g[][]- given matrix
r[][]- result matrix
copy first row n column from g[][] to r[][]
rest
for i=1 to n-1
for =1 to n-1j
if(g[][])
r[][]=min(r[i][j-1],r[i-1][j],r[i-1][j-1])+1;
else
r[][]=0;
On Wed, Jan 11, 2012 at 10:00 AM, Gene gene.ress...@gmail.com wrote:
The 1's and 0's
Already discussed
https://groups.google.com/group/algogeeks/browse_thread/thread/8a3e1a6665702f54/66bb2e804b43ae1d?hl=enlnk=gstq=MS+question+ankur#66bb2e804b43ae1d
.
On Fri, Sep 30, 2011 at 2:59 PM, sumit kumar pathak
sumitkp1...@gmail.comwrote:
*find all the zeros in first iteration and
Count number of space... Add 1 in it... that will be the no of words (say
x)...
now take two variables m n
m=x/2
n =2 //if m is odd
=3 //if m is even
now store the values in array a[] as
a[0]=m
a[1]=(m+n)mod(x)
a[2]=(m+n)mod(x)
.
.
Suppose matrix is
1 0 0 1
1 0 1 0
0 0 0 0
then we traverse the matrix for each 1 we found at a[i][j] , we will check
for i=i to irow and j=j to jcol if that contains any more 1
if it contains 1 in row then we don't make the whole row as 1..we ignore the
row and same will be for column
Take an array a[1000]input first 1000 values from the stream in it..
Here we are representing a unit time with help of space because 1000
elements in a stream will always come in constant time
So just count the occurrence of a given number in that array... then divide
the count with 1000
n=1 n=2 n=3 n=4
01010!(10!)!
1 9 9! (9!)!
2 8 8!
3 7 7!
4 6 6!
5 5 5!
6 4
mistake in last post...it was not factorialsum upto n i.e =n(n+1)/2
i.e 10! is wrong ...it will be 10(10+1)/2
On Sun, Sep 25, 2011 at 12:14 PM, Yogesh Yadav medu...@gmail.com wrote:
n=1 n=2 n=3 n=4
01010!(10!)!
1
+1 Gohana
On Sun, Sep 25, 2011 at 12:28 PM, Sanjay Rajpal srn...@gmail.com wrote:
let the number of digits be n
then answer would be ((n+9)! ) / (9! * n!)
Sanju
:)
On Sat, Sep 24, 2011 at 11:51 PM, Yogesh Yadav medu...@gmail.com wrote:
mistake in last post...it was not factorial
T1-T2-T3-T6
T1-T2-T4-T7
T1-T2-T5-T8
2 Processor:
(T1-T2) ..2 TS
(T3T4)...1 TS
(T6T5)...1 TS
(T7T8)...1 TS
4 Processor
(T1-T2) ..2 TS
(T3T4T5)...1 TS
(T6T7T8)...1 TS
.
On Sun, Sep 25, 2011 at 2:33 PM, siva viknesh sivavikne...@gmail.comwrote:
plz give detailed
we cant traverse the string twice...
if there is an error in my logic then plz tell
my logic is:
we have to take starting and ending indexes for '0','1','2' like below
0 1 2
starting_index
ending_index
now suppose our string 102112011
so we
I am not sure about these...but i guess this will do the job
In Linked List:
(rear-next==front)
In Array:
after reaching the maximum size or the end of queue ...just add one more
element in queue... and after addition just check
(queue[front]==new_element_added) ... if yes then circular queue
@algo geek:
there is some error in your logic...
consider a number 759845
according to ur logic the number greater than this will be 784559
but it should be 759854
.
On Thu, Sep 22, 2011 at 4:24 PM, algo geek geeka...@gmail.com wrote:
Keep a pointer ont the rightmost digit. Keep
digit =4
search for smaller digit than 4 towards left... it will be 3...
then answer 759453
On Thu, Sep 22, 2011 at 4:40 PM, Yogesh Yadav medu...@gmail.com wrote:
@algo geek:
there is some error in your logic...
consider a number 759845
according to ur logic the number greater than
inorder traversal will do the job... just make sure to create links...
On Thu, Sep 22, 2011 at 1:48 AM, Gene gene.ress...@gmail.com wrote:
#include stdio.h
typedef struct node_s {
int data;
struct node_s *left, *right;
} NODE;
// Convert BST to a sorted list connected by -right
759435
now am i right??
On Thu, Sep 22, 2011 at 4:55 PM, Ratan success.rata...@gmail.com wrote:
the answer should be 759435...
