refactoring a function using reduce - javascript
I've written a function to turn a string such as 'aaazeeeee' into a new string of 'aaa z eeeee'
this is the code I've tried that works
const groupCharacters = signature => {
let newSignature = "", arr = ;
let result = [...signature].reduce((accumulator, element, index) => {
// check if last element in accumulator matches current element
if (accumulator[accumulator.length -1] !== element) {
// push accumulator into array
arr.push(accumulator);
// set newSignature as new element
newSignature = element;
} else {
// else add element to newSignature
newSignature = accumulator += element;
}
// if is last item, push to array
if (index === signature.length - 1) arr.push(element);
return newSignature;
})
return arr;
}
console.log(groupCharacters('aabaaaaa'));how can I refactor it so that I don't need the new string or array? I've tried something like this
const groupCharacters = str => [...str].reduce((accumulator, element) => accumulator[accumulator.length - 1] !== element ? `${accumulator} ` : accumulator + element)it outputs 'aaa '
how do I fix it or use something like map instead?
javascript
edited Nov 13 '18 at 5:55
CertainPerformance
82.1k144067
asked Nov 13 '18 at 5:47
totalnoobtotalnoob
3261631
add a comment |
I've written a function to turn a string such as 'aaazeeeee' into a new string of 'aaa z eeeee'
this is the code I've tried that works
const groupCharacters = signature => {
let newSignature = "", arr = ;
let result = [...signature].reduce((accumulator, element, index) => {
// check if last element in accumulator matches current element
if (accumulator[accumulator.length -1] !== element) {
// push accumulator into array
arr.push(accumulator);
// set newSignature as new element
newSignature = element;
} else {
// else add element to newSignature
newSignature = accumulator += element;
}
// if is last item, push to array
if (index === signature.length - 1) arr.push(element);
return newSignature;
})
return arr;
}
console.log(groupCharacters('aabaaaaa'));how can I refactor it so that I don't need the new string or array? I've tried something like this
const groupCharacters = str => [...str].reduce((accumulator, element) => accumulator[accumulator.length - 1] !== element ? `${accumulator} ` : accumulator + element)it outputs 'aaa '
how do I fix it or use something like map instead?
javascript
edited Nov 13 '18 at 5:55
CertainPerformance
82.1k144067
asked Nov 13 '18 at 5:47
totalnoobtotalnoob
3261631
add a comment |
I've written a function to turn a string such as 'aaazeeeee' into a new string of 'aaa z eeeee'
this is the code I've tried that works
const groupCharacters = signature => {
let newSignature = "", arr = ;
let result = [...signature].reduce((accumulator, element, index) => {
// check if last element in accumulator matches current element
if (accumulator[accumulator.length -1] !== element) {
// push accumulator into array
arr.push(accumulator);
// set newSignature as new element
newSignature = element;
} else {
// else add element to newSignature
newSignature = accumulator += element;
}
// if is last item, push to array
if (index === signature.length - 1) arr.push(element);
return newSignature;
})
return arr;
}
console.log(groupCharacters('aabaaaaa'));how can I refactor it so that I don't need the new string or array? I've tried something like this
const groupCharacters = str => [...str].reduce((accumulator, element) => accumulator[accumulator.length - 1] !== element ? `${accumulator} ` : accumulator + element)it outputs 'aaa '
how do I fix it or use something like map instead?
javascript
edited Nov 13 '18 at 5:55
CertainPerformance
82.1k144067
asked Nov 13 '18 at 5:47
totalnoobtotalnoob
3261631
I've written a function to turn a string such as 'aaazeeeee' into a new string of 'aaa z eeeee'
this is the code I've tried that works
const groupCharacters = signature => {
let newSignature = "", arr = ;
let result = [...signature].reduce((accumulator, element, index) => {
// check if last element in accumulator matches current element
if (accumulator[accumulator.length -1] !== element) {
// push accumulator into array
arr.push(accumulator);
// set newSignature as new element
newSignature = element;
} else {
// else add element to newSignature
newSignature = accumulator += element;
}
// if is last item, push to array
if (index === signature.length - 1) arr.push(element);
return newSignature;
})
return arr;
}
console.log(groupCharacters('aabaaaaa'));how can I refactor it so that I don't need the new string or array? I've tried something like this
const groupCharacters = str => [...str].reduce((accumulator, element) => accumulator[accumulator.length - 1] !== element ? `${accumulator} ` : accumulator + element)it outputs 'aaa '
how do I fix it or use something like map instead?
