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$begingroup$
Today I've found /r/dailyprogrammer and I've solved an easy challenge. I'm new to coding and I'm not sure if this is good way to solve this kind of problems. Could you, please, give me some hints? How to make my code more clear and readable? Thanks!
Challange description:
A number is input in computer then a new no should get printed by
adding one to each of its digit. If you encounter a 9, insert a 10
(don't carry over, just shift things around).
For example, 998 becomes 10109.
Bonus
This challenge is trivial to do if you map it to a string to iterate
over the input, operate, and then cast it back. Instead, try doing it
without casting it as a string at any point, keep it numeric (int,
float if you need it) only.
private static int Challange(int number)
{
int digits = (int)Math.Log10(number); //this is number of digits -1
int result = 0;
for (int i=0; i <= digits; i++)
{
int tens = (int)(Math.Pow(10, digits - i));
int currentDigit = ((number / tens) % 10);
if (currentDigit == 9) result *= 10;
result += (currentDigit + 1) * tens;
}
return result;
}
c# programming-challenge
edited 2 days ago
t3chb0t
34.7k751121
asked 2 days ago
nowakasdnowakasd
434
New contributor
nowakasd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Today I've found /r/dailyprogrammer and I've solved an easy challenge. I'm new to coding and I'm not sure if this is good way to solve this kind of problems. Could you, please, give me some hints? How to make my code more clear and readable? Thanks!
Challange description:
A number is input in computer then a new no should get printed by
adding one to each of its digit. If you encounter a 9, insert a 10
(don't carry over, just shift things around).
For example, 998 becomes 10109.
Bonus
This challenge is trivial to do if you map it to a string to iterate
over the input, operate, and then cast it back. Instead, try doing it
without casting it as a string at any point, keep it numeric (int,
float if you need it) only.
private static int Challange(int number)
{
int digits = (int)Math.Log10(number); //this is number of digits -1
int result = 0;
for (int i=0; i <= digits; i++)
{
int tens = (int)(Math.Pow(10, digits - i));
int currentDigit = ((number / tens) % 10);
if (currentDigit == 9) result *= 10;
result += (currentDigit + 1) * tens;
}
return result;
}
c# programming-challenge
edited 2 days ago
t3chb0t
34.7k751121
asked 2 days ago
nowakasdnowakasd
434
New contributor
nowakasd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Is 0 a valid input? If so, what should the result be?
$endgroup$
– YawarRaza7349
yesterday
add a comment |
$begingroup$
Today I've found /r/dailyprogrammer and I've solved an easy challenge. I'm new to coding and I'm not sure if this is good way to solve this kind of problems. Could you, please, give me some hints? How to make my code more clear and readable? Thanks!
Challange description:
A number is input in computer then a new no should get printed by
adding one to each of its digit. If you encounter a 9, insert a 10
(don't carry over, just shift things around).
For example, 998 becomes 10109.
Bonus
This challenge is trivial to do if you map it to a string to iterate
over the input, operate, and then cast it back. Instead, try doing it
without casting it as a string at any point, keep it numeric (int,
float if you need it) only.
private static int Challange(int number)
{
int digits = (int)Math.Log10(number); //this is number of digits -1
int result = 0;
for (int i=0; i <= digits; i++)
{
int tens = (int)(Math.Pow(10, digits - i));
int currentDigit = ((number / tens) % 10);
if (currentDigit == 9) result *= 10;
result += (currentDigit + 1) * tens;
}
return result;
}
c# programming-challenge
edited 2 days ago
t3chb0t
34.7k751121
asked 2 days ago
nowakasdnowakasd
434
New contributor
nowakasd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Today I've found /r/dailyprogrammer and I've solved an easy challenge. I'm new to coding and I'm not sure if this is good way to solve this kind of problems. Could you, please, give me some hints? How to make my code more clear and readable? Thanks!
Challange description:
A number is input in computer then a new no should get printed by
adding one to each of its digit. If you encounter a 9, insert a 10
(don't carry over, just shift things around).
For example, 998 becomes 10109.
