@yq, didn't I ask you this question before?
On Fri, Aug 10, 2012 at 4:48 PM, yq Zhang zhangyunq...@gmail.com wrote:
@shiv, your code is correct go compute the base 26 number. However, this
question is not base 26 number obviously.
On Wed, Aug 8, 2012 at 4:46 AM, shiv narayan
public static ArrayListArrayListInteger Partition(int val, int start,
int size) {
ArrayListArrayListInteger r = new ArrayListArrayListInteger();
if (start * size val) {
return null;
}
if(size == 1)
{
r.add(new ArrayListInteger());
r.get(0).add(val);
return r;
}
for
@albert, You need to becareful when doing the divide, because there is no
ZERO. (Z - AA not Z-A0).
here is the code:
public static String ExcelMapIntToStr(int n)
{
StringBuilder sb = new StringBuilder();
while(n0)
{
sb.append((char)(('A' - 1) + (n-1)%26 + 1));
n = (n-1)/26;
}
return
.
This could be solved in liner time. finding a possible start pump use O(n)
and prove it use another O(n).
-weiq
On Sat, Jan 29, 2011 at 9:47 PM, nishaanth nishaant...@gmail.com wrote:
@Wei.Qi Can you clarify when your algorithm terminates and also what
is the running time of the algorithm
Starting with any pump A, try to finish the circle, if at pump B that can
not reach pump B+1, it means all pumps from A to B can not finish the circle
(it will go broke at pump B), then just start with B+1. After go through all
the pumps start from some pump, we got an answer. if we go back to
; should not move forward
here,
}
}
On Sat, Jan 15, 2011 at 1:55 PM, Jammy xujiayiy...@gmail.com wrote:
@Wei Please test you code on cdbbcbbca. I believe it outputs 2
instead of 8.
On Jan 14, 4:09 am, Wei.QI qiw...@gmail.com wrote:
FindStartIndex(char
x is even number probability
0.9x + x = 1
x = 1/1.9
P(2) = P(4) = P(6) = (1 / 1.9) / 3
P(4|5|6) = (P(4) + P(6)) / 0.75 = 2 / 1.9 / 3 / 0.75 = 0.4678362573099415
-weiq
On Thu, Jan 13, 2011 at 11:29 PM, snehal jain learner@gmail.com wrote:
An unbalanced dice (with 6 faces, numbered from 1 to
FindStartIndex(char[] a)
{
int start = 0;
int current = 1;
while(current a.Length)
{
if(a[current] a[start])
{
start = current;
++current;
}else if(a[current] a[start])
{
++current;
}else //a[current] ==
@juver++, can you give me a test case that will time out?
On Fri, Jan 14, 2011 at 6:44 AM, juver++ avpostni...@gmail.com wrote:
@qw
I think you solution will timed out?!
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