Re: [algogeeks] Re: Number problem
@yellow
your code turns 1000,100,10,2270,12130 to 1,1,1,27,123 repectively
simply it removes all trailing zeros ...
On Wed, Sep 22, 2010 at 8:10 PM, Yellow Sapphire pukhraj7...@gmail.comwrote:
My Solution.
Almost same as above but the flag is set by using bits.
#define SETFLAG(flag, pos) (flag=flag | (1pos-1))
#define IS_FLAG_SET(flag,pos) (flag (1pos-1))
int remove_dup(int n)
{
int temp=0, temp2=0;
unsigned int flag=0;
/*
* Reverse the number
*/
while (n0 (temp=temp*10 + n%10), n=n/10);
/*
* Find duplicates by setting the bits in the flag
*/
n=temp;
temp=0;
while (n0) {
temp2=n%10;
if(!IS_FLAG_SET(flag, temp2)) {
temp=temp*10 + temp2;
SETFLAG(flag, temp2);
}
n=n/10;
}
return temp;
}
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Re: [algogeeks] Re: Number problem
Here's the code
int rem_dup(int n)
{
int *a,i,c,t,d,count[10]={0},arr[20],j=0,sum=0;
a=(int *)malloc(sizeof(a));
for(i=0;n0;i++)//storing each digit of number in an array
{
a[i]=n%10;
n=n/10;
}
c=d=i-1;
i=0;
while(ic)//reversing the array
{
t=a[i];
a[i++]=a[c];
a[c--]=t;
}
for(i=0;i=d;i++)//setting repeated digits to -1
{if(count[a[i]]==0)
count[a[i]]++;
else
a[i]=-1;
}
for(i=0,j=0;i=d;i++) //storing all non repeating digits in anothor
array arr[]
{
if(a[i]!=-1)
{
arr[j]=a[i];
j++;
}
}
for(i=0;ij;i++)
sum=sum+arr[i]*power(10,j-i-1);
return sum;
}
On Thu, Sep 23, 2010 at 12:39 AM, Dave dave_and_da...@juno.com wrote:
@Saurabh: Doesn't this turn 10 into 1? You need to count the digits in
the number as you are reversing it, and replace the second while loop
with a for loop with that many iterations.
Dave
On Sep 21, 11:28 pm, saurabh agrawal saurabh...@gmail.com wrote:
@dave:
your code is producing 4526 for input=24526 instead of 2456
Here's corrected code :)
/CODE///
int function(int n){
int a[10]={0},temp=0,result =0;
while(n){ //reverse the number..
temp=10*temp+n%10;
n/=10;
}
n=temp;
while(n){ ///remove duplicate digits...
if(a[n%10]==0){
a[n%10]=1;
result=10*result+n%10;
}
n/=10;
}
return result;}
///END/
On Wed, Sep 22, 2010 at 8:51 AM, Dave dave_and_da...@juno.com wrote:
@Saurabh: Doesn't your code turn 123 into 321? Try this:
int function(int n)
{
int a[10]={0};
int result=0;
int place=1;
while(n){
if(a[n%10]==0){
a[n%10]=1;
result+=(n%10)*place;
place*=10;
}
n/=10;
}
return result;
}
Dave
On Sep 21, 3:12 pm, saurabh agrawal saurabh...@gmail.com wrote:
int function(int n){
int a[10]={0};
int result =0;
while(n){
if(a[n%10]==0){
a[n%10]=1;
result=10*result+n%10;
}
n/=10;}
return result;`
}
On Wed, Sep 22, 2010 at 12:39 AM, Albert alberttheb...@gmail.com
wrote:
Given a number find the number by eliminating the duplicate digits
in
the number..
for eg: 24526 ans is 2456
.
int function(int n)
{
.
.
.
}
Give all sort of solutions to this problem.
Efficiency in the code is important
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Re: [algogeeks] Re: Number problem
My code failed because it was reversing the digits and thus 0 was getting added in the front which resulted in nothing. If allowed we can use a char array else will have to find a solution which does not reverse the digits of the number. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Number problem
My solution.
int f(int n) {
if (n = 0) return n;
int digits = (int)log10(n) + 1;
int m = 0;
int flag[10] = { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 };
for (int i = digits - 1; i = 0; --i) {
int p = (int)pow(10.0, i);
int d = n / p;
n %= p;
if (flag[d] 0) continue;
flag[d] = 1;
m = 10 * m + d;
}
return m;
}
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[algogeeks] Re: Number problem
@Saurabh: Doesn't this turn 10 into 1? You need to count the digits in
the number as you are reversing it, and replace the second while loop
with a for loop with that many iterations.
