Longest common prefix
3
$begingroup$
I have made a function for finding the longest common prefix for the challenge on the leetcode site. This is the code:
var longestCommonPrefix = function(strs) {
let longestPrefix = '';
if (strs.length > 0) {
longestPrefix = strs[0];
for (let i = 1; i < strs.length; i++) {
for (let j = 0; j < longestPrefix.length; j++) {
if (strs[i][j] != longestPrefix[j]) {
longestPrefix = longestPrefix.slice(0, j);
break;
}
}
}
}
return longestPrefix;
};
I am sure there is a way to make this code better, but not sure how to do that. Would appreciate any help.
javascript algorithm strings programming-challenge
asked Dec 7 '17 at 13:59
LeffLeff
1727
$endgroup$
add a comment |
3
$begingroup$
I have made a function for finding the longest common prefix for the challenge on the leetcode site. This is the code:
var longestCommonPrefix = function(strs) {
let longestPrefix = '';
if (strs.length > 0) {
longestPrefix = strs[0];
for (let i = 1; i < strs.length; i++) {
for (let j = 0; j < longestPrefix.length; j++) {
if (strs[i][j] != longestPrefix[j]) {
longestPrefix = longestPrefix.slice(0, j);
break;
}
}
}
}
return longestPrefix;
};
I am sure there is a way to make this code better, but not sure how to do that. Would appreciate any help.
javascript algorithm strings programming-challenge
asked Dec 7 '17 at 13:59
LeffLeff
1727
$endgroup$
add a comment |
3
3
3
$begingroup$
I have made a function for finding the longest common prefix for the challenge on the leetcode site. This is the code:
var longestCommonPrefix = function(strs) {
let longestPrefix = '';
if (strs.length > 0) {
longestPrefix = strs[0];
for (let i = 1; i < strs.length; i++) {
for (let j = 0; j < longestPrefix.length; j++) {
if (strs[i][j] != longestPrefix[j]) {
longestPrefix = longestPrefix.slice(0, j);
break;
}
}
}
}
return longestPrefix;
};
I am sure there is a way to make this code better, but not sure how to do that. Would appreciate any help.
javascript algorithm strings programming-challenge
asked Dec 7 '17 at 13:59
LeffLeff
1727
$endgroup$
I have made a function for finding the longest common prefix for the challenge on the leetcode site. This is the code:
var longestCommonPrefix = function(strs) {
let longestPrefix = '';
if (strs.length > 0) {
longestPrefix = strs[0];
for (let i = 1; i < strs.length; i++) {
for (let j = 0; j < longestPrefix.length; j++) {
if (strs[i][j] != longestPrefix[j]) {
longestPrefix = longestPrefix.slice(0, j);
break;
}
}
}
}
return longestPrefix;
};
I am sure there is a way to make this code better, but not sure how to do that. Would appreciate any help.
javascript algorithm strings programming-challenge
javascript algorithm strings programming-challenge
asked Dec 7 '17 at 13:59
LeffLeff
1727
asked Dec 7 '17 at 13:59
LeffLeff
1727
edited Dec 17 '18 at 8:42
Leff
asked Dec 7 '17 at 13:59
LeffLeff
1727
asked Dec 7 '17 at 13:59
LeffLeff
1727
asked Dec 7 '17 at 13:59
LeffLeff
1727
1727
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
2
$begingroup$
From a short review;
- You should sort the strings by length ascending if you start by assigning
longestPrefix = strs[0];the prefix cannot be longer than the shortest string. I would assign
longestPrefix[j]to a variable, avoiding an array access in a nested loop
I would
returnthe found value instead of callingbreak
- Break only exits one iteration in the loop anyway
- It seems okay that if no string list is provided, that
undefinedis returned
Personal preference, I prefer
listoverstrs
function(strs)creates an anonymous function, which is terrible in stack traces, just use the good oldfunction longestCommonPrefix(strs) {
*
answered Dec 7 '17 at 15:59
konijnkonijn
27.1k455236
$endgroup$
add a comment |
2
$begingroup$
I would find the alphabetically smallest and largest string and just run your algorithm on these two. That would avoid the embedded loop.
var longestCommonPrefix = function(strs) {
if (!strs)
return '';
let smallest = strs.reduce( (min, str) => min < str ? min : str, strs[0] );
let largest = strs.reduce( (min, str) => min > str ? min : str, strs[0] );
for (let i=0; i<smallest.length; i++) {
if (smallest[i] != largest[i])
return smallest.substr(0,i);
}
return '';
};
In answer to konijn it would be minimally faster to get the smallest/largest by doing:
let smallest = strs[0];
let largest = strs[0];
for (let i=1; i<strs.length; i++) {
let s= strs[i];
if (s > largest) largest = s;
if (s < smallest) smallest = s;
}
answered Dec 7 '17 at 17:42
Marc RohloffMarc Rohloff
3,02946
$endgroup$
$begingroup$
Would asortwith apopand anunshiftnot be faster?
