# Difference between revisions of "TADM2E 4.8"

(a):

Sort S with any nlogn sorting method of your choice.
for( int i = 1; i <= n; ++i )
{
int j = x - S[i];
Binary search for j in the sub-array of S[i+1~n] and close the problem once it's been found;
}


(b):

Subtract each of S[1~n] from x to get a new array of real numbers T[1~n].
int i = 1, j = 1;
while( i <=n && j <= n )
{
if( S[i] == T[j] )
{
problem solved;
break;
}
else
{
S[i] < T[j] ? ++i : ++j;
}
}


Another Solution to part B without creating the additional Array..

i = 0;
j = n - 1;

for (i = 0; i < j; i++)
{
while( (i < j) && (S[j] + S[i] > x ) )
{ j--;
}
if (x == (S[j] + S[i]) )
{
return true;

}
}

"i" scans from left to right, "j" from right to left,
looking for the right pair...


Yet another solution to solve both parts in linear time using a hashtable.

Initialize a hash-table backed dictionary D; // O(n*lg(n)) if the underlying structure is a BST.
for each e in S:
delta = x - e;
if Search(D, delta) is not NULL:
return (e, delta);
else
Insert(D, {e => delta}); // Insert a record with key e and value delta.