31. 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];
}
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];
}32. Consider the string "abbccbba". What is the minimum number of insertions required to make the string a palindrome?
33. For which of the following, the length of the string is not equal to the length of the longest palindromic subsequence?
34. A greedy algorithm can be used to solve all the dynamic programming problems.
35. 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;
}
#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;
}36. 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;
}
#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;
}37. 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;
}
#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;
}38. 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?
39. 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?
#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?40. What is the space complexity of the above dynamic programming implementation of the assembly line scheduling problem?
Read More Section(Dynamic Programming in Data Structures)
Each Section contains maximum 100 MCQs question on Dynamic Programming in Data Structures. To get more questions visit other sections.
