Question

(1) Describe (in words) an algorithm that checks if two given lists are disjoint, meaning that they have no element in common. (2) Then write the pseudo-code for your algorithm by calling it dis- joint(0,..., An, bi, ...., bm : elements) (note that the lists can have different size).

Answer 1

• Below is the detailed explanation of the algorithm and pseudo code for the problem mentioned.

• Algorithm:

To check if two lists are disjoint or not we can:

Start traversing list1 i.e, element by element and for each element of this list1 start traversing the other list2 and check if element of list1 can be found in list2, if we found that element in list2 then we will return False(as lists are not disjoint) and otherwise we move to next element of list1 and do the same process till all the elements of list1 are checked to be in list2 or we return false somewhere.

After we come out of the loops we know that lists are disjoint as if it was not then we have returned false from there only, so return True here and end function.

• Pseudo code(function disjoint returns True if lists passed are disjoint otherwise returns False):

Procedure disjoint(a₁,.........,a_n, b₁,.........,b_m : elements):

for element1 in (a₁,.........,a_n):

  for element2 in (b₁,.........,b_m):

check if element1 == element2 then:

return False

else continue checking

return True

• If we call above function as disjoint([1,2,3], [4,5,6]) then it will return True as both lists are disjoint but if we call above function as disjoint([1,2,3], [2,5,6,8]) then it will return False as the two lists mentioned have element 2 as common so are not disjoint.

• The complexity of above mentioned algorithm is O(n*m) where n is size of first list and m is size of other list as we are checking each element of list1 for every other element of list2, so 2 ested loops are required .Hence worst/average case complexity comes out to be O(n*m)

Answer 2

• Below is the detailed explanation of the algorithm and pseudo code for the problem mentioned.
• Algorithm:

To check if two lists are disjoint or not we can:

Start traversing list1 i.e, element by element and for each element of this list1 start traversing the other list2 and check if element of list1 can be found in list2, if we found that element in list2 then we will return False(as lists are not disjoint) and otherwise we move to next element of list1 and do the same process till all the elements of list1 are checked to be in list2 or we return false somewhere.

After we come out of the loops we know that lists are disjoint as if it was not then we have returned false from there only, so return True here and end function.

• Pseudo code(function disjoint returns True if lists passed are disjoint otherwise returns False):

Procedure disjoint(a₁,.........,a_n, b₁,.........,b_m : elements):

for element1 in (a₁,.........,a_n):

  for element2 in (b₁,.........,b_m):

check if element1 == element2 then:

return False

else continue checking

return True

• If we call above function as disjoint([1,2,3], [4,5,6]) then it will return True as both lists are disjoint but if we call above function as disjoint([1,2,3], [2,5,6,8]) then it will return False as the two lists mentioned have element 2 as common so are not disjoint.
• The complexity of above mentioned algorithm is O(n*m) where n is size of first list and m is size of other list as we are checking each element of list1 for every other element of list2, so 2 ested loops are required .Hence worst/average case complexity comes out to be O(n*m).

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