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
{
thanks yogi.
it is amzing tric and solution
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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
very nice link for questions of type Q9.
http://pw1.netcom.com/~tjensen/ptr/ch9x.htm
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Q1.
First sort both the strings using counting sort perhaps.
Then starting from the first character, search for it in the second string
using binary search.
The succeeding character of A will be searched from the index in B at
which last character of A was found
Worst case: When both the
Sol for Q4
int pallindrome(char *str)
{
static length = strlen(str);
if(length 1)
return 1;
if(str[0] == str[length-1])
{
length -= 2;
return pallindrome(str+1);
}
else
return 0;
}
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Yes I am also getting the same.
On Jul 5, 12:50 am, Ritesh Srivastava riteshkumar...@gmail.com
wrote:
For Q3 .
Sum of all the digits should be 8.
I think ,
42101000 is an answer.
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what about 7?
7 occurred once right?
On Jul 6, 12:32 am, aditya kumar aditya.kumar130...@gmail.com wrote:
in 7000 : '0' occurs seven times and rst of the numbers occur zero
times. i still dint get where i am wrong . plz explain me .
On Wed, Jul 6, 2011 at 12:47 AM, L
You have given solution for number of coins required for different sum as
explained on topcoder tutorial.
But I think the question has put forward some conditions based on which it
asks us to find the denominations of the 6 coins.
You have taken the sum given(which cannot be obtained by using
On Jul 5, 4:04 am, Dumanshu duman...@gmail.com wrote:
ans1. use counting sort for character array (0 to 255) then check for
the second string if same or not.
ans2. send 1 and 2, 1 comes back, send 10 and 5, 2 comes back, send 2
and 1
ans3. As vikas said, sum of digits should b 8. In that
For Q1:
if length1!=length2 = false
else
take 2 arrays ch1[256] and ch2[256]
For every character c in each string increment the element ch[c] in
the respective array;
now traverse both arrays together.
if each ch1[] element =each ch2[] element = true
for Q5 change the expr into postfix and then build expression tree...
but is expression tree same as parse tree??
correct me if i m wrong!!
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Anyone for Q10, i wonder if it can really be solved in O(n). Very
obvious O(nlogn) is what I know
On Thu, Jul 7, 2011 at 1:25 AM, KK kunalkapadi...@gmail.com wrote:
for Q5 change the expr into postfix and then build expression tree...
but is expression tree same as parse tree??
correct me if i
My program for Q4.
// recursively find if a given string is palindrome
bool IsPalindrome(string s, int start, int start2, bool flag)
{
bool flag1 = flag;
if (start = 0 start2 (s.Length))
{
char c1 = s[start];
Q8 is a DP problem, here is the solution
#include stdio.h
#define MAX 125
#define VAL 6
int Min[MAX];
int N[VAL] = {5, 10, 20, 25, 50, 100};
int main(void)
{
int i, j;
for (i = 1; i MAX; i++)
Min[i] = 100;
for (i = 5; i MAX; i += 5)
for (j = 0; j VAL; j++)
don't give programs.. discuss the logic behind that, algorithms are
important..
On Tue, Jul 5, 2011 at 12:43 PM, Azhar Hussain azhar...@gmail.com wrote:
Q8 is a DP problem, here is the solution
#include stdio.h
#define MAX 125
#define VAL 6
int Min[MAX];
int N[VAL] = {5, 10, 20,
Q-8: 50,25,10,10,10,10
Q2-: Let A,B,C and D are the persons suct tat A takes 1min, B-2, C-5 and
D-10
* A and B go across 2 minutes
A goes back with light 1 minute
C and D go across 10 minutes
B goes back with light 2 minute
A and
q2.: Basically u must put 10min and 5 min together to save time. also,
during return path-u must have some1 with less time already there.
that's it !
On Tue, Jul 5, 2011 at 1:00 PM, udit sharma sharmaudit...@gmail.com wrote:
Q-8: 50,25,10,10,10,10
Q2-: Let A,B,C and D are the persons
I think that for Q3 answer will be unique and will be same as pointed
out by Ritesh.
On Jul 5, 1:11 pm, vikas mehta...@gmail.com wrote:
Sum of all the digits should be 8
But it was not said by the interviewer i think we must print all possible
combinations for the question
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You received
ans1. use counting sort for character array (0 to 255) then check for
the second string if same or not.
ans2. send 1 and 2, 1 comes back, send 10 and 5, 2 comes back, send 2
and 1
ans3. As vikas said, sum of digits should b 8. In that case the number
must have a zero digit because 1st digit the
@Azhar: could u plz explain the q8. problem.