const groupCharacters = signature => {
let newSignature = "", arr = ;
let result = [...signature].reduce((accumulator, element, index) => {
// check if last element in accumulator matches current element
if (accumulator[accumulator.length -1] !== element) {
// push accumulator into array
arr.push(accumulator);
// set newSignature as new element
newSignature = element;
} else {
// else add element to newSignature
newSignature = accumulator += element;
}
// if is last item, push to array
if (index === signature.length - 1) arr.push(element);
return newSignature;
})
return arr;
}
console.log(groupCharacters('aabaaaaa'));const groupCharacters = signature => {
let newSignature = "", arr = ;
let result = [...signature].reduce((accumulator, element, index) => {
// check if last element in accumulator matches current element
if (accumulator[accumulator.length -1] !== element) {
// push accumulator into array
arr.push(accumulator);
// set newSignature as new element
newSignature = element;
} else {
// else add element to newSignature
newSignature = accumulator += element;
}
// if is last item, push to array
if (index === signature.length - 1) arr.push(element);
return newSignature;
})
return arr;
}
console.log(groupCharacters('aabaaaaa'));const groupCharacters = str => [...str].reduce((accumulator, element) => accumulator[accumulator.length - 1] !== element ? `${accumulator} ` : accumulator + element)const groupCharacters = str => [...str].reduce((accumulator, element) => accumulator[accumulator.length - 1] !== element ? `${accumulator} ` : accumulator + element)javascript
javascript
edited Nov 13 '18 at 5:55
CertainPerformance
82.1k144067
asked Nov 13 '18 at 5:47
totalnoobtotalnoob
3261631
edited Nov 13 '18 at 5:55
CertainPerformance
82.1k144067
asked Nov 13 '18 at 5:47
totalnoobtotalnoob
3261631
edited Nov 13 '18 at 5:55
CertainPerformance
82.1k144067
edited Nov 13 '18 at 5:55
CertainPerformance
82.1k144067
edited Nov 13 '18 at 5:55
CertainPerformance
82.1k144067
82.1k144067
asked Nov 13 '18 at 5:47
totalnoobtotalnoob
3261631
asked Nov 13 '18 at 5:47
totalnoobtotalnoob
3261631
asked Nov 13 '18 at 5:47
totalnoobtotalnoob
3261631
3261631
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
To change your existing code but still use reduce, I'd suggest reduceing into a string, rather than array: on each iteration, concatenate with the current character, and also concatenate with a space if the next character is both defined and is not equal to the current character:
const groupCharacters = str => (
[...str].reduce((a, char, i) => a + char + (
str[i + 1] === char || str[i + 1] === undefined
? ''
: ' '
), '')
);
console.log(groupCharacters('aaazeeeee'));Or, you could use a simple regular expression - capture a word character in a group, then backreference that group as many times as possible, and replace with the whole match plus a space:
const groupCharacters = signature => signature.replace(/(w)1*(?!1|$)/g, '$& ');
console.log(groupCharacters('aaazeeeee'));To break it down:
(w)- Match any word character, capture it in the first group (so it can be backreferenced later using1)1*- Greedily repeat the character that was just matched zero or more times (repeat as much as possible)(?!1|$)- Checks that the substring that was just matched is not followed by either the end of the string, or by another of the same character. This ensures that the final repeated substring doesn't get a space appended to it as well (that is, you don't want'aaa z eeeee ').
As a side note, the regex G(w)1*+(?!$) would accomplish the same thing, is nicer to read, and is notably more efficient (the G matches the end of the last match or the beginning of the string, and the + in 1*+ makes the repetition possessive, which means that, on the final substring, the engine will be unable to backtrack, and thus fail immediately once the final full substring is checked, rather than iterating through each of its characters first). But, unfortunately, native JS doesn't support possessive quantifiers, nor the G anchor.
answered Nov 13 '18 at 5:51
CertainPerformanceCertainPerformance
82.1k144067
This looks amazing! Can you please explain in simpler words how does it work?
