Consider the following string: 7, 20, 22, 30–32, 33, 36–40, 46
I developed some code that will automatically parse said string into an array with the given ranges as follows.
Note: A typical use-case for this would be - to search for selected pages within a pdf with 100's of pages
var number_string = "7, 20, 22, 30–32, 33, 36–40, 46".toString().replace(/–/gi, '-').replace(/ /gi, '').split(',');
var new_arr = [];
$.each(number_string, function(index, value) {
if (value.match(/-/gi)) {
var range_arr = value.split('-');
var sub_arr = range(range_arr[0], range_arr[1]);
$.each(sub_arr, function(sub_index, sub_value) {
new_arr.push(parseInt(sub_value, 10));
});
} else {
new_arr.push(parseInt(value, 10));
}
});
console.log(new_arr);
function range(lowEnd, highEnd) {
var arr = [],
c = highEnd - lowEnd + 1;
while (c--) {
arr[c] = highEnd--
}
return arr;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
Was there a more streamlined non jQuery method that could have been leveraged here that is also, simple, light weight and easy to read? Please no ES6 stuff as that is Greek to me.
Note: The ToInt function is just a function that returns a valid number or 0.
The jQuery.map() method acts like a flat map (returned sub arrays are flattend). In the map's callback function, use String.search() to check if there's a dash. If not convert to number with the + operator and return. If there's a dash, split, use a for loop to convert the min and max to an array, and return the array.
function convert(str) {
var arr = jQuery.map(str.split(', '), function(s) {
if(s.search('–') === -1) return +s;
var minmax = s.split('–');
var range = [];
for(var i = +minmax[0]; i <= +minmax[1]; i++) range.push(i);
return range;
});
return arr;
}
var number_string = "7, 20, 22, 30–32, 33, 36–40, 46";
var result = convert(number_string);
console.log(result);
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
Using ESNext I would use Array.flatMap() instead of jQuery.map(), with String.inclues() to detect the dash, and Array.from() to generate the sub array.
const convert = (str) =>
str.split(', ')
.flatMap(s => {
if(!s.includes('–')) return +s;
const [min, max] = s.split('–');
return Array.from({ length: max - min + 1 }, (_, n) => n + +min);
});
var number_string = "7, 20, 22, 30–32, 33, 36–40, 46";
var result = convert(number_string, '–');
console.log(result);
I'd use .reduce. First split the initial string by commas. If the string being iterated over doesn't have -, then just push the number to the accumulator; otherwise, split by - to get a low and high number, then use a for loop to push all numbers from low to high to the accumulator:
const ToInt = Number;
const numArr = "7, 20, 22, 30–32, 33, 36–40, 46".split(', ');
const result = numArr.reduce((a, str) => {
if (!str.includes('–')) {
a.push(ToInt(str));
return a;
}
const [low, high] = str.split('–');
for (let i = Number(low); i <= high; i++) {
a.push(i);
}
return a;
}, []);
console.log(result);
If for some reason you don't want to use ES6, you can transform it to ES5 with Babel:
"use strict";
function _slicedToArray(arr, i) {
return (
_arrayWithHoles(arr) || _iterableToArrayLimit(arr, i) || _nonIterableRest()
);
}
function _nonIterableRest() {
throw new TypeError("Invalid attempt to destructure non-iterable instance");
}
function _iterableToArrayLimit(arr, i) {
var _arr = [];
var _n = true;
var _d = false;
var _e = undefined;
try {
for (
var _i = arr[Symbol.iterator](), _s;
!(_n = (_s = _i.next()).done);
_n = true
) {
_arr.push(_s.value);
if (i && _arr.length === i) break;
}
} catch (err) {
_d = true;
_e = err;
} finally {
try {
if (!_n && _i["return"] != null) _i["return"]();
} finally {
if (_d) throw _e;
}
}
return _arr;
}
function _arrayWithHoles(arr) {
if (Array.isArray(arr)) return arr;
}
var ToInt = Number;
var numArr = "7, 20, 22, 30–32, 33, 36–40, 46".split(", ");
var result = numArr.reduce(function(a, str) {
if (str.indexOf("–") === -1) {
a.push(ToInt(str));
return a;
}
var _str$split = str.split("–"),
_str$split2 = _slicedToArray(_str$split, 2),
low = _str$split2[0],
high = _str$split2[1];
for (var i = Number(low); i <= high; i++) {
a.push(i);
}
return a;
}, []);
console.log(result);
(but the ES6 version is more concise and probably a lot easier to read and understand)
I have my own implementation for this purpose
parseStringRange accepts strings like '1,2,3' and '1,2,3-5,6,7', also removes invalid chars (not numbers or , or -) and return one array with all numbers.