On Thu, Sep 22, 2011 at 4:45 PM, Yogesh Yadav medu...@gmail.com wrote:
my logic: if wrong then plz tell...
start from rightmost digit and search towards
Ans. 512
...
On Thu, Sep 22, 2011 at 5:38 PM, Rohit Upadhyaya mailtoroh...@gmail.comwrote:
int main()
{
int a=256;
char *p=a;
*++p=2;
printf(%d,a);
return(0);
}
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this
+1 abhinav
but u forgot to change array again into bits...
...
On Tue, Sep 20, 2011 at 1:02 PM, Ishan Aggarwal
ishan.aggarwal.1...@gmail.com wrote:
What are u doing in the first loop running for k=31 to k =0?
On Tue, Sep 20, 2011 at 12:50 PM, abhinav gupta guptaabhinav...@gmail.com
@ishan:
there is error in your loop bro...
for(k=31;k=0;k++)
it should be k--
.
On Tue, Sep 20, 2011 at 3:10 PM, vijay singh vijaysinghb...@gmail.comwrote:
On Tue, Sep 20, 2011 at 2:16 PM, Ishan Aggarwal
ishan.aggarwal.1...@gmail.com wrote:
Hi,
When I am running this program, I am
@saurabh:
i think may be u left some part of question to mention... the array may be
a heap array ...
or if the ques is correct then i don't think this sum can be possible in
O(log n) for previously existing array...
may be we have to make array from start...then as you mention we can use
i=0;
char *str;
while(a[i]!=NULL)
{
j=0;
while(j!=i)
{
for(k=j;k=i;k++)
{
l=0;
str[l]=a[j];
}
if(Dictionary.findword(str))
printf(str);
j++;
}
}
...
On Mon, Sep 19, 2011 at 8:20 PM, Sangeeta sangeeta15...@gmail.com wrote:
given
, Yogesh Yadav medu...@gmail.com wrote:
i=0;
char *str;
while(a[i]!=NULL)
{
j=0;
while(j!=i)
{
for(k=j;k=i;k++)
{
l=0;
str[l]=a[j];
}
if(Dictionary.findword(str))
printf(str);
j++;
}
}
...
On Mon, Sep 19, 2011 at 8
struct string_node
{
char str[];
int length;
struct string_node *ptr;
}
now traverse the strings and saving strings in str[],their length in
length and point it to next string_node
On Sun, Sep 18, 2011 at 4:47 PM, pooja pooja27tan...@gmail.com wrote:
Can some one please help me
...
...
On Sun, Sep 18, 2011 at 5:24 PM, pooja pooja27tan...@gmail.com wrote:
sorry.. i meant the longest common substring..
eg. consider 4 string.. ababa , bab , babba, abab
so the longest common substring is bab.
On Sep 18, 4:40 pm, Yogesh Yadav medu...@gmail.com wrote:
struct string_node
current position is index n (say)
largest=0;
second_largest=0;
for(i=1;i=n;i++)
{
if(a[i]a[n])
{
if(a[i]largest)
{
second_largest=largest;
largest=a[i];
}
elseif(a[i]second_largest)
second_largest=a[i];
}
}
On Mon, Sep 19, 2011 at 2:15 AM, sagar pareek sagarpar...@gmail.com wrote:
Give
x==9 is not an error...
it will be 1 when true and 0 when false...and both are executable
statements...
when we print float with %d then compiler prints 0...and when we print
integer with %f then compiler prints last assigned or printed float value to
any float variablehere it is assigned
@Ashima..
u r right...it is always printing first float variable value
On Sat, Sep 17, 2011 at 2:05 PM, Ashima . ashima.b...@gmail.com wrote:
@yogesh : but if in place of x=1 in above code, I write t=54. that means i
m assigning somevalue to a float variable.Then it should print 54
4,5
statements inside sizeof() does not get executed...it will tell just size...
int a=3;
printf(%d,sizeof(a++));
here a++ will not be executed ...it will just tell size ...