Bonus
This challenge is trivial to do if you map it to a string to iterate
over the input, operate, and then cast it back. Instead, try doing it
without casting it as a string at any point, keep it numeric (int,
float if you need it) only.
private static int Challange(int number)
{
int digits = (int)Math.Log10(number); //this is number of digits -1
int result = 0;
for (int i=0; i <= digits; i++)
{
int tens = (int)(Math.Pow(10, digits - i));
int currentDigit = ((number / tens) % 10);
if (currentDigit == 9) result *= 10;
result += (currentDigit + 1) * tens;
}
return result;
}
c# programming-challenge
c# programming-challenge
edited 2 days ago
t3chb0t
34.7k751121
asked 2 days ago
nowakasdnowakasd
434
New contributor
nowakasd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
t3chb0t
34.7k751121
asked 2 days ago
nowakasdnowakasd
434
New contributor
nowakasd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
t3chb0t
34.7k751121
edited 2 days ago
t3chb0t
34.7k751121
edited 2 days ago
t3chb0t
34.7k751121
34.7k751121
asked 2 days ago
nowakasdnowakasd
434
New contributor
nowakasd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
nowakasdnowakasd
434
asked 2 days ago
nowakasdnowakasd
434
434
New contributor
nowakasd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
nowakasd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
nowakasd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Is 0 a valid input? If so, what should the result be?
$endgroup$
– YawarRaza7349
yesterday
add a comment |
2
$begingroup$
Is 0 a valid input? If so, what should the result be?
$endgroup$
– YawarRaza7349
yesterday
2
2
$begingroup$
Is 0 a valid input? If so, what should the result be?
$endgroup$
– YawarRaza7349
yesterday
$begingroup$
Is 0 a valid input? If so, what should the result be?
$endgroup$
– YawarRaza7349
yesterday
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It's a matter of taste but I find a recursive solution to be reasonably readable :
private static int IncrementDigits(int number){
if (number == 0){
return 0;
}
int lastDigit = number % 10;
int firstDigits = number / 10;
if (lastDigit == 9){
return 10 + 100 * IncrementDigits(firstDigits);
} else {
return lastDigit + 1 + 10 * IncrementDigits(firstDigits);
}
}
Note that it only works for number > 0.
Here's a working example: https://dotnetfiddle.net/xUKs6M
answered yesterday
Eric DuminilEric Duminil
2,0611613
$endgroup$
2
$begingroup$
You could useSystem.Math.DivRemto do the division and modulo operations (on .Net Core this would be faster, the .Net Framework implementation does the naïve thing of doing the division twice)
$endgroup$
– mistertribs
yesterday
$begingroup$
@mistertribs: Thanks. I wanted to stay close enough to OP's answer. I don't really like the output argument inMath.DivRem, to be honest.
$endgroup$
– Eric Duminil
yesterday
add a comment |
$begingroup$
It works, but it's a bit tricky to follow the logic and work out why. If you work from the other end then you can avoid the Log and Pow and shifting the partial result, and I find that easier to read.
Challange is not a descriptive name. What does the method do? IncrementDigits might be a better name, for example.
answered 2 days ago
Peter TaylorPeter Taylor
17.4k2862
$endgroup$
add a comment |
$begingroup$
Computing powers repeatedly and log are not cheap operations.
You could substitute them with cheaper alternatives:
- Instead of computing powers of
tens, you could start with1and multiply in the loop steps. - Instead of computing the number of digits and using a counting loop, you could divide by 10 until reaching 0.
Something like this:
if (number == 0) {
return 1;
}
int work = number;
int tens = 1;
int result = 0;
while (work > 0) {
int digit = work % 10;
work /= 10;
result += tens * (digit + 1);
tens *= 10;
if (digit == 9) {
tens *= 10;
}
}
return result;
answered yesterday
janosjanos
98.1k12125350
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's a matter of taste but I find a recursive solution to be reasonably readable :
private static int IncrementDigits(int number){
if (number == 0){
return 0;
}
int lastDigit = number % 10;
int firstDigits = number / 10;
if (lastDigit == 9){
return 10 + 100 * IncrementDigits(firstDigits);
} else {
return lastDigit + 1 + 10 * IncrementDigits(firstDigits);
}
}
Note that it only works for number > 0.
Here's a working example: https://dotnetfiddle.net/xUKs6M
answered yesterday
Eric DuminilEric Duminil
2,0611613
$endgroup$
2
$begingroup$
You could useSystem.Math.DivRemto do the division and modulo operations (on .Net Core this would be faster, the .Net Framework implementation does the naïve thing of doing the division twice)
$endgroup$
– mistertribs
yesterday
$begingroup$
@mistertribs: Thanks. I wanted to stay close enough to OP's answer. I don't really like the output argument inMath.DivRem, to be honest.