the function also returns the numbers ordered and removes duplicates, the user also can input unordered numbers even in range, '1-5' and '5-1' will return same output.
const parseStringRange = (range = '') =>
`${range}`
?.replace(/[^0-9,-]/g, '') //remove invalid chars
?.split(',') //convert 1,2 to [1, 2]
?.flatMap(s => { //convert 1-3 to [1,2,3]
if (!s.includes('-')) return [s]
const [min, max] = s.split('-').map(e => Number(e))
if (min > max) var i = -1
else var i = 1
const r = []
for (let p = min; p != max; p += i)
r.push(p)
r.push(max)
return r;
})
?.map(e => Number(e)) //convert all string numbers to number
?.reduce((t, e, i, a) => {
if (!t)
t = [... new Set(a)]
return t
}, null) //remove duplicates
?.sort((a, b) => a - b) //sort numbers, smallest first
Related
I have an array of following strings:
['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0']
...etc.
I need a solution that will give me following ordered result
['4.5.0', '4.21.0', '4.22.0', '5.1.0', '5.5.1', '6.1.0'].
I tried to implement a sort so it first sorts by the numbers in the first position, than in case of equality, sort by the numbers in the second position (after the first dot), and so on...
I tried using sort() and localeCompare(), but if I have elements '4.5.0' and '4.11.0', I get them sorted as ['4.11.0','4.5.0'], but I need to get ['4.5.0','4.11.0'].
How can I achieve this?
You could prepend all parts to fixed size strings, then sort that, and finally remove the padding again.
var arr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
arr = arr.map( a => a.split('.').map( n => +n+100000 ).join('.') ).sort()
.map( a => a.split('.').map( n => +n-100000 ).join('.') );
console.log(arr)
Obviously you have to choose the size of the number 100000 wisely: it should have at least one more digit than your largest number part will ever have.
With regular expression
The same manipulation can be achieved without having to split & join, when you use the callback argument to the replace method:
var arr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
arr = arr.map( a => a.replace(/\d+/g, n => +n+100000 ) ).sort()
.map( a => a.replace(/\d+/g, n => +n-100000 ) );
console.log(arr)
Defining the padding function once only
As both the padding and its reverse functions are so similar, it seemed a nice exercise to use one function f for both, with an extra argument defining the "direction" (1=padding, -1=unpadding). This resulted in this quite obscure, and extreme code. Consider this just for fun, not for real use:
var arr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
arr = (f=>f(f(arr,1).sort(),-1)) ((arr,v)=>arr.map(a=>a.replace(/\d+/g,n=>+n+v*100000)));
console.log(arr);
Use the sort compare callback function
You could use the compare function argument of sort to achieve the same:
arr.sort( (a, b) => a.replace(/\d+/g, n => +n+100000 )
.localeCompare(b.replace(/\d+/g, n => +n+100000 )) );
But for larger arrays this will lead to slower performance. This is because the sorting algorithm will often need to compare a certain value several times, each time with a different value from the array. This means that the padding will have to be executed multiple times for the same number. For this reason, it will be faster for larger arrays to first apply the padding in the whole array, then use the standard sort, and then remove the padding again.
But for shorter arrays, this approach might still be the fastest. In that case, the so-called natural sort option -- that can be achieved with the extra arguments of localeCompare -- will be more efficient than the padding method:
var arr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
arr = arr.sort( (a, b) => a.localeCompare(b, undefined, { numeric:true }) );
console.log(arr);
More about the padding and unary plus
To see how the padding works, look at the intermediate result it generates:
[ "100005.100005.100001", "100004.100021.100000", "100004.100022.100000",
"100006.100001.100000", "100005.100001.100000" ]
Concerning the expression +n+100000, note that the first + is the unary plus and is the most efficient way to convert a string-encoded decimal number to its numerical equivalent. The 100000 is added to make the number have a fixed number of digits. Of course, it could just as well be 200000 or 300000. Note that this addition does not change the order the numbers will have when they would be sorted numerically.
The above is just one way to pad a string. See this Q&A for some other alternatives.
If you are looking for a npm package to compare two semver version, https://www.npmjs.com/package/compare-versions is the one.
