If a and b are the numbers then
dig = log10(a) + log10(b);
if dig has some fractional part then number of digits is dig + 1 else dig.
On Mon, Sep 20, 2010 at 11:19 AM, sumant hegde sumant@gmail.com wrote:
Adding to the partial solution, if x, y are first digits, and x*y + x + y
10, the
Just tell me wats the answer for thissequence
1, 3 ,5, -1, -2 ,0, 7 .
On 9/20/10, Dave dave_and_da...@juno.com wrote:
Krunal: If the array contains only negative numbers, shouldn't the
subsequence with the largest sum be the empty subsequence?
Dave
On Sep 19, 5:45 am, Krunal Modi
Yes ur correct...it will require some extra spacebst can be represented
in the array form ritelet me think in tht logic.
On 9/19/10, Umer Farooq the.um...@gmail.com wrote:
creating a bst would require extra space. You can do this with an array of
char dude.
On Sun, Sep 19, 2010 at
On Mon, Sep 20, 2010 at 1:15 PM, Baljeet Kumar baljeetk...@gmail.comwrote:
If a and b are the numbers then
dig = log10(a) + log10(b);
if dig has some fractional part then number of digits is dig + 1 else dig.
found this correct onw
On Mon, Sep 20, 2010 at 11:19 AM, sumant hegde
@Dave :
We have to find any element.
So whichever is larger has to be the answer.Hence, -1
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@LG JAYARAM:
It is 7.
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@Naveen:
Yes you are right, the answer is not fully correct.
ans = integral part of ((log10(a)+log10(b) +1) {floor not ceiling}.
On Mon, Sep 20, 2010 at 4:11 PM, Naveen Agrawal nav.coo...@gmail.comwrote:
@Baljeet
I think your Answer is not fully correct
It should be :
@Krunal:
can u explain hwz 7
On Mon, Sep 20, 2010 at 6:02 PM, Krunal Modi krunalam...@gmail.com wrote:
@LG JAYARAM:
It is 7.
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@Rahul. No. Considering your example 33*30, x*y + x + y = 3*3 + 3 + 3
= 15 is not 10, so, as specified by Sumant, u will need a complex
logic to solve.
Dave
On Sep 20, 5:31 am, rahul patil rahul.deshmukhpa...@gmail.com wrote:
On Mon, Sep 20, 2010 at 1:15 PM, Baljeet Kumar
use a bst and insert in it reverse of array. use a count in bst node
and incr count at each right child insertion
On Sep 14, 11:00 am, Shiv ... shivsinha2...@gmail.com wrote:
A pseudo code-
int n; //= number of inputs.
cin*a; // the inputs.
int ** invArr;
*inVArr[n-1] =
Initially we have given three point A , B, C in plane represent three nodes
of triangle, now given another point Z which lies in same plane, find out
whether that point lies on/inside the triangle or outside of triangletry
to get in min time and space complexity
--
Thanks Regards
Umesh
Initially we have given three point A , B, C in plane represent three
nodes of triangle, now given another point Z which lies in same
plane, find out whether that point lies on/inside the triangle or
outside of triangletry to get in minimum time and space
complexity
--
Thanks Regards
here is the hint
we can easily solve this
draw a triangle
draw a point is inside the triangle
connect the three vertices of the triangle with this point
you will get three small triangle
if ( area(big triangle)== sum of area of small triangles) then the point is
inside the triangle else it is
One way would be :
Create equation of three sides of the triangle
Now check for this point if it lies on left/right of the line (do it for
each of the lines)...
On Mon, Sep 20, 2010 at 9:06 PM, Praveen Baskar praveen200...@gmail.comwrote:
here is the hint
we can easily solve this
draw a
Wonder if this works:
x = A / 10^(a-1) // take it as a decimal value itself
y = B / 10^(b-1) // take it as a decimal value itself
if x * y = 10.0
return (a+b)
else
return (a+b-1)
One advantage of the above method is that it can be done mentally.
On Sep 20, 10:47 am, Dave
Two unsorted arrays are given A[n] and B[n+1]. Array A contains n integers
and B contains n+1 integers of which n are same as in array B but in
different order and one extra element x. Write an optimized algorithm to
find the value of element x. Use only one pass of both arrays A and B.
--
You
This is okay, but does more math than necessary. Here's another
approach:
// Return 0 if p is left of a-b, 2 if right of a-b, and 1 if on a-
b.
int side(PT *p, PT *a, PT *b)
{
float d = (p.x-a.x) * (b.y-a.y) - (p.y-a.y) * (b.x-a.x);
return d 0 ? 0 : d 0 ? 2 : 1;
}
// This table treats
add up all the elements in array A say sumA and array B say sumB ,substract
the sumA from sumB... You'll get the element.
On Tue, Sep 21, 2010 at 5:36 AM, Anand anandut2...@gmail.com wrote:
Two unsorted arrays are given A[n] and B[n+1]. Array A contains n integers
and B contains n+1 integers
Following up two of my postings. Sorry, but I gave the wrong system of
equations in my first posting, and then solved that incorrect system
in my second. The system of equations should be
xa * a + xb * b + xc * c = xz
ya * a + yb * b + yc * c = yz
a + b + c = 1
Then the following
There can be overflow in case of adding up all the elements. Use Xor
instead.
int result = 0;
for (int i = 0; i n ;i++){
result ^= A[i]^B[i];
}
result ^= B[i];
result is the number we need.
On Tue, Sep 21, 2010 at 9:48 AM, vishal raja vishal.ge...@gmail.com wrote:
add up all the
There can be overflow in case of adding up all the elements. Use Xor
instead.
int result = 0;
for (int i = 0; i n ;i++){
result ^= A[i]^B[i];
}
result ^= B[n]; === (correction)
result is the number we need.
On Tue, Sep 21, 2010 at 9:48 AM, vishal raja vishal.ge...@gmail.comwrote:
@gene i didn't understand the boolean thing in your algo,
Say triangle is ABC, say Point is p
Algo goes like this,
for side AB
1. calculate cross product of(p-A)*(B-A)[ Which is nothing but the side
function in gene algo] See the sign.
2. calculate the cross product of (C-A)*(B-A) . See the sign
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