43.
The dynamic programming implementation of the maximum sum rectangle problem uses which of the following algorithm?

44.
Consider the following dynamic programming implementation of the longest common subsequence problem:
#include<stdio.h>
#include<string.h>
int max_num(int a, int b)
{
      if(a > b)
        return a;
      return b;
}
int lcs(char *str1, char *str2)
{
      int i,j,len1,len2;
      len1 = strlen(str1);
      len2 = strlen(str2);
      int arr[len1 + 1][len2 + 1];
      for(i = 0; i <= len1; i++)
          arr[i][0] = 0;
      for(i = 0; i <= len2; i++)
          arr[0][i] = 0;
      for(i = 1; i <= len1; i++)
      {
            for(j = 1; j <= len2; j++)
            {
                 if(str1[i-1] == str2[j - 1])
                  ______________;
                else
                   arr[i][j] = max_num(arr[i - 1][j], arr[i][j - 1]);
            }
      }
      return arr[len1][len2];
}
int main()
{
      char str1[] = " abcedfg", str2[] = "bcdfh";
      int ans = lcs(str1,str2);
      printf("%d",ans);
      return 0;
}
Which of the following lines completes the above code?

46.
Consider the following dynamic programming implementation of the Knapsack problem:
#include<stdio.h>
int find_max(int a, int b)
{
      if(a > b)
         return a;
      return b;
}
int knapsack(int W, int *wt, int *val,int n)
{
     int ans[n + 1][W + 1];
     int itm,w;
     for(itm = 0; itm <= n; itm++)
         ans[itm][0] = 0;
     for(w = 0;w <= W; w++)
        ans[0][w] = 0;
     for(itm = 1; itm <= n; itm++)
     {
          for(w = 1; w <= W; w++)
          {
               if(wt[itm - 1] <= w)
                  ans[itm][w] = ______________;
               else
                  ans[itm][w] = ans[itm - 1][w];
          }
     }
     return ans[n][W];
}
int main()
{
     int w[] = {10,20,30}, v[] = {60, 100, 120}, W = 50;
     int ans = knapsack(W, w, v, 3);
     printf("%d",ans);
     return 0;
}
Which of the following lines completes the above code?

49.
What is the space complexity of the following dynamic programming implementation of the longest common subsequence problem where length of one string is "m" and the length of the other string is "n"?
#include<stdio.h>
#include<string.h>
int max_num(int a, int b)
{
      if(a > b)
        return a;
      return b;
}
int lcs(char *str1, char *str2)
{
      int i,j,len1,len2;
      len1 = strlen(str1);
      len2 = strlen(str2);
      int arr[len1 + 1][len2 + 1];
      for(i = 0; i <= len1; i++)
          arr[i][0] = 0;
      for(i = 0; i <= len2; i++)
          arr[0][i] = 0;
      for(i = 1; i <= len1; i++)
      {
            for(j = 1; j <= len2; j++)
            {
                 if(str1[i-1] == str2[j - 1])
                  arr[i][j] = 1 + arr[i - 1][j - 1];
                else
                   arr[i][j] = max_num(arr[i - 1][j], arr[i][j - 1]);
            }
      }
      return arr[len1][len2];
}
int main()
{
      char str1[] = " abcedfg", str2[] = "bcdfh";
      int ans = lcs(str1,str2);
      printf("%d",ans);
      return 0;
}

50.
Given a rod of length n and the selling prices of all pieces smaller than equal to n, find the most beneficial way of cutting the rod into smaller pieces. This problem is called the rod cutting problem. Which of these methods can be used to solve the rod cutting problem?

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