Tutorial(My experiences)-1-(Dynamic Programming)

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

So, most of us have heard about this term “DP” a lot and have seen people in fear as soon as they listen to it.

I used to fear DP a lot as well, but I found a way to tackle it, since then I have started to fall in love with DP.

I’ll share my way :-)

I’ll make you fall in love with this topic.

Important points:-

1. To be good at DP, all you have to do is solve classical DP-problems and their variations.
2. Recognize patterns and improve upon your thinking skills in DP
3. Love DP, it will love you back !

There are many variations of DP and I’ll discuss them all one by one so you can become a pro!

Introduction to DP:-

To find solution to a problem, we divide the problem into sub-problems, find answers to those sub-problems,combine them to get the original answer!

That’s it!

Example:- Say I ask you to calculate :- (1+2+3+4+5)

You do this:-

1. Break it into sub-problems : (1+2) + (3+4) +(5)
2. Find answers to those sub-problems: (3) + (7) + (5)
3. Combine them to get the answer to the original problem : 15 :-)

That’s what we call dynamic programming ! :-)

My personal trick :-

dp[i] usually mean the best answer to the problem till the i’th index of the array.

Obviously, final answer will be dp[n](where ’n’ is the size of the array)

We cannot calculate dp[n] directly, we first need to calculate dp[1],dp[2],… and combine their results to find the value of dp[n] :-)

Problem-1 : We are given an array of integers(a[n]) . We are given multiple queries of the form : (1, i) which means we need to output the sum of all numbers from index- ‘1’ to index ‘i’ of the array.

Analysis : Running a loop for each query and finding the sum is a good idea but not very efficient as it takes O(N*Q) time.

Lets create a dp-array of size ‘n’.

dp[1]=sum of all numbers from (1,1)

dp[2]=sum of all numbers from (1,2)…

and so on.

Say, a[5]={4,5,3,2,1}…(assume 1-based-indexing here)

So, dp[1]=4(pretty easy)…..(1)

dp[2]=4+5=9………(2)

dp[3]=4+5+3=12…..(3) and so on.

So, for any index ‘i’ ,

dp[i]=a[i]+dp[i-1],

Example:-

dp[3] = a[3] + dp[3–1] = a[3] + dp[2]= 3 + 9 =12 ….(which is same as equation…(3))

Piece of code(C++) :-

dp[0]=0;

int i=1;

while(i≤n)

{

dp[i]=a[i]+dp[i-1];

i++;

}

This took O(N) time!

Voila! We did it!

So, now for all the queries, all we gotta do is output the value of dp[i], which we have already pre-computed :-)

Time taken:- O(N+Q) [ O(N) for pre-computation, as we answered each query in O(1), total time taken was O(Q) ]

So you get the basic idea of DP is to make a recurrence relation,and then run a loop,and calculate(pre-compute) the values :-)

My aim is to start this series from very beginner level and make you reach the advanced level :-)

Tutorial-2 is coming soon!!

Link to second tutorial : https://medium.com/@karangujar43/tutorial-2-my-experiences-dynamic-programming-7254e35cec1a

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!”

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