Welcome to the Dynamic Programming MCQs Page
Dive deep into the fascinating world of Dynamic Programming with our comprehensive set of Multiple-Choice Questions (MCQs). This page is dedicated to exploring the fundamental concepts and intricacies of Dynamic Programming, a crucial aspect of Data Structures and Algorithms. In this section, you will encounter a diverse range of MCQs that cover various aspects of Dynamic Programming, from the basic principles to advanced topics. Each question is thoughtfully crafted to challenge your knowledge and deepen your understanding of this critical subcategory within Data Structures and Algorithms.
Check out the MCQs below to embark on an enriching journey through Dynamic Programming. Test your knowledge, expand your horizons, and solidify your grasp on this vital area of Data Structures and Algorithms.
Note: Each MCQ comes with multiple answer choices. Select the most appropriate option and test your understanding of Dynamic Programming. You can click on an option to test your knowledge before viewing the solution for a MCQ. Happy learning!
Dynamic Programming MCQs | Page 16 of 22
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For the optimal solution which should be the starting assembly line?
time_to_reach[2][3] = {{17, 2, 7}, {19, 4, 9}}
time_spent[2][4] = {{6, 5, 15, 7}, {5, 10, 11, 4}}
entry_time[2] = {8, 10}
exit_time[2] = {10, 7}
num_of_stations = 4For the optimal solution, which should be the exit assembly line?
time_to_reach[2][3] = {{17, 2, 7}, {19, 4, 9}}
time_spent[2][4] = {{6, 5, 15, 7}, {5, 10, 11, 4}}
entry_time[2] = {8, 10}
exit_time[2] = {10, 7}
num_of_stations = 4What is the minimum time required to build the car chassis?
time_to_reach[2][3] = {{17, 2, 7}, {19, 4, 9}}
time_spent[2][4] = {{6, 5, 15, 7}, {5, 10, 11, 4}}
entry_time[2] = {8, 10}
exit_time[2] = {10, 7}
num_of_stations = 4Which of the following lines should be inserted to complete the below code?
#include<stdio.h>
int get_min(int a, int b)
{
if(a<b)
return a;
return b;
}
int minimum_time_required(int reach[][3],int spent[][4], int *entry, int *exit, int n)
{
int t1[n], t2[n],i;
t1[0] = entry[0] + spent[0][0];
t2[0] = entry[1] + spent[1][0];
for(i = 1; i < n; i++)
{
t1[i] = get_min(t1[i-1]+spent[0][i], t2[i-1]+reach[1][i-1]+spent[0][i]);
__________;
}
return get_min(t1[n-1]+exit[0], t2[n-1]+exit[1]);
}
#include<stdio.h>
int get_min(int a, int b)
{
if(a<b)
return a;
return b;
}
int minimum_time_required(int reach[][3],int spent[][4], int *entry, int *exit, int n)
{
int t1[n], t2[n], i;
t1[0] = entry[0] + spent[0][0];
t2[0] = entry[1] + spent[1][0];
for(i = 1; i < n; i++)
{
t1[i] = get_min(t1[i-1]+spent[0][i], t2[i-1]+reach[1][i-1]+spent[0][i]);
t2[i] = get_min(t2[i-1]+spent[1][i], t1[i-1]+reach[0][i-1]+spent[1][i]);
}
return get_min(t1[n-1]+exit[0], t2[n-1]+exit[1]);
}
int main()
{
int time_to_reach[][3] = {{6, 1, 5},
{2, 4, 7}};
int time_spent[][4] = {{6, 5, 4, 7},
{5, 10, 2, 6}};
int entry_time[2] = {5, 6};
int exit_time[2] = {8, 9};
int num_of_stations = 4;
int ans = minimum_time_required(time_to_reach, time_spent, entry_time, exit_time, num_of_stations);
printf("%d",ans);
return 0;
}
#include<stdio.h>
int get_min(int a, int b)
{
if(a<b)
return a;
return b;
}
int minimum_time_required(int reach[][3],int spent[][4], int *entry, int *exit, int n)
{
int t1[n], t2[n],i;
t1[0] = entry[0] + spent[0][0];
t2[0] = entry[1] + spent[1][0];
for(i = 1; i < n; i++)
{
t1[i] = get_min(t1[i-1]+spent[0][i], t2[i-1]+reach[1][i-1]+spent[0][i]);
t2[i] = get_min(t2[i-1]+spent[1][i], t1[i-1]+reach[0][i-1]+spent[1][i]);
}
return get_min(t1[n-1]+exit[0], t2[n-1]+exit[1]);
}
int main()
{
int time_to_reach[][3] = {{6, 1, 5},
{2, 4, 7}};
int time_spent[][4] = {{6, 5, 4, 7},
{5, 10, 2, 6}};
int entry_time[2] = {5, 6};
int exit_time[2] = {8, 9};
int num_of_stations = 4;
int ans = minimum_time_required(time_to_reach, time_spent, entry_time, exit_time, num_of_stations);
printf("%d",ans);
return 0;
}
#include<stdio.h>
int get_min(int a, int b)
{
if(a<b)
return a;
return b;
}
int minimum_time_required(int reach[][3],int spent[][4], int *entry, int *exit, int n)
{
int t1[n], t2[n],i;
t1[0] = entry[0] + spent[0][0];
t2[0] = entry[1] + spent[1][0];
for(i = 1; i < n; i++)
{
t1[i] = get_min(t1[i-1]+spent[0][i], t2[i-1]+reach[1][i-1]+spent[0][i]);
t2[i] = get_min(t2[i-1]+spent[1][i], t1[i-1]+reach[0][i-1]+spent[1][i]);
}
return get_min(t1[n-1]+exit[0], t2[n-1]+exit[1]);
}
int main()
{
int time_to_reach[][3] = {{6, 1, 5},
{2, 4, 7}};
int time_spent[][4] = {{6, 5, 4, 7},
{5, 10, 2, 6}};
int entry_time[2] = {5, 6};
int exit_time[2] = {8, 9};
int num_of_stations = 4;
int ans = minimum_time_required(time_to_reach, time_spent, entry_time, exit_time, num_of_stations);
printf("%d",ans);
return 0;
}
#include<stdio.h>
int get_min(int a, int b)
{
if(a<b)
return a;
return b;
}
int minimum_time_required(int reach[][4],int spent[][5], int *entry, int *exit, int n)
{
int t1[n], t2[n], i;
t1[0] = entry[0] + spent[0][0];
t2[0] = entry[1] + spent[1][0];
for(i = 1; i < n; i++)
{
t1[i] = get_min(t1[i-1]+spent[0][i], t2[i-1]+reach[1][i-1]+spent[0][i]);
t2[i] = get_min(t2[i-1]+spent[1][i], t1[i-1]+reach[0][i-1]+spent[1][i]);
}
return get_min(t1[n-1]+exit[0], t2[n-1]+exit[1]);
}
int main()
{
int time_to_reach[][4] = {{16, 10, 5, 12},
{12, 4, 17, 8}};
int time_spent[][5] = {{13, 5, 20, 19, 9},
{15, 10, 12, 16, 13}};
int entry_time[2] = {12, 9};
int exit_time[2] = {10, 13};
int num_of_stations = 5;
int ans = minimum_time_required(time_to_reach, time_spent, entry_time, exit_time, num_of_stations);
printf("%d",ans);
return 0;
}
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