On Jul 5, 12:13 pm, Azhar Hussain azhar...@gmail.com wrote:
Q8 is a DP problem, here is the solution
#include stdio.h
#define MAX 125
#define VAL 6
int Min[MAX];
int N[VAL] = {5, 10, 20, 25, 50, 100};
int main(void)
{
int i, j;
well for question 6 I think calculating size of string will count as one
traversal?
Correct me if I am wrong?
My approach:
traverse storing each char in a string.when space encountered push the
string in stack
I am not sure how my solution is but it doesnt appears gud ryt now.
On Tue, Jul 5,
@Saurabh: u can improve upon ur solution.
char * outstr;
int c =0;
void getreversedwords(string str)
{
index = str.length()-1;
//outstr is the answer
stack st;
while(index)
{
st.push(str[index]);
if(st.top() == ' ') //top is a space
st.printstack();
index--;
}
if(st.isEmpty()==false)
I mean is the solution valid?
Because pushing will take one traversal and then popping another?
On Tue, Jul 5, 2011 at 4:58 PM, Dumanshu duman...@gmail.com wrote:
@Saurabh: u can improve upon ur solution.
char * outstr;
int c =0;
void getreversedwords(string str)
{
index = str.length()-1;
The answer to question 3 can also be
6010
right?
so no unique solution?
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@anonymous: nopes , it's wrong1 is one time there so that should
be their at index 1..
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For Q4: I think this is the optimal code
int recurPalin(char *start, char *end)
{
if (end start)
return true;
if (*start != *end)
return false;
return recurPalin(start+1, end-1);
}
-
Azhar.
On Tue, Jul 5, 2011 at 12:21 PM, vikas mehta...@gmail.com wrote:
My
boolean palindromeCheck(String string)
{
len=string.length();
if((string.length()1))
{
if((string.charAt(0)==string.charAt(len-1)))
{
str=;
str=str+string.substring(1,(len-1));
palindromeCheck(str);
}
else
{
flag=false;
}
}
return flag;
}
THis also works fine if you dont want to use pointer
Q3. ans:7000 i guess this is also a correct answer and no unique soln as
such
On Wed, Jul 6, 2011 at 12:37 AM, aditya kumar
aditya.kumar130...@gmail.comwrote:
boolean palindromeCheck(String string)
{
len=string.length();
if((string.length()1))
{
@aditya : I am wondering how many times 7 has occurred. Is it 1? Or is
it 0?
Please take a moment before posting your solution, and think whether
it is write or wrong!
On Jul 6, 12:11 am, aditya kumar aditya.kumar130...@gmail.com wrote:
Q3. ans:7000 i guess this is also a correct answer and
in 7000 : '0' occurs seven times and rst of the numbers occur zero
times. i still dint get where i am wrong . plz explain me .
On Wed, Jul 6, 2011 at 12:47 AM, L prnk.bhatna...@gmail.com wrote:
@aditya : I am wondering how many times 7 has occurred. Is it 1? Or is
it 0?
Please take a
7 occurs 1 time...
so 8th digit is 1
On Wed, Jul 6, 2011 at 1:02 AM, aditya kumar
aditya.kumar130...@gmail.comwrote:
in 7000 : '0' occurs seven times and rst of the numbers occur zero
times. i still dint get where i am wrong . plz explain me .
On Wed, Jul 6, 2011 at 12:47 AM, L
But '7' has also appeared One time, you need to take that into account.
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@piyush : thanks i thot most significant bit was take as fisrt figure .
On Wed, Jul 6, 2011 at 1:03 AM, Piyush Kapoor pkjee2...@gmail.com wrote:
7 occurs 1 time...
so 8th digit is 1
On Wed, Jul 6, 2011 at 1:02 AM, aditya kumar aditya.kumar130...@gmail.com
wrote:
in 7000 : '0' occurs
no 7000 is not possible because, '7' is also there so you have to
mention at the index of '7' that it is there 1 time..
On Jul 6, 12:32 am, aditya kumar aditya.kumar130...@gmail.com wrote:
in 7000 : '0' occurs seven times and rst of the numbers occur zero
times. i still dint get where i
@ashish thanx buddy
nice solution
On Jul 6, 7:20 am, Ashish Goel ashg...@gmail.com wrote:
Q3 : 42101000
started with 7000
then changes this to
6010
51000100 to
42101000 to
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
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