selecting pivot as a near median using order stastics method(O(n)) we can
run it in worst case O(nlogn)
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void main(int arg,char *argc)
{
char *s=argc[1];
int count=0,i;
for(i=strlen(s)-1;i=0;i--)
{
count++;
printf(%c,s[i]);
if(count==3)
{
count=0;
putchar(',');
}
}
}
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it takes O(n) and also O(n)extra space(queue)
On Wed, Dec 22, 2010 at 12:37 PM, Saurabh Koar saurabhkoar...@gmail.comwrote:
Find the first node whose left child is NULL or Right child is NULL
using BFS.(As the tree is complete,all nodes before this will have two
children).Insert at that node.
count the no of bits lets say n
then answer becomes 2^n-1
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prims check for this case [1,1,4,4]
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For
i did nt get this xor part in adithya solution
check if this works
array is valid if satisfy 2 conditions
1.max-min=n-1
2.there should be no repeatations
first one can be done in O(n)
for second
check 1xor2xor...xorn=(a[1]-min+1)xor(a[2]-min+1)xor..
if both are equal there are
sort the jobs according to their starting time and check wheather a job
starting time is less than its previous job ending time
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@rashmi u missed p7 i took it as 2
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@ravindara yeah this works perfect
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@preethika bettter to do it while inserting itself
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u check only that element and continue inserting other elements
ex m=4
elements
5 3 2 7 1 2 ...
(3,1) will be found while inseting 1 , not 3
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@preethika (3,1) is found while inserting 1 and (2,2) will be found while
inserting second 2 (which is at last position)
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