Re: [algogeeks] Re: character count in array
@mohit: that will modify the original array
and also time =O(nlogn)...
On Mon, Sep 5, 2011 at 1:01 AM, Ankuj Gupta ankuj2...@gmail.com wrote:
@mohit: that will modify the original array
On Sep 4, 6:40 pm, sarath prasath prasathsar...@gmail.com wrote:
here is my approach
where i left the non repeating characters as it is and done some good
code..
char * runlengthencode(char* str,int size)
{
int i,j,flag=0;
for(i=0,j=1;str[i]str[j]jsize;i++,j++)
{
while(str[i]==str[j])
{
j++;
flag=1;
}
if(flag)
{
j=j-1;
str[i+1]=48+(j-i+1);
flag=0;
i=j;
j++;
}
}
return str;
}
On Sat, Sep 3, 2011 at 6:54 PM, Aman Kumar amanas...@gmail.com wrote:
Hiii
if array is given like this
arr[]=aabcabbcdeadef
convert this array into like
arr[]=a4b3c2d2e2f1
how can we do this
can we do it with space complexity O(1).
reply asap
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Re: [algogeeks] Re: character count in array
+1 rahul and +1 ankuj
On Sat, Sep 3, 2011 at 11:37 AM, icy` vipe...@gmail.com wrote:
Just use a hash to count frequency of something; e.g. in ruby:
ar= %w(a a b c a b b c d e a d e f)
freq=Hash.new(0)
ar.each {|c| freq[c]+=1}
p freq
#output
#{a=4, b=3, c=2, d=2, e=2, f=1}
you could only do it in place in O(1) only if your input array is
already 2*(number of all possible characters) size, but you didnt
mention size of input array. For example, what if input was just
[a,b,c,d,e] ? The size is 5.You cant just convert the input
into [a,1,b,1,c,1,d,1,e,1] without increasing the size to 10. So
hash is the better method.
On Sep 3, 11:04 am, Ankuj Gupta ankuj2...@gmail.com wrote:
if for all inputs you array remains of same size we can take it as
constant space
On Sep 3, 7:49 pm, rajul jain rajuljain...@gmail.com wrote:
@ankuj just want to clarify that in hashing method we require array of
fixed size let say arr[26] , so is it considered as constant space or
not?
On Sat, Sep 3, 2011 at 8:02 PM, siddharam suresh
siddharam@gmail.comwrote:
sol already posted please search old thread
Thank you,
Sid.
On Sat, Sep 3, 2011 at 8:01 PM, Ankuj Gupta ankuj2...@gmail.com
wrote:
If we take our input to be characters a-z ans A-Z then we require
fixed space which can be treated as O(1).
On Sep 3, 7:10 pm, teja bala pawanjalsa.t...@gmail.com wrote:
this 'll work if u i/p the string in dis manner
aaabbcc(consecutive same)
a3b2c2
#includestdio.h
#includeconio.h
main()
{
int x=1,i=0;
char a[100];
gets(a);
while(a[i]!='\0')
{
while(a[i]==a[i+1])
{
x++;
i++;
}
printf(%c%d,a[i],x);
x=1;
i++;
}
getchar();
}
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Re: [algogeeks] Re: character count in array
hey guys
why hashing?
can't we simply sort the array of characters in-place - space complexity
O(1) .
and then simply rearrange the array in-place with character+frequency.
On Sun, Sep 4, 2011 at 3:13 PM, bharatkumar bagana
bagana.bharatku...@gmail.com wrote:
+1 rahul and +1 ankuj
On Sat, Sep 3, 2011 at 11:37 AM, icy` vipe...@gmail.com wrote:
Just use a hash to count frequency of something; e.g. in ruby:
ar= %w(a a b c a b b c d e a d e f)
freq=Hash.new(0)
ar.each {|c| freq[c]+=1}
p freq
#output
#{a=4, b=3, c=2, d=2, e=2, f=1}
you could only do it in place in O(1) only if your input array is
already 2*(number of all possible characters) size, but you didnt
mention size of input array. For example, what if input was just
[a,b,c,d,e] ? The size is 5.You cant just convert the input
into [a,1,b,1,c,1,d,1,e,1] without increasing the size to 10. So
hash is the better method.
On Sep 3, 11:04 am, Ankuj Gupta ankuj2...@gmail.com wrote:
if for all inputs you array remains of same size we can take it as
constant space
On Sep 3, 7:49 pm, rajul jain rajuljain...@gmail.com wrote:
@ankuj just want to clarify that in hashing method we require array
of
fixed size let say arr[26] , so is it considered as constant space or
not?
On Sat, Sep 3, 2011 at 8:02 PM, siddharam suresh
siddharam@gmail.comwrote:
sol already posted please search old thread
Thank you,
Sid.