Then you can sort version like this:
// ES6/TypeScript
import compareVersions from 'compare-versions';
var versions = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
var sorted = versions.sort(compareVersions);
You could split the strings and compare the parts.
function customSort(data, order) {
function isNumber(v) {
return (+v).toString() === v;
}
var sort = {
asc: function (a, b) {
var i = 0,
l = Math.min(a.value.length, b.value.length);
while (i < l && a.value[i] === b.value[i]) {
i++;
}
if (i === l) {
return a.value.length - b.value.length;
}
if (isNumber(a.value[i]) && isNumber(b.value[i])) {
return a.value[i] - b.value[i];
}
return a.value[i].localeCompare(b.value[i]);
},
desc: function (a, b) {
return sort.asc(b, a);
}
}
var mapped = data.map(function (el, i) {
return {
index: i,
value: el.split('')
};
});
mapped.sort(sort[order] || sort.asc);
return mapped.map(function (el) {
return data[el.index];
});
}
var array = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0'];
console.log('sorted array asc', customSort(array));
console.log('sorted array desc ', customSort(array, 'desc'));
console.log('original array ', array);
.as-console-wrapper { max-height: 100% !important; top: 0; }
You can check in loop if values are different, return difference, else continue
var a=['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
a.sort(function(a,b){
var a1 = a.split('.');
var b1 = b.split('.');
var len = Math.max(a1.length, b1.length);
for(var i = 0; i< len; i++){
var _a = +a1[i] || 0;
var _b = +b1[i] || 0;
if(_a === _b) continue;
else return _a > _b ? 1 : -1
}
return 0;
})
console.log(a)
Though slightly late this would be my solution;
var arr = ["5.1.1","5.1.12","5.1.2","3.7.6","2.11.4","4.8.5","4.8.4","2.10.4"],
sorted = arr.sort((a,b) => {var aa = a.split("."),
ba = b.split(".");
return +aa[0] < +ba[0] ? -1
: aa[0] === ba[0] ? +aa[1] < +ba[1] ? -1
: aa[1] === ba[1] ? +aa[2] < +ba[2] ? -1
: 1
: 1
: 1;
});
console.log(sorted);
Here's a solution I developed based on #trincot's that will sort by semver even if the strings aren't exactly "1.2.3" - they could be i.e. "v1.2.3" or "2.4"
function sortSemVer(arr, reverse = false) {
let semVerArr = arr.map(i => i.replace(/(\d+)/g, m => +m + 100000)).sort(); // +m is just a short way of converting the match to int
if (reverse)
semVerArr = semVerArr.reverse();
return semVerArr.map(i => i.replace(/(\d+)/g, m => +m - 100000))
}
console.log(sortSemVer(["1.0.1", "1.0.9", "1.0.10"]))
console.log(sortSemVer(["v2.1", "v2.0.9", "v2.0.12", "v2.2"], true))
This seems to work provided there are only digits between the dots:
var a = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0']
a = a.map(function (x) {
return x.split('.').map(function (x) {
return parseInt(x)
})
}).sort(function (a, b) {
var i = 0, m = a.length, n = b.length, o, d
o = m < n ? n : m
for (; i < o; ++i) {
d = (a[i] || 0) - (b[i] || 0)
if (d) return d
}
return 0
}).map(function (x) {
return x.join('.')
})
'use strict';
var arr = ['5.1.2', '5.1.1', '5.1.1', '5.1.0', '5.7.2.2'];
Array.prototype.versionSort = function () {
var arr = this;
function isNexVersionBigger (v1, v2) {
var a1 = v1.split('.');
var b2 = v2.split('.');
var len = a1.length > b2.length ? a1.length : b2.length;
for (var k = 0; k < len; k++) {
var a = a1[k] || 0;
var b = b2[k] || 0;
if (a === b) {
continue;
} else
return b < a;
}
}
for (var i = 0; i < arr.length; i++) {
var min_i = i;
for (var j = i + 1; j < arr.length; j++) {
if (isNexVersionBigger(arr[i], arr[j])) {
min_i = j;
}
}
var temp = arr[i];
arr[i] = arr[min_i];
arr[min_i] = temp;
}
return arr;
}
console.log(arr.versionSort());
This solution accounts for version numbers that might not be in the full, 3-part format (for example, if one of the version numbers is just 2 or 2.0 or 0.1, etc).
The custom sort function I wrote is probably mostly what you're looking for, it just needs an array of objects in the format {"major":X, "minor":X, "revision":X}:
var versionArr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
var versionObjectArr = [];
var finalVersionArr = [];
/*
split each version number string by the '.' and separate them in an
object by part (major, minor, & revision). If version number is not
already in full, 3-part format, -1 will represent that part of the
version number that didn't exist. Push the object into an array that
can be sorted.
