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}
3
$begingroup$
The task
Implement division of two positive integers without using the
division, multiplication, or modulus operators. Return the quotient as
an integer, ignoring the remainder.
My solution
const division = (dividend, divisor) => {
let remainder = null;
let quotient = 1;
const sign = ((dividend > 0 && divisor < 0) ||
(dividend < 0 && divisor > 0)) ?
~1 : 1;
let tempdividend = Math.abs(dividend);
let tempdivisor = Math.abs(divisor);
if (tempdivisor === tempdividend) {
remainder = 0;
return sign;
} else if (tempdividend < tempdivisor) {
remainder = dividend < 0 ?
sign < 0 ? ~tempdividend : tempdividend :
tempdividend;
return 0;
}
while (tempdivisor << 1 <= tempdividend) {
tempdivisor = tempdivisor << 1;
quotient = quotient << 1;
}
quotient = dividend < 0 ?
(sign < 0 ? ~quotient : quotient) + division(~(tempdividend-tempdivisor), divisor) :
(sign < 0 ? ~quotient : quotient) + division(tempdividend-tempdivisor, divisor);
return quotient;
}
javascript algorithm programming-challenge ecmascript-6 bitwise
asked Apr 18 at 16:57
thadeuszlaythadeuszlay
1,021616
$endgroup$
add a comment |
3
$begingroup$
The task
Implement division of two positive integers without using the
division, multiplication, or modulus operators. Return the quotient as
an integer, ignoring the remainder.
My solution
const division = (dividend, divisor) => {
let remainder = null;
let quotient = 1;
const sign = ((dividend > 0 && divisor < 0) ||
(dividend < 0 && divisor > 0)) ?
~1 : 1;
let tempdividend = Math.abs(dividend);
let tempdivisor = Math.abs(divisor);
if (tempdivisor === tempdividend) {
remainder = 0;
return sign;
} else if (tempdividend < tempdivisor) {
remainder = dividend < 0 ?
sign < 0 ? ~tempdividend : tempdividend :
tempdividend;
return 0;
}
while (tempdivisor << 1 <= tempdividend) {
tempdivisor = tempdivisor << 1;
quotient = quotient << 1;
}
quotient = dividend < 0 ?
(sign < 0 ? ~quotient : quotient) + division(~(tempdividend-tempdivisor), divisor) :
(sign < 0 ? ~quotient : quotient) + division(tempdividend-tempdivisor, divisor);
return quotient;
}
javascript algorithm programming-challenge ecmascript-6 bitwise
asked Apr 18 at 16:57
thadeuszlaythadeuszlay
1,021616
$endgroup$
1
$begingroup$
Your function does not work, very sloppy. I am amazed you found two arguments that would return the correct result.division(-1,1)returns-2,-2 / -1` returns3and worst any value divide 0 does not return at all.
$endgroup$
– Blindman67
Apr 18 at 23:50
1
$begingroup$
It works for positive integers (as the task ask for). But you are right about the other stuff. Will fix it. @Blindman67
$endgroup$
– thadeuszlay
Apr 19 at 8:56
add a comment |
3
3
3
$begingroup$
The task
Implement division of two positive integers without using the
division, multiplication, or modulus operators. Return the quotient as
an integer, ignoring the remainder.
My solution
const division = (dividend, divisor) => {
let remainder = null;
let quotient = 1;
const sign = ((dividend > 0 && divisor < 0) ||
(dividend < 0 && divisor > 0)) ?
~1 : 1;
let tempdividend = Math.abs(dividend);
let tempdivisor = Math.abs(divisor);
if (tempdivisor === tempdividend) {
remainder = 0;
return sign;
} else if (tempdividend < tempdivisor) {
remainder = dividend < 0 ?
sign < 0 ? ~tempdividend : tempdividend :
tempdividend;
return 0;
}
while (tempdivisor << 1 <= tempdividend) {
tempdivisor = tempdivisor << 1;
quotient = quotient << 1;
}
quotient = dividend < 0 ?
