Question

5. Design an algorithm that takes two arrays, and returns true if the arrays are disjoint, i.e. have no elements in common. Write down your algorithm as pseudocode. You don't need to write a python code. Give the asymptotic analysis for time complexity in best-case and worst-case scenario.

Answer 1

SOLUTION:

NOTE:  I have written an executable python program since I need to show an executable I have written this. There is no difference in terms of pseudocode and program in pseudo-code we will write English like statements here we will keep syntax for that there is no difference in that.

CODE:

def disjointSet(list1,list2):
##we need to iterate through all elements in the list so we have written this loop

for i in range(len(list1)):
##we need to iterate through list2 to check for matching for that
##we have written another loop
for j in range(len(list2)):
if(list1[i]==list2[j]):
##if any element is matched then it is not an disjoint set so
## we will return false
return False

##if no elements are then only this statement will execute this meansthat
## the two sets are disjoint so we will return True
return True

##call the function to test
print(disjointSet([1,2,3,4,8],[5,6,7,8]))

IMAGE:

OUTPUT:

le disjoint.py - C:/Users/vinod/AppData/Local/Programs/Python/Python37-32/disjoint.py (3.7.2) File Edit Format Run Options Window Help - 0 x def disjointSet (listi,list2): ##we need to iterate through all elements in list so we have written this loop for i in range (len (listi)): ##we need to iterate through list2 to check for matching for that ##we have written another loop for j in range (len (list2)): if(listi[i]==list2[j]): ##if any element is matched then it is not an disjoint set so ## we will return false return False ##if no elements are then only this statement will execute this meansthat ## the two sets are disjoint so we will return True return True ##call the function to test print (disjointSet([1,2,3,4,8], [5,6,7,8]))

RESTART: C:/Users/vinod/AppData/Local/Programs/Python/Python37-32/disjoint.py False

TIME COMPLEXITY:

HERE n-> Number of elements in list1

m-> Number of elements in list2

BEST CASE COMPLEXITY: O(n)-> this is when we have the first element is found in the next list we will find that the list is disjoint in only one parse of the second list its complexity is "n"

WORST CASE COMPLEXITY: $\Omega$(n*m)This is when a list is disjoint we need to check each element in the list1 with list2 so the complexity here is like each element is check for each element in list2 so it is (n*m)