@Yasir: Sort the numbers. Then
int i = 0, j = 1, m, p = 0;
while( j N )
{
m = a[j] - a[i];
if( m = K )
++p;
else if( m K )
++j;
else
++i;
}
// answer = p
Complexity = O(n log n).
Dave
On Aug 7, 12:53 pm, Yasir yasir@gmail.com wrote:
Guys,
Let's
@Dave and Piyush.
Thanks. Nice approach :)
Is further optimization possible?
(May be by increasing space complexity)
On Aug 7, 11:09 pm, Piyush Kapoor pkjee2...@gmail.com wrote:
First sort the array.Then for each element,say x ,do a binary search to find
the element y for which (y-x)=K or
On Sun, Aug 7, 2011 at 11:06 PM, Dave dave_and_da...@juno.com wrote:
@Yasir: Sort the numbers. Then
int i = 0, j = 1, m, p = 0;
while( j N )
{
m = a[j] - a[i];
if( m = K )
++p;
else if( m K )
++j;
else
++i;
}
// answer = p
Looks like an infinite
@Dave: after p++ is executed hw will ur pgm continue??
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@Shuaib and Sindhu: Right. Take out the first else, or replace ++p
with {++p ; ++i; ++j}.
Dave
On Aug 7, 1:42 pm, Shuaib Khan aries.shu...@gmail.com wrote:
On Sun, Aug 7, 2011 at 11:06 PM, Dave dave_and_da...@juno.com wrote:
@Yasir: Sort the numbers. Then
int i = 0, j = 1, m, p = 0;