*/
for(var i = 0; i < versionArr.length; i++){
var splitVersionNum = versionArr[i].split('.');
var versionObj = {};
switch(splitVersionNum.length){
case 1:
versionObj = {
"major":parseInt(splitVersionNum[0]),
"minor":-1,
"revision":-1
};
break;
case 2:
versionObj = {
"major":parseInt(splitVersionNum[0]),
"minor":parseInt(splitVersionNum[1]),
"revision":-1
};
break;
case 3:
versionObj = {
"major":parseInt(splitVersionNum[0]),
"minor":parseInt(splitVersionNum[1]),
"revision":parseInt(splitVersionNum[2])
};
}
versionObjectArr.push(versionObj);
}
//sort objects by parts, going from major to minor to revision number.
versionObjectArr.sort(function(a, b){
if(a.major < b.major) return -1;
else if(a.major > b.major) return 1;
else {
if(a.minor < b.minor) return -1;
else if(a.minor > b.minor) return 1;
else {
if(a.revision < b.revision) return -1;
else if(a.revision > b.revision) return 1;
}
}
});
/*
loops through sorted object array to recombine it's version keys to match the original string's value. If any trailing parts of the version
number are less than 0 (i.e. they didn't exist so we replaced them with
-1) then leave that part of the version number string blank.
*/
for(var i = 0; i < versionObjectArr.length; i++){
var versionStr = "";
for(var key in versionObjectArr[i]){
versionStr = versionObjectArr[i].major;
versionStr += (versionObjectArr[i].minor < 0 ? '' : "." + versionObjectArr[i].minor);
versionStr += (versionObjectArr[i].revision < 0 ? '' : "." + versionObjectArr[i].revision);
}
finalVersionArr.push(versionStr);
}
console.log('Original Array: ',versionArr);
console.log('Expected Output: ',['4.5.0', '4.21.0', '4.22.0', '5.1.0', '5.5.1', '6.1.0']);
console.log('Actual Output: ', finalVersionArr);
Inspired from the accepted answer, but ECMA5-compatible, and with regular string padding (see my comments on the answer):
function sortCallback(a, b) {
function padParts(version) {
return version
.split('.')
.map(function (part) {
return '00000000'.substr(0, 8 - part.length) + part;
})
.join('.');
}
a = padParts(a);
b = padParts(b);
return a.localeCompare(b);
}
Usage:
['1.1', '1.0'].sort(sortCallback);
const arr = ["5.1.1","5.1.12","5.1.2","3.7.6","2.11.4","4.8.5","4.8.4","2.10.4"];
const sorted = arr.sort((a,b) => {
const ba = b.split('.');
const d = a.split('.').map((a1,i)=>a1-ba[i]);
return d[0] ? d[0] : d[1] ? d[1] : d[2]
});
console.log(sorted);
This can be in an easier way using the sort method without hardcoding any numbers and in a more generic way.
enter code here
var arr = ['5.1.2', '5.1.1', '5.1.1', '5.1.0', '5.7.2.2'];
splitArray = arr.map(elements => elements.split('.'))
//now lets sort based on the elements on the corresponding index of each array
//mapped.sort(function(a, b) {
// if (a.value > b.value) {
// return 1;
// }
// if (a.value < b.value) {
// return -1;
// }
// return 0;
//});
//here we compare the first element with the first element of the next version number and that is [5.1.2,5.7.2] 5,5 and 1,7 and 2,2 are compared to identify the smaller version...In the end use the join() to get back the version numbers in the proper format.
sortedArray = splitArray.sort((a, b) => {
for (i in a) {
if (parseInt(a[i]) < parseInt(b[i])) {
return -1;
break
}
if (parseInt(a[i]) > parseInt(b[i])) {
return +1;
break
} else {
continue
}
}
}).map(p => p.join('.'))
sortedArray = ["5.1.0", "5.1.1", "5.1.1", "5.1.2", "5.7.2.2"]
sort 1.0a notation correct
use native localeCompare to sort 1.090 notation
function log(label,val){
document.body.append(label,String(val).replace(/,/g," - "),document.createElement("BR"));
}
const sortVersions = (
x,
v = s => s.match(/[a-z]|\d+/g).map(c => c==~~c ? String.fromCharCode(97 + c) : c)
) => x.sort((a, b) => (a + b).match(/[a-z]/)
? v(b) < v(a) ? 1 : -1
: a.localeCompare(b, 0, {numeric: true}))
let v=["1.90.1","1.090","1.0a","1.0.1","1.0.0a","1.0.0b","1.0.0.1","1.0a"];
log(' input : ',v);
log('sorted: ',sortVersions(v));
log('no dups:',[...new Set(sortVersions(v))]);
In ES6 you can go without regex.
const versions = ["0.4", "0.11", "0.4.1", "0.4", "0.4.2", "2.0.1","2", "0.0.1", "0.2.3"];
const splitted = versions.map(version =>
version
.split('.')