(sign < 0 ? ~quotient : quotient) + division(~(tempdividend-tempdivisor), divisor) :
(sign < 0 ? ~quotient : quotient) + division(tempdividend-tempdivisor, divisor);
return quotient;
}
javascript algorithm programming-challenge ecmascript-6 bitwise
asked Apr 18 at 16:57
thadeuszlaythadeuszlay
1,021616
$endgroup$
The task
Implement division of two positive integers without using the
division, multiplication, or modulus operators. Return the quotient as
an integer, ignoring the remainder.
My solution
const division = (dividend, divisor) => {
let remainder = null;
let quotient = 1;
const sign = ((dividend > 0 && divisor < 0) ||
(dividend < 0 && divisor > 0)) ?
~1 : 1;
let tempdividend = Math.abs(dividend);
let tempdivisor = Math.abs(divisor);
if (tempdivisor === tempdividend) {
remainder = 0;
return sign;
} else if (tempdividend < tempdivisor) {
remainder = dividend < 0 ?
sign < 0 ? ~tempdividend : tempdividend :
tempdividend;
return 0;
}
while (tempdivisor << 1 <= tempdividend) {
tempdivisor = tempdivisor << 1;
quotient = quotient << 1;
}
quotient = dividend < 0 ?
(sign < 0 ? ~quotient : quotient) + division(~(tempdividend-tempdivisor), divisor) :
(sign < 0 ? ~quotient : quotient) + division(tempdividend-tempdivisor, divisor);
return quotient;
}
javascript algorithm programming-challenge ecmascript-6 bitwise
javascript algorithm programming-challenge ecmascript-6 bitwise
asked Apr 18 at 16:57
thadeuszlaythadeuszlay
1,021616
asked Apr 18 at 16:57
thadeuszlaythadeuszlay
1,021616
edited Apr 18 at 18:00
thadeuszlay
asked Apr 18 at 16:57
thadeuszlaythadeuszlay
1,021616
asked Apr 18 at 16:57
thadeuszlaythadeuszlay
1,021616
asked Apr 18 at 16:57
thadeuszlaythadeuszlay
1,021616
1,021616
1
$begingroup$
Your function does not work, very sloppy. I am amazed you found two arguments that would return the correct result.division(-1,1)returns-2,-2 / -1` returns3and worst any value divide 0 does not return at all.
$endgroup$
– Blindman67
Apr 18 at 23:50
1
$begingroup$
It works for positive integers (as the task ask for). But you are right about the other stuff. Will fix it. @Blindman67
$endgroup$
– thadeuszlay
Apr 19 at 8:56
add a comment |
1
$begingroup$
Your function does not work, very sloppy. I am amazed you found two arguments that would return the correct result.division(-1,1)returns-2,-2 / -1` returns3and worst any value divide 0 does not return at all.
$endgroup$
– Blindman67
Apr 18 at 23:50
1
$begingroup$
It works for positive integers (as the task ask for). But you are right about the other stuff. Will fix it. @Blindman67
$endgroup$
– thadeuszlay
Apr 19 at 8:56
1
1
$begingroup$
Your function does not work, very sloppy. I am amazed you found two arguments that would return the correct result.
division(-1,1) returns -2, -2 / -1` returns 3 and worst any value divide 0 does not return at all.$endgroup$
– Blindman67
Apr 18 at 23:50
$begingroup$
Your function does not work, very sloppy. I am amazed you found two arguments that would return the correct result.
