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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.

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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 13 of 22

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Discuss
Answer: (c).When the two strings are equal
Q122.
Suppose each edit (insert, delete, replace) has a cost of one. Then, the maximum edit distance cost between the two strings is equal to the length of the larger string.
Discuss
Answer: (a).True
Q123.
Consider the strings β€œmonday” and β€œtuesday”. What is the edit distance between the two strings?

a.

3

b.

4

c.

5

d.

6

Discuss
Answer: (b).4
Q124.
Consider the two strings β€œβ€(empty string) and β€œabcd”. What is the edit distance between the two strings?
Discuss
Answer: (b).4
Q125.
Consider the following dynamic programming implementation of the edit distance problem.
Which of the following lines should be added to complete the below code?
#include<stdio.h>
#include<string.h>
int get_min(int a, int b)
{
      if(a < b)
        return a;
      return b;
}
int edit_distance(char *s1, char *s2)
{
     int len1,len2,i,j,min;
     len1 = strlen(s1);
     len2 = strlen(s2);
     int arr[len1 + 1][len2 + 1];
     for(i = 0;i <= len1; i++)
       arr[i][0] = i;
     for(i = 0; i <= len2; i++)
       arr[0][i] = i;
     for(i = 1; i <= len1; i++)
     {
         for(j = 1; j <= len2; j++)
         {
               min = get_min(arr[i-1][j],arr[i][j-1]) + 1;
               if(s1[i - 1] == s2[j - 1])
               {
                    if(arr[i-1][j-1] < min)
                       min = arr[i-1][j-1];
               }
               else
               {
                     if(arr[i-1][j-1] + 1 < min)
                         min = arr[i-1][j-1] + 1;
               }
                _____________;
         }
     }
     return arr[len1][len2];
}
int main()
{
     char s1[] = "abcd", s2[] = "defg";
     int ans = edit_distance(s1, s2);
     printf("%d",ans);
     return 0;
}

Discuss
Answer: (d).arr[i][j] = min
Q126.
What is the output of the following code?
#include<stdio.h>
#include<string.h>
int get_min(int a, int b)
{
      if(a < b)
        return a;
      return b;
}
int edit_distance(char *s1, char *s2)
{
     int len1,len2,i,j,min;
     len1 = strlen(s1);
     len2 = strlen(s2);
     int arr[len1 + 1][len2 + 1];
     for(i = 0;i <= len1; i++)
      arr[i][0] = i;
     for(i = 0; i <= len2; i++)
      arr[0][i] = i;
     for(i = 1; i <= len1; i++)
     {
         for(j = 1; j <= len2; j++)
         {
              min = get_min(arr[i-1][j],arr[i][j-1]) + 1;
              if(s1[i - 1] == s2[j - 1])
              {
                  if(arr[i-1][j-1] < min)
                      min = arr[i-1][j-1];
              }
              else
              {
                  if(arr[i-1][j-1] + 1 < min)
                     min = arr[i-1][j-1] + 1;
              }
              arr[i][j] = min;
         }
     }
     return arr[len1][len2];
}
int main()
{
     char s1[] = "abcd", s2[] = "defg";
     int ans = edit_distance(s1, s2);
     printf("%d",ans);
     return 0;
}

a.

1

b.

2

c.

3

d.

4

Discuss
Answer: (d).4
Q127.
What is the value stored in arr[2][2] when the following code is executed?
#include<stdio.h>
#include<string.h>
int get_min(int a, int b)
{
      if(a < b)
        return a;
      return b;
}
int edit_distance(char *s1, char *s2)
{
      int len1,len2,i,j,min;
      len1 = strlen(s1);
      len2 = strlen(s2);
      int arr[len1 + 1][len2 + 1];
      for(i = 0;i <= len1; i++)
        arr[i][0] = i;
      for(i = 0; i <= len2; i++)
         arr[0][i] = i;
      for(i = 1; i <= len1; i++)
      {
           for(j = 1; j <= len2; j++)
           {
                 min = get_min(arr[i-1][j],arr[i][j-1]) + 1;
                 if(s1[i - 1] == s2[j - 1])
                 {
                      if(arr[i-1][j-1] < min)
                        min = arr[i-1][j-1];
                 }
                 else
                 {
                      if(arr[i-1][j-1] + 1 < min)
                        min = arr[i-1][j-1] + 1;
                 }
                 arr[i][j] = min;
           }
      }
      return arr[len1][len2];
}
int main()
{
      char s1[] = "abcd", s2[] = "defg";
      int ans = edit_distance(s1, s2);
      printf("%d",ans);
      return 0;
}

a.

1

b.

2

c.

3

d.

4

Discuss
Answer: (b).2
Q128.
What is the output of the following code?
#include<stdio.h>
#include<string.h>
int get_min(int a, int b)
{
      if(a < b)
        return a;
      return b;
}
int edit_distance(char *s1, char *s2)
{
     int len1,len2,i,j,min;
     len1 = strlen(s1);
     len2 = strlen(s2);
     int arr[len1 + 1][len2 + 1];
     for(i = 0;i <= len1; i++)
       arr[i][0] = i;
     for(i = 0; i <= len2; i++)
       arr[0][i] = i;
     for(i = 1; i <= len1; i++)
     {
         for(j = 1; j <= len2; j++)
         {
              min = get_min(arr[i-1][j],arr[i][j-1]) + 1;
              if(s1[i - 1] == s2[j - 1])
              {
                  if(arr[i-1][j-1] < min)
                     min = arr[i-1][j-1];
              }
              else
              {
                  if(arr[i-1][j-1] + 1 < min)
                     min = arr[i-1][j-1] + 1;
              }
              arr[i][j] = min;
         }
     }
     return arr[len1][len2];
}
int main()
{
     char s1[] = "pqrstuv", s2[] = "prstuv";
     int ans = edit_distance(s1, s2);
     printf("%d",ans);
     return 0;
}

a.

1

b.

2

c.

3

d.

4

Discuss
Answer: (a).1
Q129.
Wagner–Fischer is a ____________ algorithm.
Discuss
Answer: (c).Dynamic programming
Discuss
Answer: (c).Edit distance between two strings

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