A prominent masseuse is pondering which appointment invitations to accept after receiving a string of them. She requires a 15-minute pause between visits and hence cannot take any subsequent inquiries. Find the optimal (highest total booked minutes) set the masseuse can honour given a sequence of back-to-back appointment requests (all multiples of 15 minutes, none overlap, and none may be changed). Return the time in minutes. Input EXAMPLE: 30, 15, 60, 75, 45, 15, 15, 45 180 minutes output (30, 60, 45, 45).
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A prominent masseuse is pondering which appointment invitations to accept after receiving a string of them. She requires a 15-minute pause between visits and hence cannot take any subsequent inquiries. Find the optimal (highest total booked minutes) set the masseuse can honour given a sequence of back-to-back appointment requests (all multiples of 15 minutes, none overlap, and none may be changed). Return the time in minutes.
Input EXAMPLE: 30, 15, 60, 75, 45, 15, 15, 45
180 minutes output (30, 60, 45, 45).
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- To cut an 'n' centimeter-long gold bar into 2 pieces costs $n. When a gold bar is cut into many pieces, the order in which the cuts occur can affect the total amount of costs. For example, to cut a 20 centimeter gold bar at length marks 2, 8, and 10 (numbering the length marks in ascending order from the left-hand end, starting from 1). If the cuts to occur in left-to-right order, then the first cut costs $20, the second cut costs $18 (cutting the remaining 18 centimeter bar at originally length mark 8), and the third cut costs $12, totaling $50. If the cuts to occur in right-to-left order, however, then the first cut costs $20 time, the second cut costs $10, and the third cut costs $8, totaling $38. In yet another order, the first cut is at 8 (costing $20), then the 2nd cut is at 2 (costing $8), and finally the third cut is at 10 (costing $12), for a total cost of $40. Given an 'n' centimeter-long gold bar G and an array C[1..m] containing the cutting points in ascending order): a.…A popular masseuse receives a sequence of back-to-back appointment requestsand is debating which ones to accept. She needs a 15-minute break between appointments andtherefore she cannot accept any adjacent requests. Given a sequence of back-to-back appointment requests (all multiples of 15 minutes, none overlap, and none can be moved), find the optimal(highest total booked minutes) set the masseuse can honor. Return the number of minutes.EXAMPLEInput: {30, 15, 60, 75, 45, 15, 15, 45}Output180 minutes ({30, 60, 45, 45})Consider the following snapshot of a system: Allocation: A B C DT0 0 0 1 2T1 1 0 0 0T2 1 3 5 4T3 0 6 3 2T4 0 0 1 4 Max: A B C DT0 3 0 1 2T1 1 7 5 0T2 2 3 5 6T3 0 6 5 3T4 0 6 5 6 Available: A B C D2 5 1 0 Answer using the banker’s algorithm. If you think the system is in an unsafe state, write "unsafe" in the answer box, otherwise, write the process sequence that satisfies the safe state. While answering the question write only the id of the Tasks separated with comma, (e.g., for sequence <T4,T2,T1,T0,T3> write only 4,2,1,0,3 ).
- Problem1Given a value `value`, if we want to make change for `value` cents, and we have infinitesupply of each of coins = {S1, S2, .. , Sm} valued `coins`, how many ways can we make the change?The order of `coins` doesn't matter.For example, for `value` = 4 and `coins` = [1, 2, 3], there are four solutions:[1, 1, 1, 1], [1, 1, 2], [2, 2], [1, 3].So output should be 4. For `value` = 10 and `coins` = [2, 5, 3, 6], there are five solutions: [2, 2, 2, 2, 2], [2, 2, 3, 3], [2, 2, 6], [2, 3, 5] and [5, 5].So the output should be 5. Time complexity: O(n * m) where n is the `value` and m is the number of `coins`Space complexity: O(n)""" def count(coins, value): """ Find number of combination of `coins` that adds upp to `value` Keyword arguments: coins -- int[] value -- int """ # initialize dp array and set base case as 1 dp_array = [1] + [0] * value.. (+.Consider a set of movies M1, M2, ... , Mk. There is a set of customers, each one of which indicates the two movies they would like to see this weekend. Movies are shown on Saturday evening and Sunday evening. Multiple movies may be screened at the same time. You must decide which movies should be televised on Saturday and which on Sunday, so that every customer gets to see the two movies they desire. Is there a schedule where each movie is shown at most once? Design an efficient algorithm to find such a schedule if one exists.This problem exercises the basic concepts of game playing, using tic-tac-toe (noughtsand crosses) as an example. We define Xn as the number of rows, columns, or diagonals with exactly n X’s and no O’s. Similarly, On is the number of rows, columns, or diagonals with just n O’s. The utility function assigns +1 to any position with X3 = 1 and −1 to any position with O3 = 1. All other terminal positions have utility 0. For nonterminal positions, we use a linear evaluation function defined as Eval (s) = 3X2(s)+X1(s)−(3O2(s)+O1(s))."Mark on your tree the evaluations of all the positions at depth 2."