division(-1,1) returns -2, -2 / -1` returns 3 and worst any value divide 0 does not return at all.$endgroup$
– Blindman67
Apr 18 at 23:50
1
1
$begingroup$
It works for positive integers (as the task ask for). But you are right about the other stuff. Will fix it. @Blindman67
$endgroup$
– thadeuszlay
Apr 19 at 8:56
$begingroup$
It works for positive integers (as the task ask for). But you are right about the other stuff. Will fix it. @Blindman67
$endgroup$
– thadeuszlay
Apr 19 at 8:56
add a comment |
1 Answer
1
active
oldest
votes
3
$begingroup$
Is bitwise operation mandatory? There's a much simpler way
const divide = (dividend, divisor) => {
let quotient = 0, neg = false;
if( (dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0) ){ neg = true; }
dividend = Math.abs(dividend);
divisor = Math.abs(divisor);
if(dividend < divisor) {return 0;}
else if(dividend > 0 && divisor != 0){
while(dividend >= divisor){
dividend -= divisor;
++quotient;
}
} else { // handle what you want to do for those cases..}
return neg ? -quotient : quotient;
}
You get your quotient and remainder is ignored. Just do a check if dividend is negative or divisor is 0.
answered Apr 18 at 19:05
Dblaze47Dblaze47
1564
New contributor
Dblaze47 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Welcome to Code Review! Better check dividend sign before comparing to divisor.
$endgroup$
– greybeard
Apr 18 at 20:01
1
$begingroup$
Also first if statement should be<not>and ... It may be the simple way, but divide(2**32-1, 1) may take some time to complete (On average device 11 seconds), even longer if itsdivide(Number.MAX_SAFE_INTEGER, -1)(about 22 years on average device) There is a point where complexity provides practical solutions.
$endgroup$
– Blindman67
Apr 18 at 23:49
$begingroup$
@greybeard I have fixed my answer to accommodate the negative cases. Thanks for the suggestion.
$endgroup$
– Dblaze47
Apr 19 at 4:15
$begingroup$
@Blindman67 I have fixed the comparison. As for the division case, your case is a valid example that I have taken into account before posting the answer. However, there is no one solution to solve them all, and I thought if the OP was handling smaller cases (possible range was not mentioned), and has already a solution to handle complex cases, why not a simpler one then.
$endgroup$
– Dblaze47
Apr 19 at 4:23
$begingroup$
@Dblaze47 Oh my bad, my numbers were way wrong 32Bit int is ~7 seconds, and MAX_SAFE_INTEGER is about ~5 months. Anyways besides the point, "Why not a simpler one then." Why? This is code review, the focus is the review of OP's code, its style, its logic, and with that you can add example alternative/s demonstrating the points reviewed. The OP is after all looking for feedback on their code. If alternatives were the quest then google would be the better path, don't you think.
$endgroup$
– Blindman67
Apr 19 at 15:50
|
show 1 more comment
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
Is bitwise operation mandatory? There's a much simpler way
const divide = (dividend, divisor) => {
let quotient = 0, neg = false;
if( (dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0) ){ neg = true; }
dividend = Math.abs(dividend);
divisor = Math.abs(divisor);
if(dividend < divisor) {return 0;}
else if(dividend > 0 && divisor != 0){
while(dividend >= divisor){
dividend -= divisor;
++quotient;
}
} else { // handle what you want to do for those cases..}
return neg ? -quotient : quotient;
}
You get your quotient and remainder is ignored. Just do a check if dividend is negative or divisor is 0.
answered Apr 18 at 19:05
Dblaze47Dblaze47
1564
New contributor
Dblaze47 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Welcome to Code Review! Better check dividend sign before comparing to divisor.
$endgroup$
– greybeard
Apr 18 at 20:01
1
$begingroup$
Also first if statement should be<not>and ... It may be the simple way, but divide(2**32-1, 1) may take some time to complete (On average device 11 seconds), even longer if itsdivide(Number.MAX_SAFE_INTEGER, -1)(about 22 years on average device) There is a point where complexity provides practical solutions.
$endgroup$
– Blindman67
Apr 18 at 23:49
$begingroup$
@greybeard I have fixed my answer to accommodate the negative cases. Thanks for the suggestion.
