Dynamic Programming Series

Day 3 :

Orac and Models

Problem Description

You are given n models in the shop numbered from 1 to n with sizes s1,s2......,sn.

An arrangement is good if some index j is divisible i and Sj > Si.

Problem Description

You are given n models in the shop numbered from 1 to n with sizes s1,s2......,sn.

An arrangement is good if some index j is divisible i and Sj > Si.

Note : An arrangement with one model is also considered as good

TASK : Find the maximum numbers of models he can buy?

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

Now next multiple of 4 is not available

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

Now in this combination..let us look at the values..!!

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

Now in this combination..let us look at the values..!!

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

The values are strictly increasing and hence we got one possible sequence..!!

Length = 3

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

This could be another possible sequence which is good..!!

Length = 3

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

As the value at index 6 is not strictly increasing.

Length = 2

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

This could be another possible sequence which is good..!!

Length = 2

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

This could be another possible sequence which is good..!!

Length = 2

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

This could be another possible sequence which is good..!!

Length = 2

Problem Description

Let us understand with the help of example..!!

Suppose we have an array of different sizes of models

3	6	7	7	7	7

1 2 3 4 5 6

Example 1:

The maximum length of good sequence is

So now we understood the problem..!!!

Now we will thonk how we can form a solution..!!

Can we relate the problem with sorting??

NO!!!!

Can we relate the problem with sorting??

NO!!!!

The reason is we cannot change the relative positions of the models..!!

Can we relate the problem with sorting??

NO!!!!

The reason is we cannot change the relative positions of the models..!!

So, what more can we think??

What if we know that how many previous indices are factors of the given indices??

YES!!! WE ARE ONE STEP CLOSER!!

What if we know that how many previous indices are factors of the given indices??

YES!!! WE ARE ONE STEP CLOSER!!

But how can we find that??

What if we know that how many previous indices are factors of the given indices??

YES!!! WE ARE ONE STEP CLOSER!!

But how can we find that??

Ok so let us understand..!!

LOGIC FOR FINDING THE FACTORS..!!

Let us suppose we have to find factors of : 6

We have to see values only till

\sqrt(6)

The values are 1 and 2.

So we will run a loop on these values and check..

The values are 1 and 2.

So we will run a loop on these values and check..

if n is divisible by the value.

if n/value == value

then value is a factor

else value, n/value both are factors..!!

Running this algorithm on the example we have.... i.e, 6

\sqrt(6 )

= 2

Running this algorithm on the example we have.... i.e, 6

\sqrt(6 )

= 2

1	2

Running this algorithm on the example we have.... i.e, 6

\sqrt(6 )

= 2

1	2

6 / 1 == 0

1,6 are the factors

Running this algorithm on the example we have.... i.e, 6

\sqrt(6 )

= 2

1	2

6 / 1 == 0

1,6 are the factors

6/2 == 0

2,3 are the factors

Running this algorithm on the example we have.... i.e, 6

\sqrt(6 )

= 2

1	2

6 / 1 == 0

1,6 are the factors

6/2 == 0

2,3 are the factors

So total factors are 1,2,3,and 6.

If you want more detailed explination

for finding the factors of a number then do comment in the comment section..!!

I will make seperate video content for that..!!

Now coming back to problem!!!!

We can find the factors of a particular index and know from which indices we could get the answer

Now coming back to problem!!!!

We can find the factors of a particular index and know from which indices we could get the answer

Now applying concept of dynamic programming..!!

Now coming back to problem!!!!

We can find the factors of a particular index and know from which indices we could get the answer

Now applying concept of dynamic programming..!!

Suppose that we got the maximum length at the factors of a given number than we only need to find the maximum of all these values +1 and fill at the particular position..!!

Confused ??

