Related
I have a challenge to complete where I'm given an array [-1,4,-3,5,6,9,-2] and I need to get a new array that sorts the numbers in this order: [firstGreatest, firstLowest, secondGreatest, secondLowest ...and so on]. The negative and positive numbers may be different amount, as in 4 positive, 2 negative.
This is what I tried so far, but cannot think of a better solution.
let arr = [-1, 2, -5, 3, 4, -2, 6];
function someArray(ary) {
const sorted = ary.sort((a, b) => a - b)
const highest = sorted.filter(num => num > 0).sort((a, b) => b - a)
const lowest = sorted.filter(num => num < 0).sort((a, b) => b - a)
let copy = highest
for (let i = 0; i < highest.length; i++) {
for (let j = i; j < lowest.length; j++) {
if ([i] % 2 !== 0) {
copy.splice(1, 0, lowest[j])
}
}
}
}
console.log(arr)
someArray(arr)
console.log(arr)
You can easily solve this problem with two pointers algorithm.
O(n log n) for sorting
O(n) for add the value in result.
Take two-variable i and j,
i points to the beginning of the sorted array
j points to the end of the sorted array
Now just add the value of the sorted array alternatively in final result
let arr = [-1, 2, -5, 3, 4, -2, 6];
function someArray(ary) {
const sorted = arr.sort((a, b) => b - a);
// declaration
const result = [];
let i = 0,
j = sorted.length - 1,
temp = true;
// Algorithm
while (i <= j) {
if (temp) {
result.push(sorted[i]);
i++;
} else {
result.push(sorted[j]);
j--;
}
temp = !temp;
}
return result;
}
console.log(someArray(arr));
.as-console-wrapper { max-height: 100% !important; top: 0; }
You could sort the array and pop or shift until you have no more items.
function greatestLowest(array) {
let temp = [...array].sort((a, b) => a - b),
m = 'shift',
result = [];
while (temp.length) result.push(temp[m = { pop: 'shift', shift: 'pop' }[m]]());
return result;
}
console.log(...greatestLowest([-1, 2, -5, 3, 4, -2, 6]));
The general idea is to sort the array (highest to lowest) then pick the first and the last element until the array is empty. One way of doing it could be:
const input = [-1, 2, -5, 3, 4, -2, 6];
function someArray(arr) {
// sort the original array from highest to lowest
const sorted = arr.sort((a, b) => b - a);
const output = []
while (sorted.length > 0) {
// remove the first element of the sorted array and push it into the output
output.push(...sorted.splice(0, 1));
// [check to handle arrays with an odd number of items]
// if the sorted array still contains items
// remove also the last element of the sorted array and push it into the output
if (sorted.length > 0) output.push(...sorted.splice(sorted.length - 1, 1))
}
return output;
}
// test
console.log(`input: [${input.join(',')}]`);
console.log(`input (sorted desc): [${input.sort((a, b) => b - a).join(',')}]`)
console.log(`output: [${someArray(input).join(',')}]`);
This is a simple and a shorter method:
function makearray(ar) {
ar = points.sort(function(a, b) {
return b - a
})
let newarray = []
let length = ar.length
for (let i = 0; i < length; i++) {
if (i % 2 == 0) {
newarray.push(ar[0])
ar.splice(0, 1)
} else {
newarray.push(ar[ar.length - 1])
ar.splice(ar.length - 1, 1)
}
}
return newarray
}
const points = [-1, 2, -5, 3, 4, -2, 6]
console.log(makearray(points))
Here is my attempt; a slightly-modified Fisher-Yates algorithm. I am not sure how to make sure it's random though.
const shuffleWithoutMovingFalsies = array => {
const newArray = [...array];
const getRandomValue = (i, N) => ~~(Math.random() * (N - i) + i);
newArray.forEach((elem, i, arr, j = getRandomValue(i, arr.length)) => arr[i] && arr[j] && ([arr[i], arr[j]] = [arr[j], arr[i]]));
return newArray;
}
const array = [1, 2, null, 3, null, null, 4, 5, 6, null];
const shuffledArray = shuffleWithoutMovingFalsies(array);
console.log(shuffledArray);
All I did was add arr[i] && arr[j] && as a check to make sure both elements about to be swapped are NOT falsy.