$endgroup$
– Dblaze47
Apr 19 at 4:15
$begingroup$
@Blindman67 I have fixed the comparison. As for the division case, your case is a valid example that I have taken into account before posting the answer. However, there is no one solution to solve them all, and I thought if the OP was handling smaller cases (possible range was not mentioned), and has already a solution to handle complex cases, why not a simpler one then.
$endgroup$
– Dblaze47
Apr 19 at 4:23
$begingroup$
@Dblaze47 Oh my bad, my numbers were way wrong 32Bit int is ~7 seconds, and MAX_SAFE_INTEGER is about ~5 months. Anyways besides the point, "Why not a simpler one then." Why? This is code review, the focus is the review of OP's code, its style, its logic, and with that you can add example alternative/s demonstrating the points reviewed. The OP is after all looking for feedback on their code. If alternatives were the quest then google would be the better path, don't you think.
$endgroup$
– Blindman67
Apr 19 at 15:50
|
show 1 more comment
3
$begingroup$
Is bitwise operation mandatory? There's a much simpler way
const divide = (dividend, divisor) => {
let quotient = 0, neg = false;
if( (dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0) ){ neg = true; }
dividend = Math.abs(dividend);
divisor = Math.abs(divisor);
if(dividend < divisor) {return 0;}
else if(dividend > 0 && divisor != 0){
while(dividend >= divisor){
dividend -= divisor;
++quotient;
}
} else { // handle what you want to do for those cases..}
return neg ? -quotient : quotient;
}
You get your quotient and remainder is ignored. Just do a check if dividend is negative or divisor is 0.
answered Apr 18 at 19:05
Dblaze47Dblaze47
1564
New contributor
Dblaze47 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Welcome to Code Review! Better check dividend sign before comparing to divisor.
$endgroup$
– greybeard
Apr 18 at 20:01
1
$begingroup$
Also first if statement should be<not>and ... It may be the simple way, but divide(2**32-1, 1) may take some time to complete (On average device 11 seconds), even longer if itsdivide(Number.MAX_SAFE_INTEGER, -1)(about 22 years on average device) There is a point where complexity provides practical solutions.
$endgroup$
– Blindman67
Apr 18 at 23:49
$begingroup$
@greybeard I have fixed my answer to accommodate the negative cases. Thanks for the suggestion.
$endgroup$
– Dblaze47
Apr 19 at 4:15
$begingroup$
@Blindman67 I have fixed the comparison. As for the division case, your case is a valid example that I have taken into account before posting the answer. However, there is no one solution to solve them all, and I thought if the OP was handling smaller cases (possible range was not mentioned), and has already a solution to handle complex cases, why not a simpler one then.
$endgroup$
– Dblaze47
Apr 19 at 4:23
$begingroup$
@Dblaze47 Oh my bad, my numbers were way wrong 32Bit int is ~7 seconds, and MAX_SAFE_INTEGER is about ~5 months. Anyways besides the point, "Why not a simpler one then." Why? This is code review, the focus is the review of OP's code, its style, its logic, and with that you can add example alternative/s demonstrating the points reviewed. The OP is after all looking for feedback on their code. If alternatives were the quest then google would be the better path, don't you think.
$endgroup$
– Blindman67
Apr 19 at 15:50
|
show 1 more comment
3
3
3
$begingroup$
Is bitwise operation mandatory? There's a much simpler way
const divide = (dividend, divisor) => {
let quotient = 0, neg = false;
if( (dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0) ){ neg = true; }
dividend = Math.abs(dividend);
divisor = Math.abs(divisor);
if(dividend < divisor) {return 0;}
else if(dividend > 0 && divisor != 0){
while(dividend >= divisor){
dividend -= divisor;
++quotient;
}
} else { // handle what you want to do for those cases..}
return neg ? -quotient : quotient;
}
You get your quotient and remainder is ignored. Just do a check if dividend is negative or divisor is 0.
