ok so by doing right=mid; and left=mid; in the if and else if block.
you are narrowing down the range...??
thanks for the explanation :)
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here is solution for non-decreasing number using simple recursion.
static int min=0;
static int flag=0;
int nonDec(int num)
{
if(num==0)
{
return 0;
}
nonDec(num/10);
if(flag==1)
{
return -1; // not a non-decreasing number.
}
if(min = (num%10))
{
min=num%10;
}
else
{ flag=1;
return -1;
}
ok question was different we have to find total number..
On Sat, Dec 31, 2011 at 2:08 PM, atul anand atul.87fri...@gmail.com wrote:
here is solution for non-decreasing number using simple recursion.
static int min=0;
static int flag=0;
int nonDec(int num)
{
if(num==0)
{
return 0;
merge two linked list without sorting them, the resultant list should be
sorted one...
any other better solution?
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Wed, Jan 12, 2011 at 3:15 PM, Ashish Goel ashg...@gmail.com wrote:
without
correction in Lucifier code :-
if (N 1)
{
int iter = 0;
int totalCount = 10; // this is when N = 1 ...
for (int i=0; i = 9; ++i) // earlier for (int i=0; i ** 9; ++i)
A[i] = 1;
while(++iter N)
{
for (int i = 8; i 0 ; --i)
{
A[i]+= A[i+1];// earlier
@atul
There is no. correction..
The for loop is correct.. It should be:
for (int i =0; i 9; ++i)
On 31 Dec, 16:02, atul anand atul.87fri...@gmail.com wrote:
correction in Lucifier code :-
if (N 1)
{
int iter = 0;
int totalCount = 10; // this is when N = 1 ...
for (int i=0; i = 9;
@atul..
Yes, A[i] += A[i+1] is correct..
Another correction that is required is :
for (int i = 7; i = 0 ; --i) // earlier for (int i = *8*; i 0 ; --
i)
On 31 Dec, 16:10, Lucifer sourabhd2...@gmail.com wrote:
@atul
There is no. correction..
The for loop is correct.. It should be:
for
@atul..
It seems that in the example explained above i have done it correctly
but while jolting down the code i messed up the indexing..
Thanks for bringing it up .. :)
On 31 Dec, 16:17, Lucifer sourabhd2...@gmail.com wrote:
@atul..
Yes, A[i] += A[i+1] is correct..
Another correction that
@Ashish..
I have something in mind.. but would require verification by u..
Lets say, the structure of the data node is as follows:
struct node
{
int data;
struct node *next;
};
Now, given 2 sorted linked lists we can right a O(N) time and in-place
merge process, to build a sorted merged
@atul..
Yup.. exactly.. i think the problem is itself about narrowing the
range and works really well as each node can have only one child.. :)
On 31 Dec, 13:09, atul anand atul.87fri...@gmail.com wrote:
ok so by doing right=mid; and left=mid; in the if and else if block.
you are narrowing
Nice Solution and explanation Lucifer :)
On Dec 31, 4:48 pm, Lucifer sourabhd2...@gmail.com wrote:
@atul..
Yup.. exactly.. i think the problem is itself about narrowing the
range and works really well as each node can have only one child.. :)
On 31 Dec, 13:09, atul anand
Within a 2D space, there is a batch of points(no duplicate) in the
region (0,0),(0,1),(1,0),(1,1), try to find a line which can divide
the region to 2 parts with half points in each .the input will be an
array of points and the length of the array.
struct point{
int x;
int y;
};
input :
the LL has to be a DLL as deletion would be a prob otherwise.
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Sat, Oct 1, 2011 at 9:42 AM, SAMM somnath.nit...@gmail.com wrote:
Linked List + Hash table will serve the purpose in O(1).
On 9/30/11,
@Lucifier :
i wrote for (int i=0; i = 9; ++i) because i was considering index from 1
to n
not 0 to n-1;
you are considering from 0 to n-1. so both are correct.
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O(N^2) approach..
---
Say the current start index is i..
Then starting from i keep track of the min and max element and ensure
that the diff of max and min is =K.
Say the diff exceeds at index j, then the current continuous subarray
size would be (j - i) and we
Hey
Below is a solution. Hope it works.
Let 'n' be the size of the array we have.
max = min = array(n-1)
flag=1;
for(i=(n-2) to 0)
{
if(array(i)array(i+1))
{
if(min array(i))
{
flag = 0;
break;
}
else
{
min = array(i);
}
}
else
@apoorve
say the input is 9 8..
I think it will fail...
Secondly, i see that max is not being used
On Dec 31, 8:19 pm, Apoorve Mohan apoorvemo...@gmail.com wrote:
Hey
Below is a solution. Hope it works.
Let 'n' be the size of the array we have.
max = min = array(n-1)
flag=1;
Hey
It was a typo error.
here it goes..
max = min = array(n-1)
flag=1;
for(i=(n-2) to 0)
{
if(array(i)array(i+1))
{
if(min array(i))
{
flag = 0;
break;
}
else
{
min = array(i);
}
}
else
{
if(max array(i))
{
@Apoorva..
I think i misread your code at the beginning ( apart from the typo)
If i m not wrong then you have a taken a bottom up approach to solve
the problem using the range concept..
Wherein the code that i posted is a top down approach..
Great .. :)
On 31 Dec, 21:30, Apoorve Mohan
@above
Correction..
i can be = R... hence, which can be found as part of traversal..
Another addition:
I have explained the concept using A[j] to be max.. In case A[j] is
min, then we will take a similar approach keeping in mind that R will
be closest but in terms of finding the max element..
HNY 2012(IST) in between to all members and happy coding for 2012 may all
your compilation produce ZERO errors
On Sun, Jan 1, 2012 at 12:01 AM, Lucifer sourabhd2...@gmail.com wrote:
@above
Correction..
i can be = R... hence, which can be found as part of traversal..
Another addition:
I
*if (max - min k) //* this code is unreachable in the while loop.
break;
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remove continue in while loop...then it will work fine.
On Sun, Jan 1, 2012 at 12:49 AM, atul anand atul.87fri...@gmail.com wrote:
*if (max - min k) //* this code is unreachable in the while loop.
break;
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start=0;
maxLen=0;
min=arr[0];
max=arr[0];
j=0;
for(i=1;in;i++)
{
if(arr[i] max)
max=arr[i];
else if(arr[i] min)
min=arr[i];
if( (max - min ) k maxLen ( i - j ) )
{
maxLen=i - j;
start=j;
@atul..
No need to remove continue.. 'continue' has been added for
optimization..
(max - min) K, should only be checked when either we get a new min
or max...
In case the current processed element is neither a good fit for max or
min then we need not check for the difference..
On Jan 1, 2:33
@ dave
what about the problem if it dsnt satisfy the 2 conditions(odd/even)
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