That stops it from being a fair shuffle. For example, with the array [1, null, 2], 1 should have a 50% chance of staying put and a 50% chance of swapping with 2, but instead, the split is ⅔–⅓.
As long as the auxiliary memory isn’t an issue, I’d recommend extracting the elements, shuffling them, and putting them back for simplicity:
const shuffle = arr => {
for (let i = 0; i < arr.length - 1; i++) {
const j = i + Math.floor(Math.random() * (arr.length - i));
[arr[i], arr[j]] = [arr[j], arr[i]];
}
};
const shuffleTruthy = arr => {
const truthy = arr.filter(Boolean);
shuffle(truthy);
let j = 0;
for (let i = 0; i < arr.length; i++) {
if (arr[i]) {
arr[i] = truthy[j++];
}
}
};
I've gotten stuck on this practice problem in my software engineering bootcamp, and am hoping that someone can point me in the right direction.
Write a function generatePairs that accepts an integer and generates an array containing the pairs of integers [a, b]. The pairs should be sorted by increasing values of a then increasing values of b.
here are some examples of what should be returned for different inputs:
generatePairs(3) // [ [0, 0], [0, 1], [0, 2], [0, 3], [1, 1], [1, 2], [1, 3], [2, 2], [2, 3], [3, 3] ]
generatePairs(2) // [ [0, 0], [0, 1], [0, 2], [1, 1], [1, 2], [2, 2] ]
generatePairs(1) // [ [0, 0], [0, 1], [1,1]]
generatePairs(0) // [ [0, 0]]
and here is my code so far:
function generatePairs (num){
array = [];
// 0 [0,0] [0,1]
// 1
// 2
for (i = 0; i<=num; i++){
array.push([i,i]);
if ((i+1)<=num) {
array.push([i,i+1])
}
if ( num - i <= num && i===0 && num < i ) {
array.push([i,num])
if (num + i < i) {
array.pop();
}
}
}
return array;
}
generatePairs(2) // [ [0, 0], [0, 1], [0, 2], [1, 1], [1, 2], [2, 2] ]
the issue I'm running into is that, for example, when I try it out with 2, I'm missing the [0,2] subarray. The methods I've tried to work around this have mostly consisted of additional if and else loops, but with each one I've tried I either end up with subarrays at the end that go higher than they should, or a semi-working system that would only work for 2 and not for any number that could be inputted into the function.
Isn't it much simpler than that?
function generatePairs(num) {
let arr = []
for (let i = 0; i <= num; i++) {
for (let j = i; j <= num; j++) {
arr.push([i, j])
}
}
return arr
}
console.log(generatePairs(2));
solution: (ninja code)
const generatePairs = n => Array.apply(null,{length:((n+1)*(n+2)/2)}).reduceRight((a,c,i)=>{
a.r.push([...a.p])
if(++a.p[1]>n) a.p[1]=++a.p[0]
return i?a:a.r
},{p:[0,0],r:[]})
for (let x=4;x--;) document.write(x,'->',JSON.stringify(generatePairs(x)), '<br>')
This is what you want right here. You loop through num for the first part until you get to num and for each first part loop from that first part to num.