Also, lets suppose it execute statements ...then sizeof(int),sizeof(node *)
...will always produce error because int
Ans 1. apply bubble sort. outer loop should run just two times because in
2nd time the 2nd largest element will be at 2nd last position...
comparisons=(n-1)+(n-2)
On Sat, Sep 17, 2011 at 4:27 PM, Ashima . ashima.b...@gmail.com wrote:
ans 2:
http://www.qbyte.org/puzzles/p131s.html
Ashima
)Information Systems
4th year
BITS Pilani
Rajasthan
On Sat, Sep 17, 2011 at 4:20 AM, Yogesh Yadav medu...@gmail.com wrote:
Ans 1. apply bubble sort. outer loop should run just two times because in
2nd time the 2nd largest element will be at 2nd last position...
comparisons=(n-1)+(n-2)
On Sat
, Yogesh Yadav medu...@gmail.com wrote:
Ans 1. apply bubble sort. outer loop should run just two times because in
2nd time the 2nd largest element will be at 2nd last position...
comparisons=(n-1)+(n-2)
On Sat, Sep 17, 2011 at 4:27 PM, Ashima . ashima.b...@gmail.com wrote:
ans 2:
http
the basic operation is comparision so we can express comaprision
in terms of order of O(n). newazz we still require n-1 comparisions to find
second max .
On Sat, Sep 17, 2011 at 5:06 PM, Yogesh Yadav medu...@gmail.com wrote:
@aditya:
comparing isn't complexity here... here it means
12 west ...12 south
On Sat, Sep 17, 2011 at 5:24 PM, Dheeraj Sharma dheerajsharma1...@gmail.com
wrote:
12 west and 10 south.. i thnk..is the correct one..
On Sat, Sep 17, 2011 at 5:22 PM, Yogesh Yadav medu...@gmail.com wrote:
yeah it will be 62.5%
On Sat, Sep 17, 2011 at 5:14 PM
Q3. 101 matches
..
On Sun, Sep 18, 2011 at 8:02 AM, VIHARRI viharri@gmail.com wrote:
hey i'm also thinking n + logn -2.. but couldnt able to figure out
how??? can you please explain the logic
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks
a[]= {1, 2, 3, 4,5}
a
a
a+1
a+1
a is address of whole array...so a+1 means next element after array
completion.
so ptr-1 will point to 5
...
On Fri, Sep 16, 2011 at 1:25 PM, Anup
For Stack:
just make a structure:
struct stack_with_priorityqueue
{
int num;
int priority;
struct stack_with_priorityqueue *ptr;
}
now when we add another number just increase the priority... priority++
For Queue:
do same...just decrease priority...priority--
...
*//this method can be applied only when
the value of elements in array are not very large
int count[];
for(i = 0 to n-1)
{
if(count[arr[i]] == 1)
printf( %d , arr[i]);
else
count[arr[i]]++;
}*
..
On Wed, Aug 31, 2011 at 12:38 PM, Yuchen Liao lycdra...@gmail.com wrote:
//don't know this is completely correct or not
1.find sum of elements of both arrays a[] and b[]...from 0 to n-1
2.check whichever is smaller sum_a or sum_b...say sum_a is smaller
3.generate a random number i=random(sum_a)... generates random number
between o to (sum_a)-1
//random
prime no has only 2 factors. number itself and 1.
On Wed, Sep 7, 2011 at 12:14 PM, aayush jain ajain...@gmail.com wrote:
can anybody tell me the code of find the prime no. and after finding prime
no. find its prime factore using linkes list??
--
You received this message because you
//Num is our given number obtained from date
Int num1,num2;
Int count1=0,count2=0;
void nearestpalindrome()
{
Num1=num;
For(;;)
{
If(Checkpalindrome(num1))
//Num is our given number obtained from date
1.divide num into 2 parts last 4 and first 4
ex: num=96548765
num1=9765,num2=Reverse(8765)
2.if(num1num2)
add(num1-num2) in num2
3. num3=reverse(num2)
palindrome=num1||num3
On Mon, Sep 5, 2011 at 10:51 PM, Neha Singh
@all :
Thanks for correcting me.
On Tue, Sep 6, 2011 at 3:41 AM, Piyush Grover piyush4u.iit...@gmail.comwrote:
Here's the pseudo code. I hope it should work.
d = date-string (d[0]d[1])
m = month-string (m[0]m[1])
y = year-string (y[0]y[1]y[2]y[3])
dd = d;
mm = m;
yy = y;
while(1){
@Priya:
Lvalue is not a contant value. It can be modifiable or unmodifiable. Lvalue
stands for location value that means that we can store some value here in
this location and rvalue stands for read value that means that it can be
stored at some location that is in memory having lvalue.
Here in
@Priya:
Because a is *unmodifiable* LValue. When we are using a=a+1(a++) , this
means that we are changing the *base address* of array a[], and that can not
be possible due to Library files.
On Sun, Sep 4, 2011 at 11:49 AM, priya ramesh
love.for.programm...@gmail.com wrote:
@yogesh: Thanks a
plz upload it againi cant open it...