- ProblemGiven a value `value`, if we want to make change for `value` cents, and we have infinitesupply of each of coins = {S1, S2, .. , Sm} valued `coins`, how many ways can we make the change?The order of `coins` doesn't matter.For example, for `value` = 4 and `coins` = [1, 2, 3], there are four solutions:[1, 1, 1, 1], [1, 1, 2], [2, 2], [1, 3].So output should be 4. For `value` = 10 and `coins` = [2, 5, 3, 6], there are five solutions: [2, 2, 2, 2, 2], [2, 2, 3, 3], [2, 2, 6], [2, 3, 5] and [5, 5].So the output should be 5. Time complexity: O(n * m) where n is the `value` and m is the number of `coins`Space complexity: O(n)""" def count(coins, value): """ Find number of combination of `coins` that adds upp to `value` Keyword arguments: coins -- int[] value -- int """ # initialize dp array and set base case as 1 dp_array = [1] + [0] * value) ++.There are n people who want to carpool during m days. On day i, some subset ???? of people want to carpool, and the driver di must be selected from si . Each person j has a limited number of days fj they are willing to drive. Give an algorithm to find a driver assignment di ∈ si each day i such that no person j has to drive more than their limit fj. (The algorithm should output “no” if there is no such assignment.) Hint: Use network flow. For example, for the following input with n = 3 and m = 3, the algorithm could assign Tom to Day 1 and Day 2, and Mark to Day 3. Person Day 1 Day 2 Day 3 Driving Limit 1 (Tom) x x x 2 2 (Mark) x x 1 3 (Fred) x x 0Suppose, you are working in a company ‘X’ where your job is to calculate the profit based on their investment.If the company invests 100,000 USD or less, their profit will be based on 75,000 USD as first 25,000 USD goes to set up the business in the first place. For the first 100,000 USD, the profit margin is low: 4.5%. Therefore, for every 100 dollar they spend, they get a profitof 4.5 dollar.For an investment greater than 100,000 USD, for the first 100,000 USD (actually on 75,000 USD as 25,000 is the setup cost), the profit margin is 4.5% where for the rest, it goes up to 8%. For example, if they invest 250,000 USD, they will get an 8% profit for the 150,000 USD. In addition, from the rest 100,000 USD, 25,000 is the setup cost and there will be a 4.5% profit on the rest 75,000. Investment will always be greater or equal to 25,000 and multiple of 100.Complete the RECURSIVE methods below that take an array of integers (investments)and an iterator (always sets to ZERO(‘0’) when the…
- Classes are scheduled at a school. Once students have submitted their course requests, a computer algorithm can determine the optimal schedule for everyone. The school has concluded that it will take too long to determine the best schedule. Instead, they resort to a less sophisticated method that produces a serviceable if not ideal timetable. Which guiding concept is represented here?Suppose we use the following KB (where x, y, z are variables and r1, r2, r3, goal are constants) to determine whether a particular robot can score. (a) Open(x) ∧ HasBall(x) → CanScore(x)(b) Open(x) ∧ CanAssist(y, x) ∧ HasBall(y) → CanScore(x) (c) PathClear(x,y) → CanAsist(x,y)(d) PathClear(x,z) ∧ CanAssist(z,y) → CanAssist(x,y) (e) PathClear(x,goal) → Open(x)(f) PathClear(y,x) → PathClear(x,y) (g) HasBall(r3)(h) PathClear(r1,goal) (i) PathClear(r2,r1) (j) PathClear(r3,r2) (k) PathClear(r3,goal)Suppose we use the following KB (where x, y, z are variables and r1, r2, r3, goal are constants) to determine whether a particular robot can score. (a) Open(x) ∧ HasBall(x) → CanScore(x)(b) Open(x) ∧ CanAssist(y, x) ∧ HasBall(y) → CanScore(x) (c) PathClear(x,y) → CanAsist(x,y)(d) PathClear(x,z) ∧ CanAssist(z,y) → CanAssist(x,y) (e) PathClear(x,goal) → Open(x)(f) PathClear(y,x) → PathClear(x,y) (g) HasBall(r3)(h) PathClear(r1,goal) (i) PathClear(r2,r1) (j) PathClear(r3,r2) (k) PathClear(r3,goal) Intuitively, CanScore(x) means x can score on goal. CanAssist(x, y) means there exists some series of passes that can get the ball from x to y. Open(x) means x can shoot on goal directly. And P athClear(x, y) means the path between x and y is clear. Provide a SLD-derivation for the query CanScore(x) in which the answer provided is r1. Provide a SLD-derivation for the query CanScore(x) in which the answer provided is r3. How many “distinct” derivations (i.e., involving different…