For each employee, keep track of which vacation he is assigned to, if
any.
Start with all employees unassigned.
1. Pick the first pair who have worked together and are currently
unassigned. Assign one to Trip A and the other to Trip B.
2. Loop over all pairs. If a pair are assigned to the same
@Yasir: Yups...I also have the same algo...
Just to improvise your solution, you need not do binary search on both sides
of the pivot.
Just check End points (min-max) of the both sub-array to decide which side
to do a binary search..This works in the case of duplicate elements too.
for e.g.
2 5
Yeah got it. Thanks :)
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Check this out
//But this and fails for input such as: 101 ie all nos
must be unique..
int search_element_in_rotated_array(int *a, int low, int high, int
key)
{
while(low = high)
{
int mid = low + (high - low)/2;
if(a[mid] == key)
return
nope it will not work
top kya lower part bhi sorted hai..
like 7 8 9 1 2 3 4 5 6
do it with this i/p... it will fail
ur algo will do same thing for 8 and 4
FAIL !!!
On Fri, Aug 12, 2011 at 9:27 PM, KK kunalkapadi...@gmail.com wrote:
Check this out
//But this and fails for input
how abt http://ideone.com/lN2og ??
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