function generatePairs(num) {
var output = []
for (var i = 0; i < num + 1; i++) {
for (var k = i; k < num + 1; k++) {
output.push([i, k])
}
}
return output
}
recursion
We can implement generatePairs in an wishful way -
const generatePairs = (n = 0) =>
chooseN(2, range(0, n))
range is a simple function that generates an array from a to b -
const range = (a = 0, b = 0) =>
a > b // base
? []
: [ a, ...range(a + 1, b) ] // inductive: a <= b
and chooseN is a generic that generates all n-sized samples from array a -
const chooseN = (n = 0, a = []) =>
n <= 0 // base
? [[]]
: a.length <= 0 // inductive: n > 0
? []
: chooseN(n - 1, a) // inductive: n > 0, non-empty a
.map(r => [a[0], ...r])
.concat(chooseN(n, a.slice(1)))
See generatePairs working in your own browser below -
const range = (a = 0, b = 0) =>
a > b
? []
: [ a, ...range(a + 1, b) ]
const chooseN = (n = 0, a = []) =>
n <= 0
? [[]]
: a.length <= 0
? []
: chooseN(n - 1, a)
.map(r => [a[0], ...r])
.concat(chooseN(n, a.slice(1)))
const generatePairs = (n = 0) =>
chooseN(2, range(0, n))
const log = x =>
console.log(JSON.stringify(x))
log(generatePairs(3))
// [[0,0],[0,1],[0,2],[0,3],[1,1],[1,2],[1,3],[2,2],[2,3],[3,3]]
log(generatePairs(2))
// [[0,0],[0,1],[0,2],[1,1],[1,2],[2,2]]
log(generatePairs(1))
// [[0,0],[0,1],[1,1]]
log(generatePairs(0))
// [[0,0]]
generators
Because combinatorial problems typically involve big solution spaces, it's common to generate combinations lazily. In JavaScript, we can do this using generators -
const chooseN = function* (n = 0, a = [])
{ if (n <= 0)
return yield []
if (a.length <= 0)
return
for (const r of chooseN(n - 1, a))
yield [a[0], ...r]
yield* chooseN(n, a.slice(1))
}
Notice the structurally similarity between this program and the one above -
const chooseN = function* (n = 0, a = [])
{ if (n <= 0) // if (n <= 0)
return yield [] // return [[]]
if (a.length <= 0) // else if (a.length <= 0)
return // return []
// else return:
for (const r of chooseN(n - 1, a)) // chooseN(n - 1, a).map(r =>
yield [a[0], ...r] // [a[0],...r])
yield* chooseN(n, a.slice(1)) // .concat(chooseN(n, a.slice(1)))
}
For example, we could write a solver that finds the first pair, [ a, b ] where a > 3 and 3*a is equal to 2*b. Critically, no pairs will be generated after the first solution is found -
const solver = (size = 0) =>
{ for(const [a, b] of generatePairs(size))
if (a > 3)
if (3 * a === 2 * b)
return [a, b]
}
console.log(solver(10))
// [ 4, 6 ]
And the solution a = 4, b = 6 is correct: 4 > 3 is true and 3*4 is equal to 2*6 (12).
Below, we can generate the entire array of pairs, if we wish, using Array.from -
const allPairs =
Array.from(generatePairs(3)) // <-- Array.from exhausts an iterable
console.log(allPairs)
// [[0,0],[0,1],[0,2],[0,3],[1,1],[1,2],[1,3],[2,2],[2,3],[3,3]]
Expand the snippet below to generate pairs using JavaScript's generators -
const range = (a = 0, b = 0) =>
a > b
? []
: [ a, ...range(a + 1, b) ]
const chooseN = function* (n = 0, a = [])
{ if (n <= 0)
return yield []
if (a.length <= 0)
return
for (const r of chooseN(n - 1, a))
yield [a[0], ...r]
yield* chooseN(n, a.slice(1))
}
const generatePairs = (n = 0) =>
Array.from(chooseN(2, range(0, n)))
const log = x =>
console.log(JSON.stringify(x))
log(generatePairs(3))
// [[0,0],[0,1],[0,2],[0,3],[1,1],[1,2],[1,3],[2,2],[2,3],[3,3]]
log(generatePairs(2))
// [[0,0],[0,1],[0,2],[1,1],[1,2],[2,2]]
log(generatePairs(1))
// [[0,0],[0,1],[1,1]]
log(generatePairs(0))
// [[0,0]]
Write a JS program to return an array in such a way that the first element is the first minimum and the second element is the first maximum and so on.
This program contains a function which takes one argument: an array. This function returns the array according to the requirement.
Sample Input: array=[2,4,7,1,3,8,9]. Expected Output: [1,9,2,8,3,7,4].
const arrsort=(arr)=>{
return arr.sort(function(a, b){return a - b});
}
const test=(arr)=>{
arr=arrsort(arr);
var arr2=[];
var j=0;
var k=arr.length-1;
for (var i=0;i<arr.length-1;i++){
if(i%2===0){
arr2.push(arr[j]);
j++;
}
else{
arr2.push(arr[k]);
k--;
}
}
return arr2;
}
Instead of using two indices, you could shift and pop the values of a copy of the sorted array.
