Yes Dave, you are right.
I hope algorithm below will work.
maintain a hash table with keys as the set of keywords and values as a
boolean value initialized with false.
initialize a counter count with 0
let n be the total number of keywords.
word = 1st word of the text
while( word )
{
stack1.pushe( word )
if word matches with keyword then
{
go to hash table and
if value is false then
make value true and increment count by 1.
if value is true then
do nothing
if( count == n ) then
break from while loop
word = next word in text
}
if(count == n)
{
count = 0
make all the values in hash table false
while( stack1 not empty or count != n)
{
tmp_word = stack1.pop()
queue1.enque( tmp_word )
if word matches with keyword then
{
go to hash table and
if value is false then
make value true and increment count by 1.
if value is true then
do nothing
}
}
print elements of the queue1 by poping them out.
} else { text doesnot contain subtext }
text does not have subtext
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