On Tue, Aug 16, 2011 at 8:31 PM, Nitin Nizhawan nitin.nizha...@gmail.comwrote:
sent to you ravi
On Tue, Aug 16, 2011 at 8:16 PM, ravi kumar ravikumar...@gmail.comwrote:
heyy nitin.. it says da file izz locked .. can u mail me da buk.. thanx in
Part 1:
Take 2 pointers moving at different speed...
1st with head-next
2nd with head-next-next
when 2nd will reach the end 1st will be pointing middle... so one result is
obtained
Part 2:
when 1st result is obtained then start with the node pointed by 1st as it
will be on 1/2th part of the
to 3/4th,
same way for m/n th node. But this works best for even number of nodes in
list.
For odd numbers we need to compromise like finding the middle node :)
Cheers,
Mani
On Tue, Jul 19, 2011 at 4:20 PM, Yogesh Yadav medu...@gmail.com wrote:
Part 1:
Take 2 pointers moving at different
with matrix this problem will be in O(n^2).
We don' have to just check the max number present in matrix. We have to
check as
array a[]= 1,2,3,4,5,6,8,10,12,14
diff[][]= 1 2 3 4 5 6 8 10 12 14
2 0 1 2 3 4 6 8 10 12
3 -1 0 1 2 3 5 79 11
largest size of square would be = H.C.F of width and height .
now with size known we have to just arrange squares
this can be done such that we can make a big square by adding them...
for ex 1st square can be made by just (1 square)
2nd square can be made by adding 3 sqaures around it
largest size of square would be = H.C.F of width and height .
now with size known we have to just arrange squares
this can be done such that we can make a big square by adding them...
for ex 1st square can be made by just (1 square)
2nd square can be made by adding 3 sqaures around it
array a[] stores numbers
for(i=1;in;i++)
{
diff[i-1]=a[i]-a[i-1];
}
maxcount=1;
tempcount=1;
for(j=1;jn-1;j++)
{
if(diff[j]==diff[j-1])
tempcount++;
else
{
if(tempcountmaxcount)
maxcount=tempcount;
tempcount=1;
}
}
On Sat, Jul 9, 2011
Thats wonderful para :)
On Sat, Jul 9, 2011 at 10:50 AM, John Hayes agressiveha...@gmail.comwrote:
refer KRit's clearly mentioned in it
On Sun, Jul 10, 2011 at 12:12 AM, parag khanna
khanna.para...@gmail.comwrote:
no text substitution occurs if the identifier is within the quotes
Thats wonderful parag :)
On Sun, Jul 10, 2011 at 4:56 AM, Yogesh Yadav medu...@gmail.com wrote:
Thats wonderful para :)
On Sat, Jul 9, 2011 at 10:50 AM, John Hayes agressiveha...@gmail.comwrote:
refer KRit's clearly mentioned in it
On Sun, Jul 10, 2011 at 12:12 AM, parag khanna
@raj :
array a[]= 2,3,5,6,7,8,10,12
diff[]= 1,2,1,1,1,2,2
maxcount=tempcount=1 // because am not takin in consideration of 0th index
value of diff[]
now in for loop
for j=1
check diff[j]==diff[j-1] //not equal
so check tempcountmaxcount or not //its also not
so maxcount remains same
i thinks mine is just O(n)
On Sun, Jul 10, 2011 at 5:31 AM, Yogesh Yadav medu...@gmail.com wrote:
@raj :
array a[]= 2,3,5,6,7,8,10,12
diff[]= 1,2,1,1,1,2,2
maxcount=tempcount=1 // because am not takin in consideration of 0th
index value of diff[]
now in for loop
for j=1
check
A[] = {1,2,3,4,5,6,8,10,12,14}
your algo will give output that there is an Longest AP of 6 elements which
is wrong
checkout this http://theory.cs.uiuc.edu/%7Ejeffe/pubs/pdf/arith.pdf for
an O(n^2) algorithm
On Sun, Jul 10, 2011 at 7:01 PM, Yogesh Yadav medu...@gmail.com wrote:
@raj :
array
A[] = {1,2,3,4,5,6,8,10,12,14}
your algo will give output that there is an Longest AP of 6 elements which
is wrong
checkout this http://theory.cs.uiuc.edu/%7Ejeffe/pubs/pdf/arith.pdf for
an O(n^2) algorithm
On Sun, Jul 10, 2011 at 7:01 PM, Yogesh Yadav medu...@gmail.com wrote:
@raj :
array
A[] = {1,2,3,4,5,6,8,10,12,14}
your algo will give output that there is an Longest AP of 6 elements which
is wrong
checkout this http://theory.cs.uiuc.edu/%7Ejeffe/pubs/pdf/arith.pdf for
an O(n^2) algorithm
On Sun, Jul 10, 2011 at 7:01 PM, Yogesh Yadav medu...@gmail.com wrote:
@raj :
array
work
On Sun, Jul 10, 2011 at 7:18 PM, Yogesh Yadav medu...@gmail.com wrote:
@sunny: my algo will give 6 as answer and i guess its right... if am
wrong plz explain where and why my logic is wrong
On Sun, Jul 10, 2011 at 5:37 AM, sunny agrawal
sunny816.i...@gmail.comwrote:
@Yogesh
your
yeah then it will be possible in O(n^2)
On Sun, Jul 10, 2011 at 5:57 AM, Yogesh Yadav medu...@gmail.com wrote:
@sunny : thanks ..i got it...
