Question

In terms of the programming language required, the algorithm needs to be written in pseudocode

Dynamic Programming Consider the following problem based on the transformation of a sequence (or collection) of coloured disks Assume that you have a very large collection of disks, each with an integer value representing the disk colour from the range [0, c. For example, the colour mapping might be 0-red. 1-yellow, 2-blue, 3-pink. , c-black For a given sequence of coloured disks e.g., ( 0,1,2,3,4 ), at each time step (or iteration) you are only allowed to perform one of three basic operations: . remove one disk from the sequence - the disk you select can be any of the disks eg., 0,1,2,3,4) -0,1,3,4 insert one disk at any position in the sequence the disk you select to insert can be any colour in the range 0, c . replace any single coloured disk in the sequence with a different coloured disk - the replacement disk can be any colour in the range 10. Note: after multiple operations (or iterations), the length of sequence of disks may be different to the length of the starting sequence. We suggest that you represent each sequence of disks as an array a) Write a dynamic programming implementation for the function TRANSFORM (A0..n -1], B[0.m -1) to calculate the minimum number of operations required to transform one given sequence of disks Alo..n-1) into another Bl0..m-1). For example, AO, 1, 2, 3, 4] → BO, 1, 3, 5] requires at least two operations - remove 2; replace 4 by 5 Your function must be written in clear, unambiguous pseudo code (Hint: consider approximate string search) (b) What is the time complexity of your algorithm? Justify your answer

Answer 1

C++ Program:

#include<bits/stdc++.h>
using namespace std;
int main()
{
int N,M,i,j,temp;
cout<<"Enter the no.of.Elements in A:";
cin>>N;
cout<<"Enter the no.of.Elements in B:";
cin>>M;
int A[N],B[M];
cout<<"Enter the Elements in A:"<<endl;
for(i=0;i<N;i++)
{
cin>>A[i];
}
cout<<"Enter the Elements in B:"<<endl;
for(i=0;i<M;i++)
{
cin>>B[i];
}
int count=0;
for(i=0;i<N;i++)
{
for(j=0;j<M;j++)
{
if(A[i]==B[i])
{
count++;
}
}
}
cout<<"No.of Operations:"<<abs(N-count);
}

TIME COMPLEXITY: O(N*M);