Data Structure II Practice Test 8

1. What is the space complexity of the following dynamic programming implementation of the balanced partition problem?

#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;
}

2. Which of the following methods can be used to solve the edit distance problem?

3. 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;
}

4. What is the time complexity of the following recursive implementation?

#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++)
         // subtract 1 because index starts from 0
	 max_price = max_of_two(max_price, prices[i] + rod_cut(prices,len - i - 1)); 
      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;
}

5. What is the output of the following code?

#include<stdio.h>
#include<limits.h>
int max_num(int a, int b)
{
      if(a > b)
        return a;
      return b;
}
int kadane_algo(int *arr, int len)
{
      int ans, sum, idx;
      ans =0;
      sum =0;
      for(idx =0; idx < len; idx++)
      {
          sum = max_num(0,sum + arr[idx]);
          ans = max_num(sum,ans);
      }
      return ans;
}
int max_sum_rectangle(int arr[][5],int row,int col)
{
      int left, right, tmp[row], mx_sm = INT_MIN, idx, val;
      for(left = 0; left < col; left++)
      {
           for(right = left; right < col; right++)
           {
                if(right == left)
                {
                    for(idx = 0; idx < row; idx++)
                     tmp[idx] = arr[idx][right];
                }
                else
                {
                     for(idx = 0; idx < row; idx++)
                       tmp[idx] += arr[idx][right];
                }
                val = kadane_algo(tmp,row);
                if(val > mx_sm)
                mx_sm = val;
           }
      }  
      return mx_sm;
}
int main()
{
       int arr[2][5] ={{1, 2, -1, -4, -20}, {-4, -1, 1, 7, -6}};
       int row = 2, col = 5;
       int ans = max_sum_rectangle(arr,row,col);
       printf("%d",ans);
       return 0;
}

6. What is the output of the following program?

#include<stdio.h>
#include<limits.h>
int min_jump(int *arr, int len)
{
      int j, idx, jumps[len];
      jumps[len - 1] = 0;
      for(idx = len - 2; idx >= 0; idx--)
      {	
	     int tmp_min = INT_MAX;
	     for(j = 1; j <= arr[idx] && idx + j < len; j++)
	     {
		   if(jumps[idx + j] + 1 < tmp_min)
		      tmp_min = jumps[idx + j] + 1;
	     }
	     jumps[idx] = tmp_min;
      }
      return jumps[0];
}
int main()
{
      int arr[] ={1, 1, 1, 1, 1, 1, 1, 1, 1},len = 9;
      int ans = min_jump(arr,len);
      printf("%d\n",ans);
      return 0;
}

7. Longest palindromic subsequence is an example of ______________

8. What is the output of the following code?

#include<stdio.h>
#include<limits.h>
int max_num(int a, int b)
{
      if(a > b)
        return a;
      return b;
}
int kadane_algo(int *arr, int len)
{
      int ans, sum, idx;
      ans =0;
      sum =0;
      for(idx =0; idx < len; idx++)
      {
          sum = max_num(0,sum + arr[idx]);
          ans = max_num(sum,ans);
      }
      return ans;
}
int max_sum_rectangle(int arr[][5],int row,int col)
{
      int left, right, tmp[row], mx_sm = INT_MIN, idx, val=0;
      for(left = 0; left < col; left++)
      {
           for(right = left; right < col; right++)
           {
               if(right == left)
               {
                   for(idx = 0; idx < row; idx++)
                      tmp[idx] = arr[idx][right];
               }
               else
               {
                   for(idx = 0; idx < row; idx++)
                      tmp[idx] += arr[idx][right];
               }
               val = kadane_algo(tmp,row);
               if(val > mx_sm)
               mx_sm = val;
           }
      }
      return mx_sm;
}
int main()
{
      int arr[4][5] ={{ 7,  1, -3, -6, -15},
                    { 10,  -6,  3,  -4,  11},
                    { -2, -3, -1,  2, -5},
                    { 3,  0,  1,  0,  3}};
      int row = 4, col = 5;
      int ans = max_sum_rectangle(arr,row,col);
      printf("%d",ans);
      return 0;
}

9. For any array, given that at most one element is non-zero, it is ALWAYS possible to reach the end of the array using minimum jumps.

10. What is the output of the following program?

#include<stdio.h>
int main()
{
      int coins[10]={1,3,4},lookup[100];
      int i,j,tmp,num_coins = 3,sum=10;
      lookup[0]=0;
      for(i=1;i<=sum;i++)
      {
	    int min_coins = i;
	    for(j=0;j<num_coins;j++)
	    {
	         tmp=i-coins[j];
	         if(tmp<0)
	          continue;
	         if(lookup[tmp] < min_coins)
		 min_coins=lookup[tmp];
	    }
	    lookup[i] = min_coins + 1;
      }
      printf("%d",lookup[sum]);
      return 0;
}