answered Apr 18 at 19:05
Dblaze47Dblaze47
1564
New contributor
Dblaze47 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Is bitwise operation mandatory? There's a much simpler way
const divide = (dividend, divisor) => {
let quotient = 0, neg = false;
if( (dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0) ){ neg = true; }
dividend = Math.abs(dividend);
divisor = Math.abs(divisor);
if(dividend < divisor) {return 0;}
else if(dividend > 0 && divisor != 0){
while(dividend >= divisor){
dividend -= divisor;
++quotient;
}
} else { // handle what you want to do for those cases..}
return neg ? -quotient : quotient;
}
You get your quotient and remainder is ignored. Just do a check if dividend is negative or divisor is 0.
answered Apr 18 at 19:05
Dblaze47Dblaze47
1564
New contributor
Dblaze47 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 19 at 4:14
answered Apr 18 at 19:05
Dblaze47Dblaze47
1564
New contributor
Dblaze47 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 18 at 19:05
Dblaze47Dblaze47
1564
answered Apr 18 at 19:05
Dblaze47Dblaze47
1564
1564
New contributor
Dblaze47 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dblaze47 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dblaze47 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Welcome to Code Review! Better check dividend sign before comparing to divisor.
$endgroup$
– greybeard
Apr 18 at 20:01
1
$begingroup$
Also first if statement should be<not>and ... It may be the simple way, but divide(2**32-1, 1) may take some time to complete (On average device 11 seconds), even longer if itsdivide(Number.MAX_SAFE_INTEGER, -1)(about 22 years on average device) There is a point where complexity provides practical solutions.
$endgroup$
– Blindman67
Apr 18 at 23:49
$begingroup$
@greybeard I have fixed my answer to accommodate the negative cases. Thanks for the suggestion.
$endgroup$
– Dblaze47
Apr 19 at 4:15
$begingroup$
@Blindman67 I have fixed the comparison. As for the division case, your case is a valid example that I have taken into account before posting the answer. However, there is no one solution to solve them all, and I thought if the OP was handling smaller cases (possible range was not mentioned), and has already a solution to handle complex cases, why not a simpler one then.
$endgroup$
– Dblaze47
Apr 19 at 4:23
$begingroup$
@Dblaze47 Oh my bad, my numbers were way wrong 32Bit int is ~7 seconds, and MAX_SAFE_INTEGER is about ~5 months. Anyways besides the point, "Why not a simpler one then." Why? This is code review, the focus is the review of OP's code, its style, its logic, and with that you can add example alternative/s demonstrating the points reviewed. The OP is after all looking for feedback on their code. If alternatives were the quest then google would be the better path, don't you think.
$endgroup$
– Blindman67
Apr 19 at 15:50
|
show 1 more comment
1
$begingroup$
Welcome to Code Review! Better check dividend sign before comparing to divisor.
$endgroup$
– greybeard
Apr 18 at 20:01
1
$begingroup$
Also first if statement should be<not>and ... It may be the simple way, but divide(2**32-1, 1) may take some time to complete (On average device 11 seconds), even longer if itsdivide(Number.MAX_SAFE_INTEGER, -1)(about 22 years on average device) There is a point where complexity provides practical solutions.
$endgroup$
– Blindman67
Apr 18 at 23:49
$begingroup$
@greybeard I have fixed my answer to accommodate the negative cases. Thanks for the suggestion.
$endgroup$
– Dblaze47
Apr 19 at 4:15
$begingroup$
@Blindman67 I have fixed the comparison. As for the division case, your case is a valid example that I have taken into account before posting the answer. However, there is no one solution to solve them all, and I thought if the OP was handling smaller cases (possible range was not mentioned), and has already a solution to handle complex cases, why not a simpler one then.
$endgroup$
– Dblaze47
Apr 19 at 4:23
$begingroup$
@Dblaze47 Oh my bad, my numbers were way wrong 32Bit int is ~7 seconds, and MAX_SAFE_INTEGER is about ~5 months. Anyways besides the point, "Why not a simpler one then." Why? This is code review, the focus is the review of OP's code, its style, its logic, and with that you can add example alternative/s demonstrating the points reviewed. The OP is after all looking for feedback on their code. If alternatives were the quest then google would be the better path, don't you think.