– vibhor1997a
Nov 13 '18 at 5:55
that's pretty cool, but I'd like to see how the reduce method can be refactored
– totalnoob
Nov 13 '18 at 6:11
@totalnoob See edit, reduce into a string instead of to an array
– CertainPerformance
Nov 13 '18 at 6:27
the result of the reduce version has an extra space at the end of it
– totalnoob
Nov 13 '18 at 6:44
what's the performance of these like in complexity?
– totalnoob
Nov 13 '18 at 17:03
add a comment |
I would do this using reduce with something like this as the OP wants to do with reduce:-
function myFunc(str) {
return str.split('').reduce(function (acc, cv, ci) {
if (acc.lastElem != cv) {
acc.str += (acc.lastElem == "") ? cv : ' ' + cv;
}
else {
acc.str += cv;
}
acc.lastElem = cv;
return acc;
}, { lastElem: '', str: '' }).str;
}
console.log(myFunc('aaabbbcc'));
console.log(myFunc('aaazeeeee'))How it works?
The accumulator is an object with keys lastElem(the previous element than the current element) and str(the string forming).
answered Nov 13 '18 at 6:27
vibhor1997avibhor1997a
1,5701624
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
To change your existing code but still use reduce, I'd suggest reduceing into a string, rather than array: on each iteration, concatenate with the current character, and also concatenate with a space if the next character is both defined and is not equal to the current character:
const groupCharacters = str => (
[...str].reduce((a, char, i) => a + char + (
str[i + 1] === char || str[i + 1] === undefined
? ''
: ' '
), '')
);
console.log(groupCharacters('aaazeeeee'));Or, you could use a simple regular expression - capture a word character in a group, then backreference that group as many times as possible, and replace with the whole match plus a space:
const groupCharacters = signature => signature.replace(/(w)1*(?!1|$)/g, '$& ');
console.log(groupCharacters('aaazeeeee'));To break it down:
(w)- Match any word character, capture it in the first group (so it can be backreferenced later using1)1*- Greedily repeat the character that was just matched zero or more times (repeat as much as possible)(?!1|$)- Checks that the substring that was just matched is not followed by either the end of the string, or by another of the same character. This ensures that the final repeated substring doesn't get a space appended to it as well (that is, you don't want'aaa z eeeee ').
As a side note, the regex G(w)1*+(?!$) would accomplish the same thing, is nicer to read, and is notably more efficient (the G matches the end of the last match or the beginning of the string, and the + in 1*+ makes the repetition possessive, which means that, on the final substring, the engine will be unable to backtrack, and thus fail immediately once the final full substring is checked, rather than iterating through each of its characters first). But, unfortunately, native JS doesn't support possessive quantifiers, nor the G anchor.
answered Nov 13 '18 at 5:51
CertainPerformanceCertainPerformance
82.1k144067
This looks amazing! Can you please explain in simpler words how does it work?
– vibhor1997a
Nov 13 '18 at 5:55
that's pretty cool, but I'd like to see how the reduce method can be refactored
– totalnoob
Nov 13 '18 at 6:11
@totalnoob See edit, reduce into a string instead of to an array
– CertainPerformance
Nov 13 '18 at 6:27
the result of the reduce version has an extra space at the end of it
– totalnoob
Nov 13 '18 at 6:44
what's the performance of these like in complexity?
