Dynamic Programming in Data Structure And Algorithm : Practice MCQ Questions & Answers with Explanations

The minimum time required is 43.

The path is S2,1 -> S1,2 -> S2,3 -> S2,4, where Si,j : i = line number, j = station number

For the optimal solution, the exit assembly line is line 2.

For the optimal solution, the starting assembly line is line 2.

In the dynamic programming implementation of the assembly line scheduling problem, 2 lookup tables are required one for storing the minimum time and the other for storing the assembly line number.

In the brute force algorithm, all the possible ways are calculated which are of the order of 2^{n}.

All of the above mentioned methods can be used to solve the assembly line scheduling problem.

find_max(ans[itm - 1][w - wt[itm - 1]] + val[itm - 1], ans[itm - 1][w]) completes the above code.

The Knapsack problem cannot be solved using the greedy algorithm.

In the brute force algorithm all the subsets of the items are found and the value of each subset is calculated. The subset of items with the maximum value and a weight less than equal to the maximum allowed weight gives the answer. The time taken to calculate all the subsets is O(2^{n}).

In this case, questions are used instead of items. Each question has a score which is same as each item having a value. Also, each question takes a time t which is same as each item having a weight w. You have to maximize the score in time T which is same as maximizing the value using a bag of weight W.

The maximum value you can get is 160. This can be achieved by choosing the items 1 and 3 that have a total weight of 60.

Brute force, Recursion and Dynamic Programming can be used to solve the knapsack problem.

Knapsack problem is an example of 2D dynamic programming.

The line "mx_sm = val" should be inserted to complete the above code.

The dynamic programming implementation of the maximum sum rectangle problem uses Kadane's algorithm.

The time complexity of the brute force implementation of the maximum sum rectangle problem is O(n^{4}).

The complete matrix represents the maximum sum rectangle and it's sum is 14.

Since all the elements of the 2×2 matrix are negative, the maximum sum rectangle is {-1}, a 1×1 matrix containing the largest element. The sum of elements of the maximum sum rectangle is -1.

Since all the elements of the 2×3 matrix are positive, the maximum sum rectangle is the matrix itself and the sum of elements is 12.

If a matrix contains all non-zero elements with at least one positive and at least on negative element, then the sum of elements of the maximum sum rectangle cannot be zero.