$endgroup$
– konijn
Dec 7 '17 at 18:40
1
$begingroup$
fwiw, I wouldn't useshiftsince it requires moving all the elements in the array,[0]would be quicker. Evenpoprequires modifying the array, but to answer your comment most sorts are quite expensive O(n.log n) or O(n^2) and require a lot of moving and copying. The two calculations could be combined in a loop if performance was really critical (see my addition)
$endgroup$
– Marc Rohloff
Dec 7 '17 at 22:34
$begingroup$
How you you return an answer by inspecting only the shortest and longest? Consider {'ab', 'acd', 'abcd'} -> 'a'
$endgroup$
– Pierre Menard
Dec 17 '18 at 19:57
$begingroup$
Because if the smallest and largest string share a common prefix every other string would share it too. For example 'abaaa' and 'adeee' share the prefix 'a' everything inbetween ('abbbb', 'ac', 'ada', 'adaa') will all share the same prefix.
$endgroup$
– Marc Rohloff
Dec 19 '18 at 15:17
add a comment |
0
$begingroup$
var longestCommonPrefix = function(strs) {
if (!strs || strs.length === 0) {
return '';
}
var smallest = strs.reduce((min, max) => {
return min < max ? min : max
}, strs[0]);
var longest = strs.reduce((min, max) => {
return min > max ? min : max
}, strs[0]);
var res = '';
for (var i = 0; i < smallest.length; i++) {
if (smallest[i] !== longest[i]) {
return smallest.substr(0, i);
}
else {
res += smallest[i];
}
}
return res;
};
answered 17 mins ago
1061763184qqcom1061763184qqcom
1
New contributor
1061763184qqcom 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 |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
From a short review;
- You should sort the strings by length ascending if you start by assigning
longestPrefix = strs[0];the prefix cannot be longer than the shortest string. I would assign
longestPrefix[j]to a variable, avoiding an array access in a nested loop
I would
returnthe found value instead of callingbreak
- Break only exits one iteration in the loop anyway
- It seems okay that if no string list is provided, that
undefinedis returned
Personal preference, I prefer
listoverstrs
function(strs)creates an anonymous function, which is terrible in stack traces, just use the good oldfunction longestCommonPrefix(strs) {
*
answered Dec 7 '17 at 15:59
konijnkonijn
27.1k455236
$endgroup$
add a comment |
2
$begingroup$
From a short review;
- You should sort the strings by length ascending if you start by assigning
longestPrefix = strs[0];the prefix cannot be longer than the shortest string. I would assign
longestPrefix[j]to a variable, avoiding an array access in a nested loop
I would
returnthe found value instead of callingbreak
- Break only exits one iteration in the loop anyway
- It seems okay that if no string list is provided, that
undefinedis returned
Personal preference, I prefer
listoverstrs
function(strs)creates an anonymous function, which is terrible in stack traces, just use the good oldfunction longestCommonPrefix(strs) {
*
answered Dec 7 '17 at 15:59
konijnkonijn
27.1k455236
$endgroup$
add a comment |
2
2
2
$begingroup$
From a short review;
- You should sort the strings by length ascending if you start by assigning
longestPrefix = strs[0];the prefix cannot be longer than the shortest string. I would assign
longestPrefix[j]to a variable, avoiding an array access in a nested loop
I would
returnthe found value instead of callingbreak
- Break only exits one iteration in the loop anyway
- It seems okay that if no string list is provided, that
undefinedis returned
Personal preference, I prefer
listoverstrs
function(strs)creates an anonymous function, which is terrible in stack traces, just use the good oldfunction longestCommonPrefix(strs) {
*
answered Dec 7 '17 at 15:59
konijnkonijn
27.1k455236
$endgroup$
From a short review;
- You should sort the strings by length ascending if you start by assigning
longestPrefix = strs[0];the prefix cannot be longer than the shortest string. I would assign
longestPrefix[j]to a variable, avoiding an array access in a nested loop
I would
returnthe found value instead of callingbreak
- Break only exits one iteration in the loop anyway
- It seems okay that if no string list is provided, that
undefinedis returned
Personal preference, I prefer
listoverstrs
function(strs)creates an anonymous function, which is terrible in stack traces, just use the good oldfunction longestCommonPrefix(strs) {
*
answered Dec 7 '17 at 15:59
konijnkonijn
27.1k455236
edited Dec 7 '17 at 17:36
answered Dec 7 '17 at 15:59
konijnkonijn
27.1k455236
answered Dec 7 '17 at 15:59
konijnkonijn
27.1k455236
answered Dec 7 '17 at 15:59
konijnkonijn
27.1k455236
27.1k455236
add a comment |
add a comment |
2
$begingroup$
I would find the alphabetically smallest and largest string and just run your algorithm on these two. That would avoid the embedded loop.
