Question

Assume you are given two linear lists of size n each; consider the problem of determining whether any element of one list is an element of the other (not value, element). Derive a lower bound for this problem and design an algorithm for this problem. Derive its time complexity. It should be as close the lower bound as possible.

My work so far:

//Copy list n1 into n3 int n3[n];

for(int I = 0; I < n; i++)n3[i] = n1[i];

//Modify n2 by decrementing each element by 1

for(int = 0; i < n; i++) n2[i] -= 1;

//Check that n1 and n3 bool is_equal = true;

for(int I = 0; I < n; i++)

if(n1[i] != n3[i]) is_equal = false;

Break;

//Restore n2 back to its original state

for(int I = 0; I < n; i++) n2[i] +=1;

if(is_equal)

printf(“%s”, “The two lists don’t have any common element”);

else

printf(“%s”,”The two lists have 1 or more common elements”);

This requires 4 linear loops. So, the order is O(4n) which simplifies to O(n)

Answer 1

The algorithm posted by you looks almost correct, however it needs some minor corrections.

NOTE: Since you need to check if two lists contains the same element reference, it is important to modify the memory reference for the algorithm to work.

ALGORITHM

//Copy list n1 into n3, MAKE SURE THAT THE COPY IS BY VALUE, AND NOT BY REFERENCE, OTHERWISE THE ALGORITHM WON'T WORK
int n3[n];
for(int i = 0; i < n; i++) { // this step takes O(n) time
n3[i] = n1[i];
}
//Modify n2 by decrementing each element by 1
for(int i = 0; i < n; i++) { // this step takes O(n) time
n2[i] -= 1;
}
//Check that n1 and n3 bool has_same = false;
for(int i = 0; i < n; i++) { // this step takes O(n) time in the worst case
if(n1[i] != n3[i]) {
has_same = true;
break;
}
}
//Restore n2 back to its original state
for(int i = 0; i < n; i++) { // this step takes O(n) time
n2[i] +=1;
}
if(has_same)   // this step takes O(1) time
printf(“%s”,”The two lists have 1 or more common elements”);
else
printf(“%s”, “The two lists don’t have any common element”);

Hence, the total running time of the algorithm is: 4 * O(n) + O(1) = O(n)