Dynamic Programming in Data Structures
practice.

You are given a knapsack that can carry a maximum weight of 60. There are 4 items with weights {20, 30, 40, 70} and values {70, 80, 90, 200}. What is the maximum value of the items you can carry using the knapsack?

What is the time complexity of the following dynamic programming implementation used to find the length of the longest increasing subsequence?

#include<stdio.h>
int longest_inc_sub(int *arr, int len)
{
      int i, j, tmp_max;
      int LIS[len];  // array to store the lengths of the longest increasing subsequence 
      LIS[0]=1;
      for(i = 1; i < len; i++)
      { 
           tmp_max = 0;
	   for(j = 0; j < i; j++)
	   {
	        if(arr[j] < arr[i])
	        {
		    if(LIS[j] > tmp_max)
		     tmp_max = LIS[j];  
	        }
           }
	   LIS[i] = tmp_max + 1;
      }
      int max = LIS[0];
      for(i = 0; i < len; i++)
	if(LIS[i] > max)
	   max = LIS[i];
      return max;
}
int main()
{
      int arr[] = {10,22,9,33,21,50,41,60,80}, len = 9;
      int ans = longest_inc_sub(arr, len);
      printf("%d",ans);
      return 0;
}

Consider the following assembly line problem:

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 = 4

For the optimal solution, which should be the starting assembly line?

For every non-empty string, the length of the longest palindromic subsequence is at least one.

Consider the matrices P, Q, R and S which are 20 x 15, 15 x 30, 30 x 5 and 5 x 40 matrices respectively. What is the minimum number of multiplications required to multiply the four matrices?

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?

What is the space complexity of Kadane's algorithm?

What is the output of the following code?

#include<stdio.h>
int balanced_partition(int *arr, int len)
{
     int sm = 0, i, j;
     for(i = 0;i < len; i++)
      sm += arr[i];
     if(sm % 2 != 0)
        return 0;
     int ans[sm/2 + 1][len + 1];
     for(i = 0; i <= len; i++)
      ans[0][i] = 1;
     for(i = 1; i <= sm/2; i++)
      ans[i][0] = 0;
     for(i = 1; i <= sm/2; i++)
     {
         for(j = 1;j <= len; j++)
         {
             ans[i][j] = ans[i][j-1];
             if(i >= arr[j - 1])
                 ans[i][j] = ans[i][j] || ans[i - arr[j - 1]][j - 1];
         }
     }
     return ans[sm/2][len];
}
int main()
{
     int arr[] = {5, 6, 7, 10, 3, 1}, len = 6;
     int ans = balanced_partition(arr,len);
     if(ans == 0)
       printf("false");
     else
       printf("true");
     return 0;
}

Consider the following recursive implementation of the rod cutting problem:

#include<stdio.h>
#include<limits.h>
int max_of_two(int a, int b)
{
      if(a > b)
            return a;
      return b;
}
int rod_cut(int *prices, int len)
{
      int max_price = INT_MIN; // INT_MIN is the min value an integer can take
      int i;
      if(len <= 0 )
	return 0;
      for(i = 0; i < len; i++)
	 max_price = max_of_two(_____________); // subtract 1 because index starts from 0
      return max_price;
}
int main()
{
      int prices[]={2, 5, 6, 9, 9, 17, 17, 18, 20, 22},len_of_rod = 10;
      int ans = rod_cut(prices, len_of_rod);
      printf("%d",ans);
      return 0;
}

Complete the above code.

What is the output of the following code?

#include<stdio.h>
int balanced_partition(int *arr, int len)
{
      int sm = 0, i, j;
      for(i = 0;i < len; i++)
      sm += arr[i];
      if(sm % 2 != 0)
        return 0;
      int ans[sm/2 + 1][len + 1];
      for(i = 0; i <= len; i++)
      ans[0][i] = 1;
      for(i = 1; i <= sm/2; i++)
      ans[i][0] = 0;
      for(i = 1; i <= sm/2; i++)
      {
          for(j = 1;j <= len; j++)
          {
              ans[i][j] = ans[i][j-1];
              if(i >= arr[j - 1])
                ans[i][j] = ans[i][j] || ans[i - arr[j - 1]][j - 1];
          }
      }
      return ans[sm/2][len];
}
int main()
{
      int arr[] = {3, 4, 5, 6, 7, 1}, len = 6;
      int ans = balanced_partition(arr,len);
      if(ans == 0)
         printf("false");
      else
         printf("true");
      return 0;
}

What is the time complexity of the following dynamic programming implementation of the boolean parenthesization problem?

