could you explain me the solution you gave for the 4 player client-server architecture game?

Thanks
Gowri

put them into a hash
for 1 - n item{
if n>= 10 next;
push n to hash as a key
}

for 1 - n item {
if n>= 10 next;
if hash [10-n] exisits
print n, 10-n
}

0(n)

The solution that I came up with is as follows:

Assume list is sorted (even if it isn’t, it can be sorted in O(n)) & that there are no duplicates.
Have a pointer p1 point to the beginning of the list, and have a pointer p2 point to the end of the list.

while( indexOf(p1) != indexOf(p2) || indexOf(p1) > indexOf(p2))
sum= p1+p2
if (sum < 10)
p1++
else if (sum > 10)
p2–
else
// sum == 10
print p1, p2
p1++
p2–

effectively, the algorithm runs through the list only once and prints out every pair that’s adds up to 10

o(log n) solution:

foreach item in the list //O(n)
do a binary search for 10 - item (if result is negative, skip) // O(log n)

total time O(nlogn)

o(n) solution:

foreach item in the list //O(n)
int diff = 10 - item //O(1)
search hash table for key (diff) //O(1)

total time O(n)

how does the binary search and o(n) search work?