$endgroup$
– Blindman67
Apr 19 at 15:50
1
1
$begingroup$
Welcome to Code Review! Better check dividend sign before comparing to divisor.
$endgroup$
– greybeard
Apr 18 at 20:01
$begingroup$
Welcome to Code Review! Better check dividend sign before comparing to divisor.
$endgroup$
– greybeard
Apr 18 at 20:01
1
1
$begingroup$
Also first if statement should be
< not > and ... It may be the simple way, but divide(2**32-1, 1) may take some time to complete (On average device 11 seconds), even longer if its divide(Number.MAX_SAFE_INTEGER, -1) (about 22 years on average device) There is a point where complexity provides practical solutions.$endgroup$
– Blindman67
Apr 18 at 23:49
$begingroup$
Also first if statement should be
< not > and ... It may be the simple way, but divide(2**32-1, 1) may take some time to complete (On average device 11 seconds), even longer if its divide(Number.MAX_SAFE_INTEGER, -1) (about 22 years on average device) There is a point where complexity provides practical solutions.$endgroup$
– Blindman67
Apr 18 at 23:49
$begingroup$
@greybeard I have fixed my answer to accommodate the negative cases. Thanks for the suggestion.
$endgroup$
– Dblaze47
Apr 19 at 4:15
$begingroup$
@greybeard I have fixed my answer to accommodate the negative cases. Thanks for the suggestion.
$endgroup$
– Dblaze47
Apr 19 at 4:15
$begingroup$
@Blindman67 I have fixed the comparison. As for the division case, your case is a valid example that I have taken into account before posting the answer. However, there is no one solution to solve them all, and I thought if the OP was handling smaller cases (possible range was not mentioned), and has already a solution to handle complex cases, why not a simpler one then.
$endgroup$
– Dblaze47
Apr 19 at 4:23
$begingroup$
@Blindman67 I have fixed the comparison. As for the division case, your case is a valid example that I have taken into account before posting the answer. However, there is no one solution to solve them all, and I thought if the OP was handling smaller cases (possible range was not mentioned), and has already a solution to handle complex cases, why not a simpler one then.
$endgroup$
– Dblaze47
Apr 19 at 4:23
$begingroup$
@Dblaze47 Oh my bad, my numbers were way wrong 32Bit int is ~7 seconds, and MAX_SAFE_INTEGER is about ~5 months. Anyways besides the point, "Why not a simpler one then." Why? This is code review, the focus is the review of OP's code, its style, its logic, and with that you can add example alternative/s demonstrating the points reviewed. The OP is after all looking for feedback on their code. If alternatives were the quest then google would be the better path, don't you think.
$endgroup$
– Blindman67
Apr 19 at 15:50
$begingroup$
@Dblaze47 Oh my bad, my numbers were way wrong 32Bit int is ~7 seconds, and MAX_SAFE_INTEGER is about ~5 months. Anyways besides the point, "Why not a simpler one then." Why? This is code review, the focus is the review of OP's code, its style, its logic, and with that you can add example alternative/s demonstrating the points reviewed. The OP is after all looking for feedback on their code. If alternatives were the quest then google would be the better path, don't you think.
$endgroup$
– Blindman67
Apr 19 at 15:50
|
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$begingroup$
Your function does not work, very sloppy. I am amazed you found two arguments that would return the correct result.
division(-1,1)returns-2,-2 / -1` returns3and worst any value divide 0 does not return at all.$endgroup$
– Blindman67
Apr 18 at 23:50
1
$begingroup$
It works for positive integers (as the task ask for). But you are right about the other stuff. Will fix it. @Blindman67
$endgroup$
– thadeuszlay
Apr 19 at 8:56