ObjectiveMcq
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157 practice sets · Page 5 of 8
In dynamic programming, the technique of storing the previously calculated values is called ___________
Saving value property
Storing value property
Memoization
Mapping
Memoization is the technique in which previously calculated values are stored, so that, these values can be used to solve other subproblems.
A greedy algorithm can be used to solve all the dynamic programming problems.
True
False
A greedy algorithm gives optimal solution for all subproblems, but when these locally optimal solutions are combined it may NOT result into a globally optimal solution. Hence, a greedy algorithm CANNOT be used to solve all the dynamic programming problems.
When dynamic programming is applied to a problem, it takes far less time as compared to other methods that don't take advantage of overlapping subproblems.
True
False
Dynamic programming calculates the value of a subproblem only once, while other methods that don't take advantage of the overlapping subproblems property may calculate the value of the same subproblem several times. So, dynamic programming saves the time of recalculation and takes far less time as compared to other methods that don't take advantage of the overlapping subproblems property.
If a problem can be solved by combining optimal solutions to non-overlapping problems, the strategy is called _____________
Dynamic programming
Greedy
Divide and conquer
Recursion
In divide and conquer, the problem is divided into smaller non-overlapping subproblems and an optimal solution for each of the subproblems is found. The optimal solutions are then combined to get a global optimal solution. For example, mergesort uses divide and conquer strategy.
If a problem can be broken into subproblems which are reused several times, the problem possesses ____________ property.
Overlapping subproblems
Optimal substructure
Memoization
Greedy
Overlapping subproblems is the property in which value of a subproblem is used several times.
If an optimal solution can be created for a problem by constructing optimal solutions for its subproblems, the problem possesses ____________ property.
Overlapping subproblems
Optimal substructure
Memoization
Greedy
Optimal substructure is the property in which an optimal solution is found for the problem by constructing optimal solutions for the subproblems.
Which of the following is/are property/properties of a dynamic programming problem?
Optimal substructure
Overlapping subproblems
Greedy approach
Both optimal substructure and overlapping subproblems
A problem that can be solved using dynamic programming possesses overlapping subproblems as well as optimal substructure properties.
Which of the following lines should be added to complete the above code?
arr[num_of_dice][S]
arr[dice][sm]
arr[dice][S]
arr[S][dice]
The line arr[num_of_dice][S] completes the above code.
There are 10 dice having 5 faces. The faces are numbered from 1 to 5. What is the number of ways in which a sum of 4 can be achieved?
0
2
4
8
Since there are 10 dice and the minimum value each die can take is 1, the minimum possible sum is 10. Hence, a sum of 4 cannot be achieved.
There are n dice with f faces. The faces are numbered from 1 to f. What is the maximum possible sum that can be obtained when the n dice are rolled together?
1
f*f
n*n
n*f
The sum will be maximum when all the faces show a value f. The sum in this case will be n*f.
There are n dice with f faces. The faces are numbered from 1 to f. What is the minimum possible sum that can be obtained when the n dice are rolled together?
1
f
n
n*f
The sum will be minimum when all the faces show a value 1. The sum in this case will be n.
You have 2 dice each of them having 6 faces numbered from 1 to 6. What is the number of ways in which a sum of 11 can be achieved?
0
1
2
3
The sum of 11 can be achieved when the dice show {6, 5} or {5, 6}.
You have 3 dice each having 6 faces. What is the number of permutations that can be obtained when you roll the 3 dice together?
27
36
216
81
The total number of permutations that can be obtained is 6 * 6 * 6 = 216.
You have n dice each having f faces. What is the number of permutations that can be obtained when you roll the n dice together?
nnn...f times
fff...n times
nnn...n times
fff...f times
Each die can take f values and there are n dice. So, the total number of permutations is fff...n times.
You are given n dice each having f faces. You have to find the number of ways in which a sum of S can be achieved. This is the dice throw problem. Which of the following methods can be used to solve the dice throw problem?
Brute force
Recursion
Dynamic programming
Brute force, Recursion and Dynamic Programming
Brute force, Recursion and Dynamic Programming can be used to solve the dice throw problem.
Which of the following lines should be added to complete the "if(op[pos] == '|')" part of the code?
False[row][col] += True[row][pos] * False[pos+1][col]; True[row][col] += t_row_pos * t_pos_col + False[row][pos] * False[pos+1][col];
False[row][col] += False[row][pos] * True[pos+1][col]; True[row][col] += t_row_pos * t_pos_col - True[row][pos] * True[pos+1][col];
False[row][col] += True[row][pos] * True[pos+1][col]; True[row][col] += t_row_pos * t_pos_col + True[row][pos] * True[pos+1][col];
False[row][col] += False[row][pos] * False[pos+1][col]; True[row][col] += t_row_pos * t_pos_col - False[row][pos] * False[pos+1][col];
The following lines should be added:
False[row][col] += False[row][pos] * False[pos+1][col];
True[row][col] += t_row_pos * t_pos_col + False[row][pos] * False[pos+1][col];
What is the maximum number of ways in which a boolean expression with n + 1 terms can be parenthesized, such that the output is true?
nth catalan number
n factorial
n cube
n square
The number of ways will be maximum when all the possible parenthesizations result in a true value. The number of possible parenthesizations is given by the nth catalan number.
Which of the following gives the total number of ways of parenthesizing an expression with n + 1 terms?
n factorial
n square
n cube
nth catalan number
The nth Catalan number gives the total number of ways of parenthesizing an expression with n + 1 terms.
Consider the expression T | F ∧ T. In how many ways can the expression be parenthesized so that the output is F (false)?
0
1
2
3
The expression can be parenthesized as (T | F) ∧ T, so that the output is F (false).
Consider the expression T & F ∧ T. What is the number of ways in which the expression can be parenthesized so that the output is T (true)?
0
1
2
3
The expression can be parenthesized as (T & F) ∧ T or T & (F ∧ T), so that the output is T.