– totalnoob
Nov 13 '18 at 17:03
add a comment |
To change your existing code but still use reduce, I'd suggest reduceing into a string, rather than array: on each iteration, concatenate with the current character, and also concatenate with a space if the next character is both defined and is not equal to the current character:
const groupCharacters = str => (
[...str].reduce((a, char, i) => a + char + (
str[i + 1] === char || str[i + 1] === undefined
? ''
: ' '
), '')
);
console.log(groupCharacters('aaazeeeee'));Or, you could use a simple regular expression - capture a word character in a group, then backreference that group as many times as possible, and replace with the whole match plus a space:
const groupCharacters = signature => signature.replace(/(w)1*(?!1|$)/g, '$& ');
console.log(groupCharacters('aaazeeeee'));To break it down:
(w)- Match any word character, capture it in the first group (so it can be backreferenced later using1)1*- Greedily repeat the character that was just matched zero or more times (repeat as much as possible)(?!1|$)- Checks that the substring that was just matched is not followed by either the end of the string, or by another of the same character. This ensures that the final repeated substring doesn't get a space appended to it as well (that is, you don't want'aaa z eeeee ').
As a side note, the regex G(w)1*+(?!$) would accomplish the same thing, is nicer to read, and is notably more efficient (the G matches the end of the last match or the beginning of the string, and the + in 1*+ makes the repetition possessive, which means that, on the final substring, the engine will be unable to backtrack, and thus fail immediately once the final full substring is checked, rather than iterating through each of its characters first). But, unfortunately, native JS doesn't support possessive quantifiers, nor the G anchor.
answered Nov 13 '18 at 5:51
CertainPerformanceCertainPerformance
82.1k144067
This looks amazing! Can you please explain in simpler words how does it work?
– vibhor1997a
Nov 13 '18 at 5:55
that's pretty cool, but I'd like to see how the reduce method can be refactored
– totalnoob
Nov 13 '18 at 6:11
@totalnoob See edit, reduce into a string instead of to an array
– CertainPerformance
Nov 13 '18 at 6:27
the result of the reduce version has an extra space at the end of it
– totalnoob
Nov 13 '18 at 6:44
what's the performance of these like in complexity?
– totalnoob
Nov 13 '18 at 17:03
add a comment |
To change your existing code but still use reduce, I'd suggest reduceing into a string, rather than array: on each iteration, concatenate with the current character, and also concatenate with a space if the next character is both defined and is not equal to the current character:
const groupCharacters = str => (
[...str].reduce((a, char, i) => a + char + (
str[i + 1] === char || str[i + 1] === undefined
? ''
: ' '
), '')
);
console.log(groupCharacters('aaazeeeee'));Or, you could use a simple regular expression - capture a word character in a group, then backreference that group as many times as possible, and replace with the whole match plus a space:
const groupCharacters = signature => signature.replace(/(w)1*(?!1|$)/g, '$& ');
console.log(groupCharacters('aaazeeeee'));To break it down:
(w)- Match any word character, capture it in the first group (so it can be backreferenced later using1)1*- Greedily repeat the character that was just matched zero or more times (repeat as much as possible)(?!1|$)- Checks that the substring that was just matched is not followed by either the end of the string, or by another of the same character. This ensures that the final repeated substring doesn't get a space appended to it as well (that is, you don't want'aaa z eeeee ').
As a side note, the regex G(w)1*+(?!$) would accomplish the same thing, is nicer to read, and is notably more efficient (the G matches the end of the last match or the beginning of the string, and the + in 1*+ makes the repetition possessive, which means that, on the final substring, the engine will be unable to backtrack, and thus fail immediately once the final full substring is checked, rather than iterating through each of its characters first). But, unfortunately, native JS doesn't support possessive quantifiers, nor the G anchor.
answered Nov 13 '18 at 5:51
CertainPerformanceCertainPerformance
82.1k144067
To change your existing code but still use reduce, I'd suggest reduceing into a string, rather than array: on each iteration, concatenate with the current character, and also concatenate with a space if the next character is both defined and is not equal to the current character:
const groupCharacters = str => (
[...str].reduce((a, char, i) => a + char + (
str[i + 1] === char || str[i + 1] === undefined
? ''
: ' '
), '')
);
console.log(groupCharacters('aaazeeeee'));Or, you could use a simple regular expression - capture a word character in a group, then backreference that group as many times as possible, and replace with the whole match plus a space:
const groupCharacters = signature => signature.replace(/(w)1*(?!1|$)/g, '$& ');
console.log(groupCharacters('aaazeeeee'));To break it down:
(w)- Match any word character, capture it in the first group (so it can be backreferenced later using1)1*- Greedily repeat the character that was just matched zero or more times (repeat as much as possible)(?!1|$)- Checks that the substring that was just matched is not followed by either the end of the string, or by another of the same character. This ensures that the final repeated substring doesn't get a space appended to it as well (that is, you don't want'aaa z eeeee ').
