this finds first and last index in pair structure in logn
void repindxs(int array[],int start, int end, int k, pairint, int *p, int
n/*last index*/)
{
if(startend) return ;
int m = (start+end)/2;
if( array[m] == k (m-10?-1:array[m-1] k) )
p-first = m;
else if(array[m] k || (array[m]==k
How to find the indexes of a repeated element in the sorted array in
*log n*time..
e.g.
a: 1 3 4 4 5 6 7 8 8 9 9 9 10
x=9
ouput is 10 and 12 (indexing from 1)
--
Regards
Kumar Raja
M.Tech(SIT)
IIT Kharagpur,
10it60...@iitkgp.ac.in
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Using binary search, first find the first occurrence of x. Using this first
occurrence run another binary search to find the last occurrence of x.
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Suppose the array is not sorted and we have to find if an element has
occurred earlier or not; and if yes, then remove
it.what is the best achievable time and how?
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