Tutorial-3-(My Experiences)-Dynamic Programming

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

Tutorial “3” has come.

Its party time.

If you haven’t read the second part yet, then here it is :

Tutorial-2-(My Experiences)-Dynamic Programming

Aim of this series:- “ To make you fall in love with Dynamic Programming. To show you the actual beauty of this topic.To go from beginner level to master level.”

Some of the common tricks we have learned as of now :

1. Usually, we create a dp-array , and dp[i] means the maximum sum(best-answer) we could get till index-’i’ of the array.
2. We need to find a recurrence relation, in simple words, we need to create a formula for dp[i] in terms of dp[i-1],dp[i-2],….etc…
3. We run a for loop, which calculates dp[1],dp[2]…..and finally dp[n].
4. dp[n] means the answer for the whole array.

Last time, we discussed this problem

:-

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}

Now, lets solve a harder variation of this problem,which will make our understanding strong.

Modified Version : We are given ‘2’ arrays . Select some elements from both of these arrays (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.(Note:-If you select, say 5'th element from Array-1, then you are not allowed to select 4'th element from either Array-1 or Array-2 :-) )

Example:-

Array 1(a1) : {1,5,3,21234}

Array 2(a2) : {-4509,200,3,40}

Answer:- (200+21234=21434)

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

dp[2]= 200(200),[This is the best answer you could get if size of the array was two]

dp[3]=200(200), [This is the best answer you could get if size of the array was three]………lets call this equation-(1)…

dp[4]=21434(200+21234), [This is the best answer you could get if size of the array was four]

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 (of either array-1 or array-2, if I do this, I can’t include the 3rd element(from both the arrays) as consecutive elements are not allowed, so,

dp[4]= (max(a1[4],a2[4])) + dp[2])…..(answer will be 4th element plus the best answer till index ‘2’ of the arrays)

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(max(a1[4],a2[4])+dp[2],dp[3])

=maximum(max(40,21234)+200,200)

=maximum(21234+200,200)

=maximum(21434,200)

=21434…..(verify this with equation…(1))

Voila, we did it !!!

So,the recurrence relation for this problem is,

for any ‘i’ ,

dp[i]=max( max(a1[i],a2[i]) +dp[i-2],dp[i-1])

Some C++ code:-

//dp[1] and dp[2] is calculated by observation.

//we run a loop from i=3 to i=n , dp[n] is the final answer.

i=3;

while(i≤n)

{

dp[i]=max( max(a1[i],a2[i])+dp[i-2],dp[i-1]);

i++;

}

cout<<dp[n];

Tutorial-4 is coming soon ;) ❤ !!!!

Link to 4th tutorial : https://medium.com/@karangujar43/tutorial-4-my-experiences-dynamic-programming-9df183219b12

One of the reviews of my buyers:-

“ Totally intuitive and beginner-friendly tutorials. Over the course of a month, I’ve tried to start with DP but eventually had to give up every time due to the difficult-to-grab explanations. Thought of giving your tutorial a chance and guess what? I could get it effortlessly due to its lucid nature. Thanks!”