.map(i => +i))
.map(i => {
let items;
if (i.length === 1) {
items = [0, 0]
i.push(...items)
}
if (i.length === 2) {
items = [0]
i.push(...items)
}
return i
})
.sort((a, b) => {
for(i in a) {
if (a[i] < b[i]) {
return -1;
}
if (a[i] > b[i]) {
return +1;
}
}
})
.map(item => item.join('.'))
const sorted = [...new Set(splitted)]
If ES6 I do this:
versions.sort((v1, v2) => {
let [, major1, minor1, revision1 = 0] = v1.match(/([0-9]+)\.([0-9]+)(?:\.([0-9]+))?/);
let [, major2, minor2, revision2 = 0] = v2.match(/([0-9]+)\.([0-9]+)(?:\.([0-9]+))?/);
if (major1 != major2) return parseInt(major1) - parseInt(major2);
if (minor1 != minor2) return parseInt(minor1) - parseInt(major2);
return parseInt(revision1) - parseInt(revision2);
});
**Sorted Array Object by dotted version value**
var sampleData = [
{ name: 'Edward', value: '2.1.2' },
{ name: 'Sharpe', value: '2.1.3' },
{ name: 'And', value: '2.2.1' },
{ name: 'The', value: '2.1' },
{ name: 'Magnetic', value: '2.2' },
{ name: 'Zeros', value: '0' },
{ name: 'Zeros', value: '1' }
];
arr = sampleData.map( a => a.value).sort();
var requireData = [];
arr.forEach(function(record, index){
var findRecord = sampleData.find(arr => arr.value === record);
if(findRecord){
requireData.push(findRecord);
}
});
console.log(requireData);
[check on jsfiddle.net][1]
[1]: https://jsfiddle.net/jx3buswq/2/
It is corrected now!!!
i am trying to use recursion to return each character in a string. However, the output is not
//We define a function with input parameter.
function countCharInString(string) {
//vi Define an empty objec
const result = {};
//we loop through the length of string
for (let i = 0; i < string.length; i++) {
//create another variable for each element in string
const ch = string[i];
//BASE CASE: if string is empty, return Object with nothing
if (!result[ch]) {
return result[ch]=0;
} else {
//RECURSION: 1 plus whatever the length of the substring from the next character onwards is
return countCharInString(result[ch] + 1)
}
}
}
console.log(countCharInString("Vi skal tælle bogstaver"))
the output should be the following:
var result = {
l : 3,
a : 2,
e : 2,
s : 2,
t : 2,
v : 2,
b: 1,
i : 1,
k : 1,
o : 1,
r : 1,
æ : 1
};
i would suggest to do it with a simple reduce like so
var inputString = 'donald duck';
var result = inputString.split('').reduce((acc, char, index) => {
if (acc[char] !== undefined) {
acc[char] = acc[char] + 1;
}
else {
acc = { ...acc, [char]: 1 }
}
return acc
}, {})
see: https://jsfiddle.net/yswu91zh/21/
Only recursion would not give you the output that you are asking for. After recursively counting character you have to sort it by frequency and then by character. I have excluded a bunch of punctuation with space from counting, if you want exclude more just add it to the punctuation string. You have to use String.prototype.localeCompare() method to compare the characters. This method compares two strings in the current locale. As you are using Danish language you have to specify locale as da.
const punctuations = '.,:;!? ';
const countCharInString = (str, p = {}) => {
if (str.length === 0) return p;
const key = str[0].toLowerCase();
if (!punctuations.includes(key)) {
if (!p[key]) p[key] = 1;
else p[key] += 1;
}
return countCharInString(str.slice(1), p);
};
const cmp = (x, y) => {
if (x[1] === y[1]) {
return x[0].localeCompare(y[0], 'da');
}
return x[1] < y[1] ? 1 : -1;
};
const ret = Object.fromEntries(
Object.entries(countCharInString('Vi skal tælle bogstaver')).sort(cmp)
);
console.log(ret);
please can somebody help?