int count_bool_parenthesization(char *sym, char *op)
{
      int str_len = strlen(sym);
      int True[str_len][str_len],False[str_len][str_len];
      int row,col,length,l;
      for(row = 0, col = 0; row < str_len; row++,col++)
      {
          if(sym[row] == 'T')
          {
              True[row][col] = 1;
              False[row][col] = 0;
          }
          else
          {
              True[row][col] = 0;
              False[row][col] = 1;
          }
      }
      for(length = 1; length < str_len; length++)
      {
          for(row = 0, col = length; col < str_len; col++, row++)
          {
              True[row][col] = 0;
              False[row][col] = 0;
              for(l = 0; l < length; l++)
              {
                  int pos = row + l;
                  int t_row_pos = True[row][pos] + False[row][pos];
                  int t_pos_col = True[pos+1][col] + False[pos+1][col];
                  if(op[pos] == '|')
                  {
                      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];;
                  }
                  if(op[pos] == '&')
                  {
                     True[row][col] += True[row][pos] * True[pos+1][col];
                     False[row][col] += t_row_pos * t_pos_col - True[row][pos] * True[pos+1][col];
                  }
                  if(op[pos] == '^')
                  {
                     True[row][col] += True[row][pos] * False[pos+1][col] 
                                      + False[row][pos] * True[pos + 1][col];
                     False[row][col] += True[row][pos] * True[pos+1][col] 
                                       + False[row][pos] * False[pos+1][col];
                  }
              }
          }
      }
      return True[0][str_len-1];
}

Consider the string "abbccbba". What is the minimum number of insertions required to make the string a palindrome?

For which of the following, the length of the string is not equal to the length of the longest palindromic subsequence?

A greedy algorithm can be used to solve all the dynamic programming problems.

What is the space complexity of the following dynamic programming implementation of the rod cutting problem?

#include<stdio.h>
#include<limits.h>
int rod_cut(int *prices, int len)
{
      int max_val[len + 1];
      int i,j,tmp_price,tmp_idx;
      max_val[0] = 0;
      for(i = 1; i <= len; i++)
      {
	   int tmp_max = INT_MIN; // minimum value an integer can hold
	   for(j = 1; j <= i; j++)
	   {
	         tmp_idx = i - j;
                 //subtract 1 because index of prices starts from 0
	         tmp_price = prices[j-1] + max_val[tmp_idx]; 
	         if(tmp_price > tmp_max)
	           tmp_max = tmp_price;
	    }
	    max_val[i] = tmp_max;
       }
       return max_val[len];
}
int main()
{
       int prices[]={2, 5, 6, 9, 9, 17, 17, 18, 20, 22},len_of_rod = 5;
       int ans = rod_cut(prices, len_of_rod);
       printf("%d",ans);
       return 0;
}

What is the space complexity of the following naive method used to find the maximum sub-array sum in an array containing n elements?

#include<stdio.h>
int main()
{
     int arr[1000]={2, -1, 3, -4, 1, -2, -1, 5, -4}, len=9;
     int cur_max, tmp_max, strt_idx, sub_arr_idx;
     cur_max = arr[0];
     for(strt_idx = 0; strt_idx < len; strt_idx++)
     {
	  tmp_max=0;
	  for(sub_arr_idx = strt_idx; sub_arr_idx < len; sub_arr_idx++)
	  {
	       tmp_max +=arr[sub_arr_idx];
	       if(tmp_max > cur_max)
		 _____________;
	  }
     }
     printf("%d",cur_max);
     return 0;
}

What is the value stored in arr[2][3] when the following code is executed?

#include<stdio.h>
#include<limits.h>
int mat_chain_multiplication(int *mat, int n)
{
     int arr[n][n];
     int i,k,row,col,len;
     for(i=1;i<n;i++)
         arr[i][i] = 0;
     for(len = 2; len < n; len++)
     {
          for(row = 1; row <= n - len + 1; row++)
          {
               col = row + len - 1;
               arr[row][col] = INT_MAX;
               for(k = row; k <= col - 1; k++)
               {
                     int tmp = arr[row][k] + arr[k + 1][col] + mat[row - 1] * mat[k] * mat[col];
                     if(tmp < arr[row][col])
                     arr[row][col] = tmp;
               }
          }
     }
     return arr[1][n - 1];
}
int main()
{
     int mat[6] = {20,30,40,50};
     int ans = mat_chain_multiplication(mat,4);
     printf("%d",ans);
     return 0;
}

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?

Consider the following dynamic programming implementation of the rod cutting problem:

#include<stdio.h>
#include<limits.h>
int rod_cut(int *prices, int len)
{
      int max_val[len + 1];
      int i,j,tmp_price,tmp_idx;
      max_val[0] = 0;
      for(i = 1; i <= len; i++)
      {
	   int tmp_max = INT_MIN; // minimum value an integer can hold
	   for(j = 1; j <= i; j++)
	   {
	         tmp_idx = i - j;
                 //subtract 1 because index of prices starts from 0
	         tmp_price = _____________; 
	         if(tmp_price > tmp_max)
	           tmp_max = tmp_price;
	    }
	    max_val[i] = tmp_max;
       }
       return max_val[len];
}
int main()
{
       int prices[]={2, 5, 6, 9, 9, 17, 17, 18, 20, 22},len_of_rod = 5;
       int ans = rod_cut(prices, len_of_rod);
       printf("%d",ans);
       return 0;
}

Which line will complete the ABOVE code?

What is the space complexity of the above dynamic programming implementation of the assembly line scheduling problem?