var array = [2, 4, 7, 1, 3, 8, 9]
const arrsort = arr => arr.sort((a, b) => a - b);
const test = (arr) => {
var copy = arrsort(arr.slice()),
result = [],
fn = 'pop';
while (copy.length) {
fn = { pop: 'shift', shift: 'pop' }[fn];
result.push(copy[fn]());
}
return result;
}
console.log(test(array));
You can first sort() the array in ascending order and then loop through half of the array. And push() the values at corresponding indexes.
let arr = [2,4,7,1,3,8,9];
function order(arr){
let res = [];
arr = arr.slice().sort((a,b) => a-b);
for(let i = 0; i < Math.floor(arr.length/2); i++){
res.push(arr[i],arr[arr.length - 1 - i]);
}
return arr.length % 2 ? res.concat(arr[Math.floor((arr.length - 1)/2)]) : res;
}
console.log(order(arr))
You could sort the array, then copy and reverse and push to another array
const a = [2,4,7,1,3,8,9];
a.sort();
const b = a.slice().reverse();
const res = [];
for (let i = 0; i < a.length; i++) {
if (res.length < a.length) res.push(a[i]);
if (res.length < a.length) res.push(b[i]);
}
console.log(res);
Or use a Set
const a = [2,4,7,1,3,8,9];
a.sort();
const b = a.slice().reverse();
const res = new Set();
a.forEach((e, i) => (res.add(e), res.add(b[i])));
console.log(Array.from(res));
There are many ways are available to do this. And my solution is one of themm i hope.
Find max and min value, push them into another array. And delete max, min from actual array.
let array=[2,4,7,1,3,8,9];
let finalArray = [];
let max, min;
for(let i = 0; i < array.length; i++) {
max = Math.max(...array);
min = Math.min(...array);
finalArray.push(min);
finalArray.push(max);
array = array.filter(function(el) {
return el != max && el != min;
})
}
console.log(finalArray);
After sorting array this would work
myarr = [1,2,3,4,5,6];
let lastindex = myarr.length-1;
for(let i = 1 ; i <= myarr.length/ 2; i = i+2) {
ele = myarr[i];
myarr[i] = myarr[lastindex];
myarr[lastindex] = ele;
lastindex--;
}
Final Output will be: [1, 6, 3, 5, 4, 2]
You can use two iterators after sorting your array, one goes ascending and the other goes descending, until they cross each other.
Here's the code:
const array = [2, 4, 7, 1, 3, 8, 9];
const test = arr => {
const result = [];
const sortedArr = array.sort((a, b) => a - b);
for (let i = 0, j = sortedArr.length - 1; i <= j; i++, j--) {
result.push(sortedArr[i]);
i == j || result.push(sortedArr[j]);
}
return result;
};
console.log(test(array));
You can easily achieve the result using two pointer algorithm
function getValue(arr) {
const result = [];
let start = 0,
end = arr.length - 1;
while (start < end) result.push(arr[start++], arr[end--]);
if (start === end) result.push(arr[start]);
return result;
}
const array = [2, 4, 7, 1, 3, 8, 9];
const sorted = array.sort();
console.log(getValue(sorted));
I want to split an array into pairs of arrays.
var arr = [2, 3, 4, 5, 6, 4, 3, 5, 5]
would be
var newarr = [
[2, 3],
[4, 5],
[6, 4],
[3, 5],
[5]
]
You can use js reduce
initialArray.reduce(function(result, value, index, array) {
if (index % 2 === 0)
result.push(array.slice(index, index + 2));
return result;
}, []);
Lodash has a method for this: https://lodash.com/docs/4.17.10#chunk
_.chunk([2,3,4,5,6,4,3,5,5], 2);
// => [[2,3],[4,5],[6,4],[3,5],[5]]
There's no pre-baked function to do that, but here's a simple solution:
var splitPairs = function(arr) {
var pairs = [];
for (var i=0 ; i<arr.length ; i+=2) {
if (arr[i+1] !== undefined) {
pairs.push ([arr[i], arr[i+1]]);
} else {
pairs.push ([arr[i]]);
}
}
return pairs;
};
Yet another that's a bit of a mish-mash of the already-posted answers. Adding it because having read the answers I still felt things could be a little easier to read:
var groups = [];
for(var i = 0; i < arr.length; i += 2)
{
groups.push(arr.slice(i, i + 2));
}
There is now the flexible Array#flatMap(value, index, array):
const pairs = arr.flatMap((_, i, a) => i % 2 ? [] : [a.slice(i, i + 2)]);
And the possibly more efficient, but goofy looking Array.from(source, mapfn?):
const pairs = Array.from({ length: arr.length / 2 }, (_, i) => arr.slice(i * 2, i * 2 + 2))
It's possible to group an array into pairs/chunks in one line without libraries:
function chunks(arr, size = 2) {
return arr.map((x, i) => i % size == 0 && arr.slice(i, i + size)).filter(x => x)
}
console.log(chunks([1, 2, 3, 4, 5, 6, 7])) // -> [[1, 2], [3, 4], [5, 6], [7]]
Here's a good generic solution:
function splitInto(array, size, inplace) {
var output, i, group;
if (inplace) {
output = array;
for (i = 0; i < array.length; i++) {
group = array.splice(i, size);
output.splice(i, 0, group);
}
} else {
output = [];
for (i = 0; i < array.length; i += size) {
output.push(array.slice(i, size + i));
}
}
return output;
}
For your case, you can call it like this:
var arr= [2,3,4,5,6,4,3,5,5];
var newarr = splitInto(arr, 2);
The inplace argument determines whether the operation is done in-place or not.