var longestCommonPrefix = function(strs) {
if (!strs)
return '';
let smallest = strs.reduce( (min, str) => min < str ? min : str, strs[0] );
let largest = strs.reduce( (min, str) => min > str ? min : str, strs[0] );
for (let i=0; i<smallest.length; i++) {
if (smallest[i] != largest[i])
return smallest.substr(0,i);
}
return '';
};
In answer to konijn it would be minimally faster to get the smallest/largest by doing:
let smallest = strs[0];
let largest = strs[0];
for (let i=1; i<strs.length; i++) {
let s= strs[i];
if (s > largest) largest = s;
if (s < smallest) smallest = s;
}
answered Dec 7 '17 at 17:42
Marc RohloffMarc Rohloff
3,02946
$endgroup$
$begingroup$
Would asortwith apopand anunshiftnot be faster?
$endgroup$
– konijn
Dec 7 '17 at 18:40
1
$begingroup$
fwiw, I wouldn't useshiftsince it requires moving all the elements in the array,[0]would be quicker. Evenpoprequires modifying the array, but to answer your comment most sorts are quite expensive O(n.log n) or O(n^2) and require a lot of moving and copying. The two calculations could be combined in a loop if performance was really critical (see my addition)
$endgroup$
– Marc Rohloff
Dec 7 '17 at 22:34
$begingroup$
How you you return an answer by inspecting only the shortest and longest? Consider {'ab', 'acd', 'abcd'} -> 'a'
$endgroup$
– Pierre Menard
Dec 17 '18 at 19:57
$begingroup$
Because if the smallest and largest string share a common prefix every other string would share it too. For example 'abaaa' and 'adeee' share the prefix 'a' everything inbetween ('abbbb', 'ac', 'ada', 'adaa') will all share the same prefix.
$endgroup$
– Marc Rohloff
Dec 19 '18 at 15:17
add a comment |
2
$begingroup$
I would find the alphabetically smallest and largest string and just run your algorithm on these two. That would avoid the embedded loop.
var longestCommonPrefix = function(strs) {
if (!strs)
return '';
let smallest = strs.reduce( (min, str) => min < str ? min : str, strs[0] );
let largest = strs.reduce( (min, str) => min > str ? min : str, strs[0] );
for (let i=0; i<smallest.length; i++) {
if (smallest[i] != largest[i])
return smallest.substr(0,i);
}
return '';
};
In answer to konijn it would be minimally faster to get the smallest/largest by doing:
let smallest = strs[0];
let largest = strs[0];
for (let i=1; i<strs.length; i++) {
let s= strs[i];
if (s > largest) largest = s;
if (s < smallest) smallest = s;
}
answered Dec 7 '17 at 17:42
Marc RohloffMarc Rohloff
3,02946
$endgroup$
$begingroup$
Would asortwith apopand anunshiftnot be faster?
$endgroup$
– konijn
Dec 7 '17 at 18:40
1
$begingroup$
fwiw, I wouldn't useshiftsince it requires moving all the elements in the array,[0]would be quicker. Evenpoprequires modifying the array, but to answer your comment most sorts are quite expensive O(n.log n) or O(n^2) and require a lot of moving and copying. The two calculations could be combined in a loop if performance was really critical (see my addition)
$endgroup$
– Marc Rohloff
Dec 7 '17 at 22:34
$begingroup$
How you you return an answer by inspecting only the shortest and longest? Consider {'ab', 'acd', 'abcd'} -> 'a'
$endgroup$
– Pierre Menard
Dec 17 '18 at 19:57
$begingroup$
Because if the smallest and largest string share a common prefix every other string would share it too. For example 'abaaa' and 'adeee' share the prefix 'a' everything inbetween ('abbbb', 'ac', 'ada', 'adaa') will all share the same prefix.
