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You are given a list of positive integers along with a sequence of operations from the set $\left \{ *,+\right \}$ .You construct expressions from these two lists so that:

  • The numbers in the expression are drawn from the first list, without repetition and without altering their order.
  • All the operators in the second list are used in the expression, in the same order.

For example, if the two lists are $[1,3,2,1,4]$ and  $[′∗′,′+′]$  the set of possible expressions you can form are 
$1∗3+2,1∗3+1,1∗3+4,1∗2+1,1∗2+4,\ldots,2∗1+4,1∗3+2,1∗3+1,1∗3+4,1∗2+1,1∗2+4,\ldots,2∗1+4$
For each expression, the value is computed by bracketing operators from the right. That is, the expressions above are evaluated as

$1∗(3+2),  1∗(3+1),  1∗(3+4),  1∗(2+1),  1∗(2+4),…,2∗(1+4),  1∗(3+2),  1∗(3+1),  1∗(3+4),  1∗(2+1),  1∗(2+4),…,2∗(1+4)$
The aim is to determine maximum value among these expressions. In this example, the maximum value is $18$, from the expression $3*2+4,$ which is evaluated as $3∗(2+4) = 3*6 =18$. 
You may assume that the length of the first list is more than the length of the second list. 
Describe an algorithm to solve this problem.

asked in Algorithms by (435 points)
edited by | 203 views
0
I think the answer should be O(n log^k (n))
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WHY!!!
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the bracketing done in example is only applicable in that example right?
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In the example they are showing how to bracket the elements of the first list.

Suppose the first list is {1,2,3,4,5} and the second list is {+,*,+}

Then one expression will be 1+(2*(3+4))

I hope this helps.
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I have done the solution below. Check it tell me if anything is wrong.
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I think the answer is correct THANK YOU

1 Answer

+2 votes

Solution

 

 

answered by Loyal (8.4k points)
edited by
+1
You are not taking into account that the numbers from the list should be in the same order, in that case max heap won't work out as it changes the order.

Your algorithm is wrong ,

Let's take an example : Numbers list 2,3,1,4 and operator sequence *,+. In this case the maximum value possible is 3*(1+4)=15 and this does not use the 3 largest number so clearly your solution is wrong.

@Arjun Suresh please the change the accepted answer to not accepted.
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Vinay thanks a lot for pointing out my mistake. Without it I would be happy with my wrong explanation. I have edited my answer can u please check it once more and tell me whether I am wrong or not.
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Nice explaination @Kushagra

According to me, here computing the list A1 should take O(n2) time (since we're comparing the sum of every element of list A every element having index greater than it). Similarly for A2, A3, A4 also (since the question says we can assume that the length of the first list is more than the length of the second list).

Correct me if i am wrong.

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what is the correct answer?  can someone please tell.
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The above answer is correct except the time complexity should be O(n^2 k). As pointed by @Nirmal in the previous comment
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You have assumed that list B contains,

+ , * , * ,+  as operators. But it is given that list B is a set. So it cant have repetitions.

Am I right?
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neither of the lists are sets look at the 1st list of the example given by the question setter [1,3,2,1,4]  1 is repeated in it

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