On Sat, Sep 3, 2011 at 8:01 PM, Ankuj Gupta ankuj2...@gmail.com
wrote:
If we take our input to be characters a-z ans A-Z then we require
fixed space which can be treated as O(1).
On Sep 3, 7:10 pm, teja bala pawanjalsa.t...@gmail.com wrote:
this 'll work if u i/p the string in dis manner
aaabbcc(consecutive same)
a3b2c2
#includestdio.h
#includeconio.h
main()
{
int x=1,i=0;
char a[100];
gets(a);
while(a[i]!='\0')
{
while(a[i]==a[i+1])
{
x++;
i++;
}
printf(%c%d,a[i],x);
x=1;
i++;
}
getchar();
}
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Re: [algogeeks] Re: character count in array
@mohit :good one
On Sun, Sep 4, 2011 at 4:34 PM, mohit verma mohit89m...@gmail.com wrote:
hey guys
why hashing?
can't we simply sort the array of characters in-place - space complexity
O(1) .
and then simply rearrange the array in-place with character+frequency.
On Sun, Sep 4, 2011 at 3:13 PM, bharatkumar bagana
bagana.bharatku...@gmail.com wrote:
+1 rahul and +1 ankuj
On Sat, Sep 3, 2011 at 11:37 AM, icy` vipe...@gmail.com wrote:
Just use a hash to count frequency of something; e.g. in ruby:
ar= %w(a a b c a b b c d e a d e f)
freq=Hash.new(0)
ar.each {|c| freq[c]+=1}
p freq
#output
#{a=4, b=3, c=2, d=2, e=2, f=1}
you could only do it in place in O(1) only if your input array is
already 2*(number of all possible characters) size, but you didnt
mention size of input array. For example, what if input was just
[a,b,c,d,e] ? The size is 5.You cant just convert the input
into [a,1,b,1,c,1,d,1,e,1] without increasing the size to 10. So
hash is the better method.
On Sep 3, 11:04 am, Ankuj Gupta ankuj2...@gmail.com wrote:
if for all inputs you array remains of same size we can take it as
constant space
On Sep 3, 7:49 pm, rajul jain rajuljain...@gmail.com wrote:
@ankuj just want to clarify that in hashing method we require array
of
fixed size let say arr[26] , so is it considered as constant space or
not?
On Sat, Sep 3, 2011 at 8:02 PM, siddharam suresh
siddharam@gmail.comwrote:
sol already posted please search old thread
Thank you,
Sid.
On Sat, Sep 3, 2011 at 8:01 PM, Ankuj Gupta ankuj2...@gmail.com
wrote:
If we take our input to be characters a-z ans A-Z then we require
fixed space which can be treated as O(1).
On Sep 3, 7:10 pm, teja bala pawanjalsa.t...@gmail.com wrote:
this 'll work if u i/p the string in dis manner
aaabbcc(consecutive same)
a3b2c2
#includestdio.h
#includeconio.h
main()
{
int x=1,i=0;
char a[100];
gets(a);
while(a[i]!='\0')
{
while(a[i]==a[i+1])
{
x++;
i++;
}
printf(%c%d,a[i],x);
x=1;
i++;
}
getchar();
}
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*
Mobile +91 8056127652*
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[algogeeks] Re: character count in array
@mohit: that will modify the original array
On Sep 4, 6:40 pm, sarath prasath prasathsar...@gmail.com wrote:
here is my approach
where i left the non repeating characters as it is and done some good code..
char * runlengthencode(char* str,int size)
{
int i,j,flag=0;
for(i=0,j=1;str[i]str[j]jsize;i++,j++)
{
while(str[i]==str[j])
{
j++;
flag=1;
}
if(flag)
{
j=j-1;
str[i+1]=48+(j-i+1);
flag=0;
i=j;
j++;
}
}
return str;
}
On Sat, Sep 3, 2011 at 6:54 PM, Aman Kumar amanas...@gmail.com wrote:
Hiii
if array is given like this
arr[]=aabcabbcdeadef
convert this array into like
arr[]=a4b3c2d2e2f1
how can we do this
can we do it with space complexity O(1).
reply asap
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[algogeeks] Re: character count in array
If we take our input to be characters a-z ans A-Z then we require
fixed space which can be treated as O(1).
On Sep 3, 7:10 pm, teja bala pawanjalsa.t...@gmail.com wrote:
this 'll work if u i/p the string in dis manner
aaabbcc(consecutive same)
a3b2c2
#includestdio.h
#includeconio.h
main()
{
int x=1,i=0;
char a[100];
gets(a);
while(a[i]!='\0')
{
while(a[i]==a[i+1])
{
x++;
i++;
}
printf(%c%d,a[i],x);
x=1;
i++;
}
getchar();
}
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Re: [algogeeks] Re: character count in array
sol already posted please search old thread
Thank you,
Sid.