If i have a total or a sum for instance 91
How can I create an array of the least amount of elements needed to get to the total value?
[50, 20, 10 , 5, 3, 2, 1] totaling this array will provide 91.
I know how to perform the opposite function using reduce or like so:
<script>
var numbers = [65, 44, 12, 4];
function getSum(total, num) {
return total + num;
}
function myFunction(item) {
document.getElementById("demo").innerHTML = numbers.reduce(getSum);
}
</script>
Greedy algorithm
Here is a solution using greedy algorithm. Note that this solution will work correctly in case when all the smaller numbers are divisors of all the bigger numbers such as in case [50, 10, 5, 1]. (see dynamic algorithm below this one for solution that can handle any input)
50 mod 10 = 0
50 mod 5 = 0
50 mod 1 = 0
10 mod 5 = 0
10 mod 1 = 0
5 mod 1 = 0
const sum = xs => xs.reduce((acc, v) => acc + v, 0);
function pickSubset(options, total, currentPick) {
if (sum(currentPick) === total) { return currentPick; }
if (options.length === 0) { return null; }
const firstVal = options[0];
let res = null;
if (sum(currentPick) + firstVal > total) {
res = pickSubset(options.slice(1), total, currentPick);
} else {
let opt1 = pickSubset(options, total, currentPick.concat(options[0]));
let opt2 = pickSubset(options.slice(1), total, currentPick.concat(options[0]));
if (opt1 && opt2) {
opt1.length < opt2.length ? res = opt1 : res = opt2
} else if (opt1) {
res = opt1;
} else {
res = opt2;
}
}
return res;
}
const total = 73;
const options = [50, 25, 10, 5, 2, 1];
console.log(pickSubset(options, total, []));
To handle unsorted input you can wrap it in another function and sort it prior to passing it to the main function.
const sum = xs => xs.reduce((acc, v) => acc + v, 0);
function pickSubset(options, total, currentPick) {
const sortedOptions = options.sort((a, b) => b - a);
function _pickSubset(options, total, currentPick) {
if (sum(currentPick) === total) { return currentPick; }
if (options.length === 0) { return null; }
const firstVal = options[0];
let res = null;
if (sum(currentPick) + firstVal > total) {
res = pickSubset(options.slice(1), total, currentPick);
} else {
let opt1 = pickSubset(options, total, currentPick.concat(options[0]));
let opt2 = pickSubset(options.slice(1), total, currentPick.concat(options[0]));
if (opt1 && opt2) {
opt1.length < opt2.length ? res = opt1 : res = opt2
} else if (opt1) {
res = opt1;
} else {
res = opt2;
}
}
return res;
}
return _pickSubset(sortedOptions, total, currentPick);
}
const total = 73;
const options = [50, 25, 10, 5, 2, 1].reverse();
console.log(pickSubset(options, total, []));
Dynamic programming (bottom-up natural ordering approach)
This solution works correctly for any type of input.
function pickSubset(options, total) {
function _pickSubset(options, change, minNums, numsUsed) {
for (let i = 0; i < change + 1; i++) {
let count = i;
let newNum = 1;
let arr = options.filter(v => v <= i);
for (let j of arr) {
if (minNums[i - j] + 1 < count) {
count = minNums[i - j] + 1;
newNum = j;
}
}
minNums[i] = count;
numsUsed[i] = newNum;
}
return minNums[change];
}
function printNums(numsUsed, change) {
const res = [];
let num = change;
while (num > 0) {
let thisNum = numsUsed[num];
res.push(thisNum);
num = num - thisNum;
}
return res;
}
const numsUsed = [];
const numsCount = [];
_pickSubset(options, total, numsCount, numsUsed);
return printNums(numsUsed, total);
}
const options = [50, 10, 5, 2, 1];
console.log(pickSubset(options, 73));
Dynamic programming (top-down memoization approach)
// helper function that generates all the possible solutions
// meaning, all the possible ways in which we can pay the provided amount
// and caches those solutions;
// returns the number of possible solutions but that is not neccessary
// in this case
const _pickSubset = (toPay, options, currentPick, cache) => {
if (toPay < 0) { return 0; }
if (toPay === 0) {
cache.add(currentPick);
return 1;
}
if (options.length === 0) { return 0; }
return _pickSubset(toPay - options[0], options, currentPick.concat(options[0]), cache)
+ _pickSubset(toPay, options.slice(1), currentPick, cache);
};
// memoize only with respect to the first two arguments - toPay, bills
// the other two are not necessary in this case
const memoizeFirstTwoArgs = fn => {
const cache = new Map();
return (...args) => {
const key = JSON.stringify(args.slice(0, 2));
if (cache.has(key)) { return cache.get(key); }
const res = fn(...args);
cache.set(key, res);
return res;
};
};
// uses memoized version of makeChange and provides cache to that function;
// after cache has been populated, by executing memoized version of makeChange,
// find the option with smallest length and return it
const pickSubset = (toPay, options) => {
const cache = new Set();
const memoizedPickSubset = memoizeFirstTwoArgs(_pickSubset);
memoizedPickSubset(toPay, options, [], cache);
let minLength = Infinity;
let resValues;
for (const value of cache) {
if (value.length < minLength) {
minLength = value.length;
resValues = value;
}
}
return resValues;
}
const options = [50, 25, 10, 5, 2, 1];
const toPay = 73;
console.log(pickSubset(toPay, options));
I need help. Can you help me to solve this task.