As a side note, the regex G(w)1*+(?!$) would accomplish the same thing, is nicer to read, and is notably more efficient (the G matches the end of the last match or the beginning of the string, and the + in 1*+ makes the repetition possessive, which means that, on the final substring, the engine will be unable to backtrack, and thus fail immediately once the final full substring is checked, rather than iterating through each of its characters first). But, unfortunately, native JS doesn't support possessive quantifiers, nor the G anchor.
const groupCharacters = str => (
[...str].reduce((a, char, i) => a + char + (
str[i + 1] === char || str[i + 1] === undefined
? ''
: ' '
), '')
);
console.log(groupCharacters('aaazeeeee'));const groupCharacters = str => (
[...str].reduce((a, char, i) => a + char + (
str[i + 1] === char || str[i + 1] === undefined
? ''
: ' '
), '')
);
console.log(groupCharacters('aaazeeeee'));const groupCharacters = signature => signature.replace(/(w)1*(?!1|$)/g, '$& ');
console.log(groupCharacters('aaazeeeee'));const groupCharacters = signature => signature.replace(/(w)1*(?!1|$)/g, '$& ');
console.log(groupCharacters('aaazeeeee'));answered Nov 13 '18 at 5:51
CertainPerformanceCertainPerformance
82.1k144067
edited Nov 13 '18 at 6:45
answered Nov 13 '18 at 5:51
CertainPerformanceCertainPerformance
82.1k144067
answered Nov 13 '18 at 5:51
CertainPerformanceCertainPerformance
82.1k144067
answered Nov 13 '18 at 5:51
CertainPerformanceCertainPerformance
82.1k144067
82.1k144067
This looks amazing! Can you please explain in simpler words how does it work?
– vibhor1997a
Nov 13 '18 at 5:55
that's pretty cool, but I'd like to see how the reduce method can be refactored
– totalnoob
Nov 13 '18 at 6:11
@totalnoob See edit, reduce into a string instead of to an array
– CertainPerformance
Nov 13 '18 at 6:27
the result of the reduce version has an extra space at the end of it
– totalnoob
Nov 13 '18 at 6:44
what's the performance of these like in complexity?
– totalnoob
Nov 13 '18 at 17:03
add a comment |
This looks amazing! Can you please explain in simpler words how does it work?
– vibhor1997a
Nov 13 '18 at 5:55
that's pretty cool, but I'd like to see how the reduce method can be refactored
– totalnoob
Nov 13 '18 at 6:11
@totalnoob See edit, reduce into a string instead of to an array
– CertainPerformance
Nov 13 '18 at 6:27
the result of the reduce version has an extra space at the end of it
– totalnoob
Nov 13 '18 at 6:44
what's the performance of these like in complexity?
– totalnoob
Nov 13 '18 at 17:03
This looks amazing! Can you please explain in simpler words how does it work?
– vibhor1997a
Nov 13 '18 at 5:55
This looks amazing! Can you please explain in simpler words how does it work?
– vibhor1997a
Nov 13 '18 at 5:55
that's pretty cool, but I'd like to see how the reduce method can be refactored
– totalnoob
Nov 13 '18 at 6:11
that's pretty cool, but I'd like to see how the reduce method can be refactored
– totalnoob
Nov 13 '18 at 6:11
@totalnoob See edit, reduce into a string instead of to an array
– CertainPerformance
Nov 13 '18 at 6:27
@totalnoob See edit, reduce into a string instead of to an array
– CertainPerformance
Nov 13 '18 at 6:27
the result of the reduce version has an extra space at the end of it
– totalnoob
Nov 13 '18 at 6:44
the result of the reduce version has an extra space at the end of it
– totalnoob
Nov 13 '18 at 6:44
what's the performance of these like in complexity?