On Sat, Sep 3, 2011 at 8:01 PM, Ankuj Gupta ankuj2...@gmail.com wrote:
If we take our input to be characters a-z ans A-Z then we require
fixed space which can be treated as O(1).
On Sep 3, 7:10 pm, teja bala pawanjalsa.t...@gmail.com wrote:
this 'll work if u i/p the string in dis manner
aaabbcc(consecutive same)
a3b2c2
#includestdio.h
#includeconio.h
main()
{
int x=1,i=0;
char a[100];
gets(a);
while(a[i]!='\0')
{
while(a[i]==a[i+1])
{
x++;
i++;
}
printf(%c%d,a[i],x);
x=1;
i++;
}
getchar();
}
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Re: [algogeeks] Re: character count in array
@ankuj just want to clarify that in hashing method we require array of
fixed size let say arr[26] , so is it considered as constant space or not?
On Sat, Sep 3, 2011 at 8:02 PM, siddharam suresh siddharam@gmail.comwrote:
sol already posted please search old thread
Thank you,
Sid.
On Sat, Sep 3, 2011 at 8:01 PM, Ankuj Gupta ankuj2...@gmail.com wrote:
If we take our input to be characters a-z ans A-Z then we require
fixed space which can be treated as O(1).
On Sep 3, 7:10 pm, teja bala pawanjalsa.t...@gmail.com wrote:
this 'll work if u i/p the string in dis manner
aaabbcc(consecutive same)
a3b2c2
#includestdio.h
#includeconio.h
main()
{
int x=1,i=0;
char a[100];
gets(a);
while(a[i]!='\0')
{
while(a[i]==a[i+1])
{
x++;
i++;
}
printf(%c%d,a[i],x);
x=1;
i++;
}
getchar();
}
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[algogeeks] Re: character count in array
if for all inputs you array remains of same size we can take it as
constant space
On Sep 3, 7:49 pm, rajul jain rajuljain...@gmail.com wrote:
@ankuj just want to clarify that in hashing method we require array of
fixed size let say arr[26] , so is it considered as constant space or not?
On Sat, Sep 3, 2011 at 8:02 PM, siddharam suresh
siddharam@gmail.comwrote:
sol already posted please search old thread
Thank you,
Sid.
On Sat, Sep 3, 2011 at 8:01 PM, Ankuj Gupta ankuj2...@gmail.com wrote:
If we take our input to be characters a-z ans A-Z then we require
fixed space which can be treated as O(1).
On Sep 3, 7:10 pm, teja bala pawanjalsa.t...@gmail.com wrote:
this 'll work if u i/p the string in dis manner
aaabbcc(consecutive same)
a3b2c2
#includestdio.h
#includeconio.h
main()
{
int x=1,i=0;
char a[100];
gets(a);
while(a[i]!='\0')
{
while(a[i]==a[i+1])
{
x++;
i++;
}
printf(%c%d,a[i],x);
x=1;
i++;
}
getchar();
}
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[algogeeks] Re: character count in array
Just use a hash to count frequency of something; e.g. in ruby:
ar= %w(a a b c a b b c d e a d e f)
freq=Hash.new(0)
ar.each {|c| freq[c]+=1}
p freq
#output
#{a=4, b=3, c=2, d=2, e=2, f=1}
you could only do it in place in O(1) only if your input array is
already 2*(number of all possible characters) size, but you didnt
mention size of input array. For example, what if input was just
[a,b,c,d,e] ? The size is 5.You cant just convert the input
into [a,1,b,1,c,1,d,1,e,1] without increasing the size to 10. So
hash is the better method.
On Sep 3, 11:04 am, Ankuj Gupta ankuj2...@gmail.com wrote:
if for all inputs you array remains of same size we can take it as
constant space
On Sep 3, 7:49 pm, rajul jain rajuljain...@gmail.com wrote:
@ankuj just want to clarify that in hashing method we require array of
fixed size let say arr[26] , so is it considered as constant space or not?
On Sat, Sep 3, 2011 at 8:02 PM, siddharam suresh
siddharam@gmail.comwrote:
sol already posted please search old thread
Thank you,
Sid.
On Sat, Sep 3, 2011 at 8:01 PM, Ankuj Gupta ankuj2...@gmail.com wrote:
If we take our input to be characters a-z ans A-Z then we require
fixed space which can be treated as O(1).
On Sep 3, 7:10 pm, teja bala pawanjalsa.t...@gmail.com wrote:
this 'll work if u i/p the string in dis manner
aaabbcc(consecutive same)
a3b2c2
#includestdio.h
#includeconio.h
main()
{
int x=1,i=0;
char a[100];
gets(a);
while(a[i]!='\0')
{
while(a[i]==a[i+1])
{
x++;
i++;
}
printf(%c%d,a[i],x);
x=1;
i++;
}
getchar();
}
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