$endgroup$
– Eric Duminil
yesterday
add a comment |
$begingroup$
It's a matter of taste but I find a recursive solution to be reasonably readable :
private static int IncrementDigits(int number){
if (number == 0){
return 0;
}
int lastDigit = number % 10;
int firstDigits = number / 10;
if (lastDigit == 9){
return 10 + 100 * IncrementDigits(firstDigits);
} else {
return lastDigit + 1 + 10 * IncrementDigits(firstDigits);
}
}
Note that it only works for number > 0.
Here's a working example: https://dotnetfiddle.net/xUKs6M
answered yesterday
Eric DuminilEric Duminil
2,0611613
$endgroup$
2
$begingroup$
You could useSystem.Math.DivRemto do the division and modulo operations (on .Net Core this would be faster, the .Net Framework implementation does the naïve thing of doing the division twice)
$endgroup$
– mistertribs
yesterday
$begingroup$
@mistertribs: Thanks. I wanted to stay close enough to OP's answer. I don't really like the output argument inMath.DivRem, to be honest.
$endgroup$
– Eric Duminil
yesterday
add a comment |
$begingroup$
It's a matter of taste but I find a recursive solution to be reasonably readable :
private static int IncrementDigits(int number){
if (number == 0){
return 0;
}
int lastDigit = number % 10;
int firstDigits = number / 10;
if (lastDigit == 9){
return 10 + 100 * IncrementDigits(firstDigits);
} else {
return lastDigit + 1 + 10 * IncrementDigits(firstDigits);
}
}
Note that it only works for number > 0.
Here's a working example: https://dotnetfiddle.net/xUKs6M
answered yesterday
Eric DuminilEric Duminil
2,0611613
$endgroup$
It's a matter of taste but I find a recursive solution to be reasonably readable :
private static int IncrementDigits(int number){
if (number == 0){
return 0;
}
int lastDigit = number % 10;
int firstDigits = number / 10;
if (lastDigit == 9){
return 10 + 100 * IncrementDigits(firstDigits);
} else {
return lastDigit + 1 + 10 * IncrementDigits(firstDigits);
}
}
Note that it only works for number > 0.
Here's a working example: https://dotnetfiddle.net/xUKs6M
answered yesterday
Eric DuminilEric Duminil
2,0611613
edited yesterday
answered yesterday
Eric DuminilEric Duminil
2,0611613
answered yesterday
Eric DuminilEric Duminil
2,0611613
answered yesterday
Eric DuminilEric Duminil
2,0611613
2,0611613
2
$begingroup$
You could useSystem.Math.DivRemto do the division and modulo operations (on .Net Core this would be faster, the .Net Framework implementation does the naïve thing of doing the division twice)
$endgroup$
– mistertribs
yesterday
$begingroup$
@mistertribs: Thanks. I wanted to stay close enough to OP's answer. I don't really like the output argument inMath.DivRem, to be honest.
$endgroup$
– Eric Duminil
yesterday
add a comment |
2
$begingroup$
You could useSystem.Math.DivRemto do the division and modulo operations (on .Net Core this would be faster, the .Net Framework implementation does the naïve thing of doing the division twice)
$endgroup$
– mistertribs
yesterday
$begingroup$
@mistertribs: Thanks. I wanted to stay close enough to OP's answer. I don't really like the output argument inMath.DivRem, to be honest.
$endgroup$
– Eric Duminil
yesterday
2
2
$begingroup$
You could use
System.Math.DivRem to do the division and modulo operations (on .Net Core this would be faster, the .Net Framework implementation does the naïve thing of doing the division twice)$endgroup$
– mistertribs
yesterday
$begingroup$
You could use
System.Math.DivRem to do the division and modulo operations (on .Net Core this would be faster, the .Net Framework implementation does the naïve thing of doing the division twice)$endgroup$
– mistertribs
yesterday
$begingroup$
@mistertribs: Thanks. I wanted to stay close enough to OP's answer. I don't really like the output argument in
Math.DivRem, to be honest.$endgroup$
– Eric Duminil
yesterday
$begingroup$
@mistertribs: Thanks. I wanted to stay close enough to OP's answer. I don't really like the output argument in
Math.DivRem, to be honest.$endgroup$
– Eric Duminil
yesterday
add a comment |
$begingroup$
It works, but it's a bit tricky to follow the logic and work out why. If you work from the other end then you can avoid the Log and Pow and shifting the partial result, and I find that easier to read.
