Consider the algorithm shown below:

count = 0
for i in range(0, len(X), 1):
	for j in range(0, len(Y), 1):
		if(X[i] == Y[j]):
			count++

Both X and Y are strings of lengths n and m respectively. Write
the total (theoretical) runtime of the algorithm above. Your
anwser would be a function of n and m. Follow the approach shown in
class. For each line write down each time as a variable and then 
add them up to obtain the total time. (5 pts)
 
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Convert the following pseudocode to Python (10 pts)

V[0][0] = 0

## Initialization:
for i = 1 to length of seq1 { V[i][0] = i*g; }
For i = 1 to length of seq2 { V[0][i] = i*g; }

## Recurrence:
for i = 1 to length of seq1{
	for j = 1 to length of seq2{

				V[i-1][j-1] + m(or mm)
		V[i][j] = max {	V[i-1][j] + g
				V[i][j-1] + g

		if(maximum is V[i-1][j-1] + m(or mm)) then T[i][j] = D
		else if (maximum is V[i-1][j] + g)    then T[i][j] = U
		else			        then T[i][j] = L	
	}
}

Submit hardcopies in class on Feb 23rd.