return true;
bool wb[size+1];
memset(wb, 0, sizeof(wb)); // Initialize all values as false.
for (int i=1; i<=size; i++)
{
// if wb[i] is false, then check if current prefix can make it true.
// Current prefix is "str.substr(0, i)"
if (wb[i] == false && dictionaryContains( str.substr(0, i) )){
wb[i] = true;
//cout<<str.substr(0, i)<<wb[i]<<"\n";
}
// wb[i] is true, then check for all substrings starting from
// (i+1)th character and store their results.
if (wb[i] == true)
{
// If we reached the last prefix
if (i == size)
return true;
for (int j = i+1; j <= size; j++)
{
if (wb[j] == false && dictionaryContains( str.substr(i, j-i) )){
wb[j] = true;
}
// If we reached the last character
if (j == size && wb[j] == true)
return true;
}
}
}
for (int i = 1; i <= size; i++)
cout<<str.substr(0, i) << "-" <<wb[i]<<"\n";
// If we have tried all prefixes and none of them worked
return false;
}
// Driver program to test above functions
int main()
{
wordBreak("ilikemngo")? cout <<"Yes\n": cout << "No\n";
return 0;
}
OutPut of Above Programm
You can see working code and understand more at : http://ideone.com/rvQZdI
Code courtesy : geeksforgeeks.org
This is a famous Google interview question, also being asked by many other companies now a days.
Consider the following dictionary
{ i, like, go, hired, gohired, site, sweet,fruit, man, go, mango}
Input: ilike
Output: Yes
The string can be segmented as "i like".
Input: ilikemango
Output: Yes
The string can be segmented as "i like mango" or "i like man go".
Recursive implementation:
Very easy method to keep dictionary in Hash.
Recursively traverse each substring and find whether dictionary contains it or not
example :
SubString : i , restString :likemango
for likemango
SubString : l, restString :ikemango
i-likemango
l-ikemango
li-kemango
lik-emango
like-mango
m-ango
ma-ango
man-go
g-o
go-
Finally as for I , Like and Mango , it returns true, it can will finally result true.
Code :
#include <iostream>
using namespace std;
int dictionaryContains(string word)
{
string dictionary[] = {"i", "like", "go", "hired", "gohired", "site", "sweet","fruit", "man", "go", "mango"};
int size = sizeof(dictionary)/sizeof(dictionary[0]);
for (int i = 0; i < size; i++)
if (dictionary[i].compare(word) == 0)
return true;
return false;
}
bool wordBreak(string str)
{
int size = str.size();
if (size == 0) return true;
for (int i=1; i<=size; i++)
{
cout << str.substr(0, i) <<"-" <<str.substr(i, size-i)<<"\n";
if (dictionaryContains( str.substr(0, i) ) &&
wordBreak( str.substr(i, size-i) ))
return true;
}
return false;
}
// Driver program to test above functions
int main()
{
wordBreak("ilikemango")? cout <<"Yes\n": cout << "No\n";
return 0;
}
Working Code : http://ideone.com/rGnFmp
As this Approach can be done with Recursive way, IT can be done with Dynamic Programming..
W'll Put Dynamic Programming solution later..
Now lets see when We input String ilikemngo What can be output trace?
i-likemngo
l-ikemngo
li-kemngo
lik-emngo
like-mngo
m-ngo
mn-go
mng-o
mngo-
likem-ngo
likemn-go
likemng-o
likemngo-
il-ikemngo
ili-kemngo
ilik-emngo
ilike-mngo
ilikem-ngo
ilikemn-go
ilikemng-o
ilikemngo-
Hear we have found that "ilike" can be created in sentence, then once "mngo" iteration completes, we again check for likemngo, and ilikemango .... as we can see in output trace.
So in order to reduce recursion and re-computation we can store required output.
that can be done with the help of Dynamic Programming.
Now if What we can store and how to proceed
We need to store information in order to reduce computation of substring can be broken into sentence or not... so
we can store like
i - TRUE
il - False
ili - False
ilik - False
ilike - TRUE
.
.
. and so on.
So we can store such values in a Array Arr[i], which will store the value 1 if string [0 .. i] can form a sentence.
#include <iostream>
#include <string.h>
using namespace std;
int dictionaryContains(string word)
{
string dictionary[] = {"i", "like", "go", "hired", "gohired", "site", "sweet","fruit", "man", "go", "mango"};
int size = sizeof(dictionary)/sizeof(dictionary[0]);
for (int i = 0; i < size; i++)
if (dictionary[i].compare(word) == 0)
return true;
return false;
}
bool wordBreak(string str)
{
int size = str.size();
if (size == 0) return true;
bool wb[size+1];
memset(wb, 0, sizeof(wb)); // Initialize all values as false.
for (int i=1; i<=size; i++)
{
// if wb[i] is false, then check if current prefix can make it true.
// Current prefix is "str.substr(0, i)"
if (wb[i] == false && dictionaryContains( str.substr(0, i) )){
wb[i] = true;
//cout<<str.substr(0, i)<<wb[i]<<"\n";
}
// wb[i] is true, then check for all substrings starting from
// (i+1)th character and store their results.
if (wb[i] == true)
{
// If we reached the last prefix
if (i == size)
return true;
for (int j = i+1; j <= size; j++)
{
if (wb[j] == false && dictionaryContains( str.substr(i, j-i) )){
wb[j] = true;
}
// If we reached the last character
if (j == size && wb[j] == true)
return true;
}
}
}
for (int i = 1; i <= size; i++)
cout<<str.substr(0, i) << "-" <<wb[i]<<"\n";
// If we have tried all prefixes and none of them worked
return false;
}
// Driver program to test above functions
int main()
{
wordBreak("ilikemngo")? cout <<"Yes\n": cout << "No\n";
return 0;
}
OutPut of Above Programm
i-1
il-0
ili-0
ilik-0
ilike-1
ilikem-0
ilikemn-0
ilikemng-0
ilikemngo-0
You can see working code and understand more at : http://ideone.com/rvQZdI
Code courtesy : geeksforgeeks.org