$endgroup$
– Marc Rohloff
Dec 19 '18 at 15:17
add a comment |
2
2
2
$begingroup$
I would find the alphabetically smallest and largest string and just run your algorithm on these two. That would avoid the embedded loop.
var longestCommonPrefix = function(strs) {
if (!strs)
return '';
let smallest = strs.reduce( (min, str) => min < str ? min : str, strs[0] );
let largest = strs.reduce( (min, str) => min > str ? min : str, strs[0] );
for (let i=0; i<smallest.length; i++) {
if (smallest[i] != largest[i])
return smallest.substr(0,i);
}
return '';
};
In answer to konijn it would be minimally faster to get the smallest/largest by doing:
let smallest = strs[0];
let largest = strs[0];
for (let i=1; i<strs.length; i++) {
let s= strs[i];
if (s > largest) largest = s;
if (s < smallest) smallest = s;
}
answered Dec 7 '17 at 17:42
Marc RohloffMarc Rohloff
3,02946
$endgroup$
I would find the alphabetically smallest and largest string and just run your algorithm on these two. That would avoid the embedded loop.
var longestCommonPrefix = function(strs) {
if (!strs)
return '';
let smallest = strs.reduce( (min, str) => min < str ? min : str, strs[0] );
let largest = strs.reduce( (min, str) => min > str ? min : str, strs[0] );
for (let i=0; i<smallest.length; i++) {
if (smallest[i] != largest[i])
return smallest.substr(0,i);
}
return '';
};
In answer to konijn it would be minimally faster to get the smallest/largest by doing:
let smallest = strs[0];
let largest = strs[0];
for (let i=1; i<strs.length; i++) {
let s= strs[i];
if (s > largest) largest = s;
if (s < smallest) smallest = s;
}
answered Dec 7 '17 at 17:42
Marc RohloffMarc Rohloff
3,02946
edited Dec 7 '17 at 22:37
answered Dec 7 '17 at 17:42
Marc RohloffMarc Rohloff
3,02946
answered Dec 7 '17 at 17:42
Marc RohloffMarc Rohloff
3,02946
answered Dec 7 '17 at 17:42
Marc RohloffMarc Rohloff
3,02946
3,02946
$begingroup$
Would asortwith apopand anunshiftnot be faster?
$endgroup$
– konijn
Dec 7 '17 at 18:40
1
$begingroup$
fwiw, I wouldn't useshiftsince it requires moving all the elements in the array,[0]would be quicker. Evenpoprequires modifying the array, but to answer your comment most sorts are quite expensive O(n.log n) or O(n^2) and require a lot of moving and copying. The two calculations could be combined in a loop if performance was really critical (see my addition)
$endgroup$
– Marc Rohloff
Dec 7 '17 at 22:34
$begingroup$
How you you return an answer by inspecting only the shortest and longest? Consider {'ab', 'acd', 'abcd'} -> 'a'
$endgroup$
– Pierre Menard
Dec 17 '18 at 19:57
$begingroup$
Because if the smallest and largest string share a common prefix every other string would share it too. For example 'abaaa' and 'adeee' share the prefix 'a' everything inbetween ('abbbb', 'ac', 'ada', 'adaa') will all share the same prefix.
$endgroup$
– Marc Rohloff
Dec 19 '18 at 15:17
add a comment |
$begingroup$
Would asortwith apopand anunshiftnot be faster?
$endgroup$
– konijn
Dec 7 '17 at 18:40
1
$begingroup$
fwiw, I wouldn't useshiftsince it requires moving all the elements in the array,[0]would be quicker. Evenpoprequires modifying the array, but to answer your comment most sorts are quite expensive O(n.log n) or O(n^2) and require a lot of moving and copying. The two calculations could be combined in a loop if performance was really critical (see my addition)
$endgroup$
– Marc Rohloff
Dec 7 '17 at 22:34
$begingroup$
How you you return an answer by inspecting only the shortest and longest? Consider {'ab', 'acd', 'abcd'} -> 'a'
$endgroup$
– Pierre Menard
Dec 17 '18 at 19:57
$begingroup$
Because if the smallest and largest string share a common prefix every other string would share it too. For example 'abaaa' and 'adeee' share the prefix 'a' everything inbetween ('abbbb', 'ac', 'ada', 'adaa') will all share the same prefix.