On Sun, Jul 10, 2011 at 5:55 AM, sunny agrawal sunny816.i...@gmail.comwrote:
Longest AP for that Example is of 7 elements
2,4,6,8,10,12,14
Step 1: first make a diff[][]
Step 2: search the diff in matrix
Complexity will be O(n^2)
On Sun, Jul 10, 2011 at 6:24 AM, JAIDEV YADAV jaid...@gmail.com wrote:
try Matrix search...
On Sun, Jul 10, 2011 at 7:34 PM, Yogesh Yadav medu...@gmail.com wrote:
yeah then it will be possible in O
91,110,134161 i guess
6,24,60,120210
On Sat, Jul 9, 2011 at 5:46 PM, Aman Goyal aman.goya...@gmail.com wrote:
210 for the last one you posted
On Sat, Jul 9, 2011 at 4:33 PM, amit the cool amitthecoo...@gmail.comwrote:
6,24,60,120,_
--
You received this message because you are
int count1=count2=0;
given number=n;
int m=n; //save in temp variable m
while(m!=0) //count number of bits in given number
{
m=m (m-1);
count1++;
}
m=n; //save in temp variable m
while(count1!=count2) //check no. of bits same or not
{
m=m+1;
@gopi: i didnt really understand what u want to say... what start,end and
destination denotes here??
u said it should start with 1 but in result it is starting with 9...plz
explain ur question again
On Sat, Jul 9, 2011 at 7:21 PM, Gopi kodaligopi...@gmail.com wrote:
Hi Geeks
Can anyone
size of a=m size of b =n
a[]={1,3,77,78,90} and b[]={2,5,79,81}
lr
while(l=m r=n)
{
if(b[r]a[l])
{
swap(a[l],b[r]);
l++;
sort(b[]);
}
elseif(a[l]b[r])
l++;
}
On Fri, Jul 8,
a[]={2,2,3,3,3,4,4,5,6,6,7}
size of array = n
if not sorted than sort
for(i=1;in-1;i++) //check for elements with index 1 to n-2
{
if(a[i]!=a[i-1] a[i]!=a[i+1])
print(a[i]);
}
//now check for index 0 and n-1
if(a[0]!=a[1])
print(a[0]);
if(a[n-1]!=a[n-2])
let 10 digit phone number stored in array a[]
take a array b[10] and initialise its elements with 0
int gp2[][2]; // group with 2 elements will store here
int gp3[][3]; // group with 3 elements will store here
int l=0,r=0,count=0;
int m[],n; //l,m,n is global
for(i=0;i10;i++)
++;
n--;
}
On Sat, Jul 9, 2011 at 2:27 AM, Yogesh Yadav medu...@gmail.com wrote:
let 10 digit phone number stored in array a[]
take a array b[10] and initialise its elements with 0
int gp2[][2]; // group with 2 elements will store here
int gp3[][3]; // group with 3 elements
Q10.
array a[]={19 16 18 3 50 70 12 11 14}
array b[];
sum required=30
3 11 12 14 16 18 19 50 70 //sort
0 1 2 3 4 5 6 7 8 //index
lr
count=0,i=0;
while(lr)
{
if(a[l]+a[r])sum
r--;
elseif(a[l]+a[r])sum
l++;
else
{
Q3.
start with 7000 //8 digit number
7000
7001 //as 7 is 1 time so 8th place is 1
6001 //as due to 1 on 8th place no. of zeros =6
6010 //as 6 is present so 7th place is 1 and 7 is not present so 8th
place is again changed to 0
6110 //as 1 is present on 7th place so at 2nd
69 matches
Mail list logo