Challange is not a descriptive name. What does the method do? IncrementDigits might be a better name, for example.
answered 2 days ago
Peter TaylorPeter Taylor
17.4k2862
$endgroup$
add a comment |
$begingroup$
It works, but it's a bit tricky to follow the logic and work out why. If you work from the other end then you can avoid the Log and Pow and shifting the partial result, and I find that easier to read.
Challange is not a descriptive name. What does the method do? IncrementDigits might be a better name, for example.
answered 2 days ago
Peter TaylorPeter Taylor
17.4k2862
$endgroup$
add a comment |
$begingroup$
It works, but it's a bit tricky to follow the logic and work out why. If you work from the other end then you can avoid the Log and Pow and shifting the partial result, and I find that easier to read.
Challange is not a descriptive name. What does the method do? IncrementDigits might be a better name, for example.
answered 2 days ago
Peter TaylorPeter Taylor
17.4k2862
$endgroup$
It works, but it's a bit tricky to follow the logic and work out why. If you work from the other end then you can avoid the Log and Pow and shifting the partial result, and I find that easier to read.
Challange is not a descriptive name. What does the method do? IncrementDigits might be a better name, for example.
answered 2 days ago
Peter TaylorPeter Taylor
17.4k2862
answered 2 days ago
Peter TaylorPeter Taylor
17.4k2862
answered 2 days ago
Peter TaylorPeter Taylor
17.4k2862
answered 2 days ago
Peter TaylorPeter Taylor
17.4k2862
17.4k2862
add a comment |
add a comment |
$begingroup$
Computing powers repeatedly and log are not cheap operations.
You could substitute them with cheaper alternatives:
- Instead of computing powers of
tens, you could start with1and multiply in the loop steps. - Instead of computing the number of digits and using a counting loop, you could divide by 10 until reaching 0.
Something like this:
if (number == 0) {
return 1;
}
int work = number;
int tens = 1;
int result = 0;
while (work > 0) {
int digit = work % 10;
work /= 10;
result += tens * (digit + 1);
tens *= 10;
if (digit == 9) {
tens *= 10;
}
}
return result;
answered yesterday
janosjanos
98.1k12125350
$endgroup$
add a comment |
$begingroup$
Computing powers repeatedly and log are not cheap operations.
You could substitute them with cheaper alternatives:
- Instead of computing powers of
tens, you could start with1and multiply in the loop steps. - Instead of computing the number of digits and using a counting loop, you could divide by 10 until reaching 0.
Something like this:
if (number == 0) {
return 1;
}
int work = number;
int tens = 1;
int result = 0;
while (work > 0) {
int digit = work % 10;
work /= 10;
result += tens * (digit + 1);
tens *= 10;
if (digit == 9) {
tens *= 10;
}
}
return result;
answered yesterday
janosjanos
98.1k12125350
$endgroup$
add a comment |
$begingroup$
Computing powers repeatedly and log are not cheap operations.
You could substitute them with cheaper alternatives:
- Instead of computing powers of
tens, you could start with1and multiply in the loop steps. - Instead of computing the number of digits and using a counting loop, you could divide by 10 until reaching 0.
Something like this:
if (number == 0) {
return 1;
}
int work = number;
int tens = 1;
int result = 0;
while (work > 0) {
int digit = work % 10;
work /= 10;
result += tens * (digit + 1);
tens *= 10;
if (digit == 9) {
tens *= 10;
}
}
return result;
answered yesterday
janosjanos
98.1k12125350
$endgroup$
Computing powers repeatedly and log are not cheap operations.
You could substitute them with cheaper alternatives:
- Instead of computing powers of
tens, you could start with1and multiply in the loop steps. - Instead of computing the number of digits and using a counting loop, you could divide by 10 until reaching 0.
Something like this:
if (number == 0) {
return 1;
}
int work = number;
int tens = 1;
int result = 0;
while (work > 0) {
int digit = work % 10;
work /= 10;
result += tens * (digit + 1);
tens *= 10;
if (digit == 9) {
tens *= 10;
}
}
return result;
answered yesterday
janosjanos
98.1k12125350
answered yesterday
janosjanos
98.1k12125350
answered yesterday
janosjanos
98.1k12125350
answered yesterday
janosjanos
98.1k12125350
98.1k12125350
add a comment |
add a comment |
nowakasd is a new contributor. Be nice, and check out our Code of Conduct.
nowakasd is a new contributor. Be nice, and check out our Code of Conduct.
nowakasd is a new contributor. Be nice, and check out our Code of Conduct.
nowakasd is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Is 0 a valid input? If so, what should the result be?
$endgroup$
– YawarRaza7349
yesterday