$endgroup$
– Marc Rohloff
Dec 19 '18 at 15:17
$begingroup$
Would a
sort with a pop and an unshift not be faster?$endgroup$
– konijn
Dec 7 '17 at 18:40
$begingroup$
Would a
sort with a pop and an unshift not be faster?$endgroup$
– konijn
Dec 7 '17 at 18:40
1
1
$begingroup$
fwiw, I wouldn't use
shift since it requires moving all the elements in the array, [0] would be quicker. Even pop requires modifying the array, but to answer your comment most sorts are quite expensive O(n.log n) or O(n^2) and require a lot of moving and copying. The two calculations could be combined in a loop if performance was really critical (see my addition)$endgroup$
– Marc Rohloff
Dec 7 '17 at 22:34
$begingroup$
fwiw, I wouldn't use
shift since it requires moving all the elements in the array, [0] would be quicker. Even pop requires modifying the array, but to answer your comment most sorts are quite expensive O(n.log n) or O(n^2) and require a lot of moving and copying. The two calculations could be combined in a loop if performance was really critical (see my addition)$endgroup$
– Marc Rohloff
Dec 7 '17 at 22:34
$begingroup$
How you you return an answer by inspecting only the shortest and longest? Consider {'ab', 'acd', 'abcd'} -> 'a'
$endgroup$
– Pierre Menard
Dec 17 '18 at 19:57
$begingroup$
How you you return an answer by inspecting only the shortest and longest? Consider {'ab', 'acd', 'abcd'} -> 'a'
$endgroup$
– Pierre Menard
Dec 17 '18 at 19:57
$begingroup$
Because if the smallest and largest string share a common prefix every other string would share it too. For example 'abaaa' and 'adeee' share the prefix 'a' everything inbetween ('abbbb', 'ac', 'ada', 'adaa') will all share the same prefix.
$endgroup$
– Marc Rohloff
Dec 19 '18 at 15:17
$begingroup$
Because if the smallest and largest string share a common prefix every other string would share it too. For example 'abaaa' and 'adeee' share the prefix 'a' everything inbetween ('abbbb', 'ac', 'ada', 'adaa') will all share the same prefix.
$endgroup$
– Marc Rohloff
Dec 19 '18 at 15:17
add a comment |
0
$begingroup$
var longestCommonPrefix = function(strs) {
if (!strs || strs.length === 0) {
return '';
}
var smallest = strs.reduce((min, max) => {
return min < max ? min : max
}, strs[0]);
var longest = strs.reduce((min, max) => {
return min > max ? min : max
}, strs[0]);
var res = '';
for (var i = 0; i < smallest.length; i++) {
if (smallest[i] !== longest[i]) {
return smallest.substr(0, i);
}
else {
res += smallest[i];
}
}
return res;
};
answered 17 mins ago
1061763184qqcom1061763184qqcom
1
New contributor
1061763184qqcom 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 |
0
$begingroup$
var longestCommonPrefix = function(strs) {
if (!strs || strs.length === 0) {
return '';
}
var smallest = strs.reduce((min, max) => {
return min < max ? min : max
}, strs[0]);
var longest = strs.reduce((min, max) => {
return min > max ? min : max
}, strs[0]);
var res = '';
for (var i = 0; i < smallest.length; i++) {
if (smallest[i] !== longest[i]) {
return smallest.substr(0, i);
}
else {
res += smallest[i];
}
}
return res;
};
answered 17 mins ago
1061763184qqcom1061763184qqcom
1
New contributor
1061763184qqcom 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 |
0
0
0
$begingroup$
var longestCommonPrefix = function(strs) {
if (!strs || strs.length === 0) {
return '';
}
var smallest = strs.reduce((min, max) => {
return min < max ? min : max
}, strs[0]);
var longest = strs.reduce((min, max) => {
return min > max ? min : max
}, strs[0]);
var res = '';
for (var i = 0; i < smallest.length; i++) {
if (smallest[i] !== longest[i]) {
return smallest.substr(0, i);
}
else {
res += smallest[i];
}
}
return res;
};
answered 17 mins ago
1061763184qqcom1061763184qqcom
1
New contributor
1061763184qqcom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
var longestCommonPrefix = function(strs) {
if (!strs || strs.length === 0) {
return '';
}
var smallest = strs.reduce((min, max) => {
return min < max ? min : max
}, strs[0]);
var longest = strs.reduce((min, max) => {
return min > max ? min : max
}, strs[0]);
var res = '';
for (var i = 0; i < smallest.length; i++) {
if (smallest[i] !== longest[i]) {
return smallest.substr(0, i);
}
else {
res += smallest[i];
}
}
return res;
};
answered 17 mins ago
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answered 17 mins ago
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answered 17 mins ago
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answered 17 mins ago
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1061763184qqcom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
1061763184qqcom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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