Dont be !!! Let's see with example..!!

we have reached at n = 3

3	4	5	7	8

Suppose we take index i = 4

1 2 3 4 5

we have reached at n = 3

3	4	5	7	8

i = 4

1 2 3 4 5

i=1 and i=2 are factors

we have reached at n = 3

3	4	5	7	8

i = 4

1 2 3 4 5

i=1 and i=2 are factors

1	2

DP array

we have reached at n = 3

3	4	5	7	8

i = 4

1 2 3 4 5

i=1 and i=2 are factors

1	2

DP array

MAX (1,2) = 2

we have reached at n = 3

3	4	5	7	8

i = 4

1 2 3 4 5

i=1 and i=2 are factors

1	2		3

DP array

MAX (1,2) = 2

Now total length including ith position will be 2+1 = 3

i = 4

we have reached at n = 3

{
	int t;
	cin >> t;
	while(t--)
	{
		int n;
		cin >> n;
		long long int* arr = new long long int[n+1]();
		for(int  i =1; i<=n; i++)
			cin >> arr[i];
		long long int* dp = new long long int[n+1]();
		for(int  i =1; i<=n; i++)
			dp[i] = 1;
		long long int ans = 1;
		for(int  i = 1; i<=n; i++)
		{
             for(int j = 1; j*j<=i; j++)
             {
             	if(i%j == 0)
             	{
             		if(i/j == i  && arr[i] > arr[j])
             			dp[i] = max(dp[i],dp[j]+1);
             		else
             		{
             			if(arr[i] > arr[j])
             			dp[i] = max(dp[j]+1,dp[i]);
             		    if(arr[i] > arr[i/j])
             		    dp[i] = max(dp[i/j]+1,dp[i]);
             		}
             	}
             }
             ans = max(ans,dp[i]);
		}
		cout << ans << endl;
	}



return 0;
}

we have reached at n = 3

Example!!!

1	4	2	3	6	4	9

1	1	1	1	1	1	1

For i = 1 No factors

we have reached at n = 3

Example!!!

1	4	2	3	6	4	9

1	1	1	1	1	1	1

For i = 2 , only factor is 1

we have reached at n = 3

Example!!!

1	4	2	3	6	4	9

1	1	1	1	1	1	1

For i = 2 , only factor is 1

Check if arr[1] < arr[2]

we have reached at n = 3

Example!!!

1	4	2	3	6	4	9

1	1	1	1	1	1	1

For i = 2 , only factor is 1

Check if arr[1] < arr[2] YESS

we have reached at n = 3

Example!!!

1	4	2	3	6	4	9

1	1	1	1	1	1	1

For i = 2 , only factor is 1

Check if arr[1] < arr[2] YESS

DP[2] = MAX(DP[2],DP[1]+1)

MAX(1,2) = 2

we have reached at n = 3

Example!!!

1	4	2	3	6	4	9

1	2	1	1	1	1	1

Similary for i = 3, factor is 1

Therefore DP[3] = MAX(1,2)

DP[3] = 2

we have reached at n = 3

Example!!!

1	4	2	3	6	4	9

1	2	2	1	1	1	1

Factors of 4 are 1,2.

Check arr[1] and arr[2] < arr[4]

Only arr[1] is less than arr[4]

DP[4] = MAX(DP[4], DP[1]+1) = 2

we have reached at n = 3

Example!!!

1	4	2	3	6	4	9

1	2	2	2	1	1	1

Similarly for 5 DP[5] = 2

we have reached at n = 3

Example!!!

1	4	2	3	6	4	9

1	2	2	2	2	1	1

Factors are 1,2,3

Only arr[1] and arr[3] is less than arr[6]

DP[6] = MAX(DP[6], DP[1]+1, DP[3]+1) = MAX(1,2,3) = 3

we have reached at n = 3

Example!!!

1	4	2	3	6	4	9

1	2	2	2	2	3	1

Now we can find for i = 7

DP[7] = 2 as 1 is only factor

we have reached at n = 3

Example!!!

1	4	2	3	6	4	9

1	2	2	2	2	3	2

We are getting max length for good sequence at i = 6

we have reached at n = 3

Example!!!

1	4	2	3	6	4	9

1	2	2	2	2	3	2

Answer = 3

So that's It for Day 3

Concepts that we have learnt

So that's It for Day 1

We have taken a very first step in Dynamic Programming.

Concepts that we have learnt

1. Decision making in dynamic programming.

So that's It for Day 1

We have taken a very first step in Dynamic Programming.

Concepts that we have learnt

1. Decision making in dynamic programming.

2. Using previous store results to calculate optimal present result..!!

Still, Have any Doubts???

Comment it You will get a reply.

Send your doubt to email provided in description.

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By Chirayu Jain

Dynamic Programming Series

Dynamic Programming Series

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