Dave
On Sep 21, 11:28 pm, saurabh agrawal saurabh...@gmail.com wrote:
@dave:
your code is producing 4526 for input=24526 instead of 2456
Here's corrected code :)
/CODE///
int function(int n){
int a[10]={0},temp=0,result =0;
while(n){ //reverse the number..
temp=10*temp+n%10;
n/=10;
}
n=temp;
while(n){ ///remove duplicate digits...
if(a[n%10]==0){
a[n%10]=1;
result=10*result+n%10;
}
n/=10;
}
return result;}
///END/
On Wed, Sep 22, 2010 at 8:51 AM, Dave dave_and_da...@juno.com wrote:
@Saurabh: Doesn't your code turn 123 into 321? Try this:
int function(int n)
{
int a[10]={0};
int result=0;
int place=1;
while(n){
if(a[n%10]==0){
a[n%10]=1;
result+=(n%10)*place;
place*=10;
}
n/=10;
}
return result;
}
Dave
On Sep 21, 3:12 pm, saurabh agrawal saurabh...@gmail.com wrote:
int function(int n){
int a[10]={0};
int result =0;
while(n){
if(a[n%10]==0){
a[n%10]=1;
result=10*result+n%10;
}
n/=10;}
return result;`
}
On Wed, Sep 22, 2010 at 12:39 AM, Albert alberttheb...@gmail.com
wrote:
Given a number find the number by eliminating the duplicate digits in
the number..
for eg: 24526 ans is 2456
.
int function(int n)
{
.
.
.
}
Give all sort of solutions to this problem.
Efficiency in the code is important
--
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[algogeeks] Re: Number problem
@Saurabh: Doesn't your code turn 123 into 321? Try this:
int function(int n)
{
int a[10]={0};
int result=0;
int place=1;
while(n){
if(a[n%10]==0){
a[n%10]=1;
result+=(n%10)*place;
place*=10;
}
n/=10;
}
return result;
}
Dave
On Sep 21, 3:12 pm, saurabh agrawal saurabh...@gmail.com wrote:
int function(int n){
int a[10]={0};
int result =0;
while(n){
if(a[n%10]==0){
a[n%10]=1;
result=10*result+n%10;
}
n/=10;}
return result;`
}
On Wed, Sep 22, 2010 at 12:39 AM, Albert alberttheb...@gmail.com wrote:
Given a number find the number by eliminating the duplicate digits in
the number..
for eg: 24526 ans is 2456
.
int function(int n)
{
.
.
.
}
Give all sort of solutions to this problem.
Efficiency in the code is important
--
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Re: [algogeeks] Re: Number problem
@dave:
your code is producing 4526 for input=24526 instead of 2456
Here's corrected code :)
/CODE///
int function(int n){
int a[10]={0},temp=0,result =0;
while(n){ //reverse the number..
temp=10*temp+n%10;
n/=10;
}
n=temp;
while(n){ ///remove duplicate digits...
if(a[n%10]==0){
a[n%10]=1;
result=10*result+n%10;
}
n/=10;
}
return result;
}
///END/
On Wed, Sep 22, 2010 at 8:51 AM, Dave dave_and_da...@juno.com wrote:
@Saurabh: Doesn't your code turn 123 into 321? Try this:
int function(int n)
{
int a[10]={0};
int result=0;
int place=1;
while(n){
if(a[n%10]==0){
a[n%10]=1;
result+=(n%10)*place;
place*=10;
}
n/=10;
}
return result;
}
Dave
On Sep 21, 3:12 pm, saurabh agrawal saurabh...@gmail.com wrote:
int function(int n){
int a[10]={0};
int result =0;
while(n){
if(a[n%10]==0){
a[n%10]=1;
result=10*result+n%10;
}
n/=10;}
return result;`
}
On Wed, Sep 22, 2010 at 12:39 AM, Albert alberttheb...@gmail.com
wrote:
Given a number find the number by eliminating the duplicate digits in
the number..
for eg: 24526 ans is 2456
.
int function(int n)
{
.
.
.
}
Give all sort of solutions to this problem.
Efficiency in the code is important
--
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