Tutorial-2-(My Experiences)-Dynamic Programming

My video tutorials on competitive programming:-https://www.youtube.com/watch?v=Y4AK8Q7_wcM

Tutorial-2 is out!!!!!

Fasten your seat-belt!!!

If you haven’t read the first part yet, then here it is : https://medium.com/@karangujar43/tutorial-my-experiences-1-dynamic-programming-5fa956967c6d

I like to explain and express things in such a manner so that everybody can easily understand.That’s the reason I am able to connect with so many competitive programmers !!!

So, let’s dive deep into DP as promised :-)

Problem-2:- Understanding this problem and its solution properly will make a strong foundation for you in the DP world .(This worked for me :-) )

Here we go- Given an array of integers(positive as well as negative) ,select some elements from this array(select a subset) such that:-

1. Sum of those elements is maximum(Sum of the subset is maximum) .
2. No 2 elements in the subset should be consecutive.

Example :- {2,4,6,7,8}

Answer:- {2+6+8=16}

Common Trick : We create a dp-array , and dp[i] means the maximum sum we could get till index-’i’ of the array.

For the above example,

dp[1] = 2 (2), [This is the best answer you could get if size of the array was one]

dp[2]= 4(4),[This is the best answer you could get if size of the array was two]

dp[3]=8(6+2),[This is the best answer you could get if size of the array was three]………lets call this equation-(1)…

dp[4]=11(7+4),[This is the best answer you could get if size of the array was four]

dp[5]=16(2+6+8),[[This is the best answer you could get if size of the array was five]

Now , the next thing that I am going to say, you gotta feel it deeply . If you are able to do it, you win!!

Say, I have calculated dp[1],dp[2] and dp[3] by pure observation.

Now, I have to calculate dp[4].

So I have only 2 choices:-

i)Include the 4'th element , if I do this, I can’t include the 3rd element as consecutive elements are not allowed, so ,

dp[4]= a[4]+dp[2]…..(answer will be 4th element plus the best answer till index ‘2’ of the array)

ii)Exclude the 4th element, we don’t select it! So the previous answer is the current answer, we don’t change anything,

dp[4]=dp[3],

Final answer is the maximum of two choices :->

dp[4]=maximum(a[4]+dp[2],dp[3])

=maximum(7+4,8)

=maximum(11,8)

=11…..(verify this with equation…(1))

Voila, we did it !!!

So,the recurrence relation for this problem is,

for any ‘i’ ,

dp[i]=max(a[i]+dp[i-2],dp[i-1])

We will calculate dp[i] for all ‘i’ from (1 to n) using the above formula.

And dp[n] will be the maximum possible sum for the whole array!!

Woohooo!!!!

Some code:-

dp[0] = 0;

dp[1] = max(a[1],dp[0]);

i = 2;

while(i≤n)

{

dp[i] = max(dp[i-2]+a[i], dp[i-1]);

i++;

}

Tutorial-3 is coming soon!! ❤ ;)

Link to third tutorial : https://medium.com/@karangujar43/tutorial-3-my-experiences-dynamic-programming-c711b9ff1071