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AysviL [449]
3 years ago
11

Suppose you have two arrays: Arr1 and Arr2. Arr1 will be sorted values. For each element v in Arr2, you need to write a pseudo c

ode that will print the number of elements in Arr1 that is less than or equal to v. For example: suppose you are given two arrays of size 5 and 3 respectively. 5 3 [size of the arrays] Arr1 = 1 3 5 7 9 Arr2 = 6 4 8 The output should be 3 2 4 Explanation: Firstly, you should search how many numbers are there in Arr1 which are less than 6. There are 1, 3, 5 which are less than 6 (total 3 numbers). Therefore, the answer for 6 will be 3. After that, you will do the same thing for 4 and 8 and output the corresponding answers which are 2 and 4. Your searching method should not take more than O (log n) time. Sample Input Sample Output 5 5 1 1 2 2 5 3 1 4 1 5 4 2 4 2 5
Engineering
1 answer:
brilliants [131]3 years ago
6 0

Answer:

The algorithm is as follows:

1. Declare Arr1 and Arr2

2. Get Input for Arr1 and Arr2

3. Initialize count to 0

4. For i in Arr2

4.1 For j in Arr1:

4.1.1 If i > j Then

4.1.1.1 count = count + 1

4.2 End j loop

4.3 Print count

4.4 count = 0

4.5 End i loop

5. End

Explanation:

This declares both arrays

1. Declare Arr1 and Arr2

This gets input for both arrays

2. Get Input for Arr1 and Arr2

This initializes count to 0

3. Initialize count to 0

This iterates through Arr2

4. For i in Arr2

This iterates through Arr1 (An inner loop)

4.1 For j in Arr1:

This checks if current element is greater than current element in Arr1

4.1.1 If i > j Then

If yes, count is incremented by 1

4.1.1.1 count = count + 1

This ends the inner loop

4.2 End j loop

Print count and set count to 0

<em>4.3 Print count</em>

<em>4.4 count = 0</em>

End the outer loop

4.5 End i loop

End the algorithm

5. End

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A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t
Nataly [62]

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes

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