Write the function that returns the summ of all numbers even string numbers that are passed as arguments and the amount of these numbers is unlimited.
If one of the parameters is an array, then
Sum is also added to the sum of the values of this array (if any of the
Values of this array is also an array, then the result is added
Also the sum of its values, and so on).
var sum = getSum (1, '1', 'one', [2, '2', 'two']);
console.log(sum);
I've tried to write something like this:
function getSum(){
var separeted = string.split(",");
var sum =0;
for(var i=0;i<arguments.length;i++){
sum += parseInt(arguments[i].toString()..match(/(\d+)/));
}
return sum;
}
var sum = getSum(1,"1","one",[2,'2', 'two']);
console.log(sum);
and here I've stacked. I think that I will need split and filter but how to combine that all in one function I can't find.
Thanks.
You could define a recursive function like so:
function getSum () {
return Array.from(arguments).reduce((sum, value) => {
if (Array.isArray(value)) {
sum += getSum.apply(this, value)
} else {
sum += Number(value)
}
return sum
}, 0)
}
var sum = getSum (1, '1', '3', [2, '2', '10']);
console.log(sum);
In order to account for string numbers, you must define a mapping from words to numbers and iteratively traverse the string of words to generate the number. Here's a possible implementation of that:
var stringToNumber = (function () {
const primary = new Map([
['zero', 0],
['one', 1],
['two', 2],
['three', 3],
['four', 4],
['five', 5],
['six', 6],
['seven', 7],
['eight', 8],
['nine', 9]
])
const secondary = new Map([
['ten', 10],
['eleven', 11],
['twelve', 12],
['thirteen', 13],
['fourteen', 14],
['fifteen', 15],
['sixteen', 16],
['seventeen', 17],
['eighteen', 18],
['nineteen', 19]
])
const prefix = new Map([
['twenty', 20],
['thirty', 30],
['forty', 40],
['fifty', 50],
['sixty', 60],
['seventy', 70],
['eighty', 80],
['ninety', 90]
])
const magnitude = new Map([
['hundred', 1e2],
['thousand', 1e3],
['million', 1e6],
['billion', 1e9],
['trillion', 1e12],
['quadrillion', 1e15],
['quintillion', 1e18],
['sextillion', 1e21],
['septillion', 1e24],
['octillion', 1e27],
['nonillion', 1e30],
['decillion', 1e33]
])
const types = { primary, secondary, prefix, magnitude }
class Parser {
static parse(word) {
if (isNaN(word)) {
const [type = null] = Object.keys(types)
.filter(type => types[type].has(word))
const value = types[type] ? types[type].get(word) : NaN
return { type, sign: 1, value }
} else {
const value = Math.abs(word)
const sign = Math.sign(word)
const [type = 'primary'] = Object.keys(types)
.filter(type => Array.from(types[type].values()).includes(value))
return { type, sign, value }
}
}
constructor() {
this.words = []
}
push(word) {
const parsed = Parser.parse(word)
if (parsed.type === null) {
return this.words.length
}
return this.words.push(parsed)
}
valueOf() {
if (this.words.length === 0) {
return NaN
}
const words = this.words
let total = 0
let { type: lastType, sign, value: run } = words[0]
let maxMagnitude = lastType === 'magnitude' ? run : 1
for (const { type, value } of words.slice(1)) {
switch (type) {
case 'magnitude':
if (value > maxMagnitude) {
run = (total + run) * value
total = 0
maxMagnitude = value
} else {
run *= value
}
break
case 'secondary':
case 'prefix':
switch (lastType) {
case 'magnitude':
total += run
run = value
break
case 'primary':
case 'secondary':
case 'prefix':
run = Number(String(run) + String(value))
}
break
case 'primary':
switch (lastType) {
case 'magnitude':
total += run
run = value
break
case 'prefix':
run += value
break
case 'primary':
case 'secondary':
run = Number(String(run) + String(value))
}
}
lastType = type
}
return sign * (total + run)
}
}
return function stringToNumber (string) {
const words = string
.trim()
.toLowerCase()
.split(/\s+/g)
const parser = new Parser()
for (const word of words) {