– totalnoob
Nov 13 '18 at 17:03
what's the performance of these like in complexity?
– totalnoob
Nov 13 '18 at 17:03
add a comment |
I would do this using reduce with something like this as the OP wants to do with reduce:-
function myFunc(str) {
return str.split('').reduce(function (acc, cv, ci) {
if (acc.lastElem != cv) {
acc.str += (acc.lastElem == "") ? cv : ' ' + cv;
}
else {
acc.str += cv;
}
acc.lastElem = cv;
return acc;
}, { lastElem: '', str: '' }).str;
}
console.log(myFunc('aaabbbcc'));
console.log(myFunc('aaazeeeee'))How it works?
The accumulator is an object with keys lastElem(the previous element than the current element) and str(the string forming).
answered Nov 13 '18 at 6:27
vibhor1997avibhor1997a
1,5701624
add a comment |
I would do this using reduce with something like this as the OP wants to do with reduce:-
function myFunc(str) {
return str.split('').reduce(function (acc, cv, ci) {
if (acc.lastElem != cv) {
acc.str += (acc.lastElem == "") ? cv : ' ' + cv;
}
else {
acc.str += cv;
}
acc.lastElem = cv;
return acc;
}, { lastElem: '', str: '' }).str;
}
console.log(myFunc('aaabbbcc'));
console.log(myFunc('aaazeeeee'))How it works?
The accumulator is an object with keys lastElem(the previous element than the current element) and str(the string forming).
answered Nov 13 '18 at 6:27
vibhor1997avibhor1997a
1,5701624
add a comment |
I would do this using reduce with something like this as the OP wants to do with reduce:-
function myFunc(str) {
return str.split('').reduce(function (acc, cv, ci) {
if (acc.lastElem != cv) {
acc.str += (acc.lastElem == "") ? cv : ' ' + cv;
}
else {
acc.str += cv;
}
acc.lastElem = cv;
return acc;
}, { lastElem: '', str: '' }).str;
}
console.log(myFunc('aaabbbcc'));
console.log(myFunc('aaazeeeee'))How it works?
The accumulator is an object with keys lastElem(the previous element than the current element) and str(the string forming).
answered Nov 13 '18 at 6:27
vibhor1997avibhor1997a
1,5701624
I would do this using reduce with something like this as the OP wants to do with reduce:-
function myFunc(str) {
return str.split('').reduce(function (acc, cv, ci) {
if (acc.lastElem != cv) {
acc.str += (acc.lastElem == "") ? cv : ' ' + cv;
}
else {
acc.str += cv;
}
acc.lastElem = cv;
return acc;
}, { lastElem: '', str: '' }).str;
}
console.log(myFunc('aaabbbcc'));
console.log(myFunc('aaazeeeee'))How it works?
The accumulator is an object with keys lastElem(the previous element than the current element) and str(the string forming).
function myFunc(str) {
return str.split('').reduce(function (acc, cv, ci) {
if (acc.lastElem != cv) {
acc.str += (acc.lastElem == "") ? cv : ' ' + cv;
}
else {
acc.str += cv;
}
acc.lastElem = cv;
return acc;
}, { lastElem: '', str: '' }).str;
}
console.log(myFunc('aaabbbcc'));
console.log(myFunc('aaazeeeee'))function myFunc(str) {
return str.split('').reduce(function (acc, cv, ci) {
if (acc.lastElem != cv) {
acc.str += (acc.lastElem == "") ? cv : ' ' + cv;
}
else {
acc.str += cv;
}
acc.lastElem = cv;
return acc;
}, { lastElem: '', str: '' }).str;
}
console.log(myFunc('aaabbbcc'));
console.log(myFunc('aaazeeeee'))answered Nov 13 '18 at 6:27
vibhor1997avibhor1997a
1,5701624
answered Nov 13 '18 at 6:27
vibhor1997avibhor1997a
1,5701624
answered Nov 13 '18 at 6:27
vibhor1997avibhor1997a
1,5701624
answered Nov 13 '18 at 6:27
vibhor1997avibhor1997a
1,5701624
1,5701624
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