parser.push(word)
}
return parser.valueOf()
}
})()
function getSum () {
return Array.from(arguments).reduce((sum, value) => {
if (Array.isArray(value)) {
sum += getSum.apply(this, value)
} else if (!isNaN(value)) {
sum += Number(value)
} else {
sum += stringToNumber(String(value))
}
return sum
}, 0)
}
var sum = getSum (1, '1', 'one hundred thousand', [2, '2', 'twenty six hundred']);
console.log(sum);
Thanks to everyone I found the solution in the old "O'reilly Javascript by David Flanagan"
so as I thought in my loop I had to check if it's not null than change the element into Number()
Here is the code from the book
function flexisum(a) {
var total = 0;
for(var i = 0; i < arguments.length; i++) {
var element = arguments[i], n;
if (element == null) continue; // Ignore null & undefined
if (isArray(element)) // if argument is array
n = flexisum.apply(this, element); // count sum of all elements
else if (typeof element === "function") // if it's a function
n = Number(element()); // call and convert it to the 'number'
else n = Number(element); // or to convert it anyway
if (isNaN(n)) // If it was not possible to convert to a number, initiate exc.
throw Error("flexisum(): can't convert " + element +
"into the number");
total += n; // Otherwise, add n to the total
}
return total;
}
but in my case I've chaged it for my needs and took away the exc. so it could count sum without string words.
if (isNaN(n)) continue;
total += n;
}
So, thanks to everybody.
Books are the good sourse))))
i think need to iterate any number of label ,so you can use following method
function jsArrSum(arr) {
var sum = 0;
for (var i = 0; i < arr.length; i++) {
if (typeof arr[i] == 'object')
sum += jsArrSum(arr[i]);
else
sum += arr[i];
}
return sum;
}
jsArrSum(array);
I was able to pull all single integers after 'reduce', but not working when there's all duplicates and output should be 0, not hitting my else or else if - code keeps outputting 0 vs the single integers
var singleNumber = function(nums) {
var sorted_array = nums.sort();
for (var i=0; i < sorted_array.length; i++){
var previous = sorted_array[i-1];
var next = sorted_array[i+1];
var singles = {key: 0};
var singlesArray = [];
if (sorted_array[i] !== previous && sorted_array[i] !== next){
singlesArray.push(sorted_array[i]);
singlesArray.reduce(function(singles, key){
singles.key = key;
//console.log('key', key);
return singles.key;
},{});
}
else if(singlesArray.length === 0) {
singles.key = 0;
return singles.key;
}
}
console.log('singles.key', singles.key);
return singles.key;
};
console.log(singleNumber([2,1,3,4,4]));
// tests
const n1 = [1,2,3,4,4] //[1,2,3]
const n2 = [1] //[1]
const n3 = [1,1] //0
const n4 = [1,1,1] //0
const n5 = [1,5,3,4,5] //[1,3,4]
const n6 = [1,2,3,4,5] //[1,2,3,4,5]
const n7 = [1,5,3,4,5,6,7,5] //[1,3,4,6,7]
const singleNumber = numbers => {
const reducer = (acc, val) => {
// check to see if we have this key
if (acc[val]) {
// yes, so we increment its value by one
acc[val] = acc[val] + 1
} else {
// no, so it's a new key and we assign 1 as default value
acc[val] = 1
}
// return the accumulator
return acc
}
// run the reducer to group the array into objects to track the count of array elements
const grouped = numbers.reduce(reducer, {})
const set = Object.keys(grouped)
// return only those keys where the value is 1, if it's not 1, we know its a duplicate
.filter(key => {
if (grouped[key] == 1) {
return true
}
})
// object.keys makes our keys strings, so we need run parseInt to convert the string back to integer
.map(key => parseInt(key))
// check to array length. If greater than zero, return the set. If it is zero, then all the values were duplicates
if (set.length == 0) {
return 0
} else {
// we return the set
return set
}
}
console.log(singleNumber(n7))
https://jsbin.com/sajibij/edit?js,console