Here's a demo below:
function splitInto(array, size, inplace) {
var output, i, group;
if (inplace) {
output = array;
for (i = 0; i < array.length; i++) {
group = array.splice(i, size);
output.splice(i, 0, group);
}
} else {
output = [];
for (i = 0; i < array.length; i += size) {
output.push(array.slice(i, size + i));
}
}
return output;
}
var arr= [2,3,4,5,6,4,3,5,5];
var newarr = splitInto(arr, 2);
disp(newarr);
// or we can do it in-place...
splitInto(arr, 3, true);
disp(arr);
function disp(array) {
var json = JSON.stringify(array);
var text = document.createTextNode(json);
var pre = document.createElement('pre');
pre.appendChild(text);
document.body.appendChild(pre);
}
A slightly different approach than using a for loop for comparison. To avoid modifying the original array slice makes a shallow copy since JS passes objects by reference.
function pairArray(a) {
var temp = a.slice();
var arr = [];
while (temp.length) {
arr.push(temp.splice(0,2));
}
return arr;
}
var array = [2,3,4,5,6,4,3,5,5];
var newArr = pairArray(array);
function pairArray(a) {
var temp = a.slice();
var arr = [];
while (temp.length) {
arr.push(temp.splice(0,2));
}
return arr;
}
document.write('<pre>' + JSON.stringify(newArr) + '</pre>');
I would use lodash for situations like this.
Here is a solution using _.reduce:
var newArr = _(arr).reduce(function(result, value, index) {
if (index % 2 === 0)
result.push(arr.slice(index, index + 2));
return result;
}, []);
var arr = [2,3,4,5,6,4,3,5,5];
var newArr = _(arr).reduce(function(result, value, index) {
if (index % 2 === 0)
result.push(arr.slice(index, index + 2));
return result;
}, []);
document.write(JSON.stringify(newArr)); // [[2,3],[4,5],[6,4],[3,5],[5]]
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/3.10.0/lodash.min.js"></script>
Here's another solution using lodash helpers:
function toPairs(array) {
const evens = array.filter((o, i) => i % 2);
const odds = array.filter((o, i) => !(i % 2));
return _.zipWith(evens, odds, (e, o) => e ? [o, e] : [o]);
}
console.log(toPairs([2,3,4,5,6,4,3,5,5]));
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.13.1/lodash.min.js"></script>
const items = [1, 2, 3, 4, 5];
const createBucket = (bucketItems, bucketSize) => buckets => {
return bucketItems.length === 0 ? buckets : [...buckets, bucketItems.splice(0, bucketSize)];
};
const bucketWithItems = items.reduce(createBucket([...items], 4), []);
Here is a short and more generic solution:
function splitArrayIntoPairs(arr, n) {
var len = arr.length
var pairs = []
for (let i = 0; i < len; i += n) {
var temp = []
for (var j = i; j < (i + n); j++) {
if (arr[j] !== undefined) {
temp.push(arr[j])
}
}
pairs.push(temp)
}
return pairs
}
Where arr is your array and n is no of pairs
This combines some of the answers above but without Object.fromEntires. The output is similar to what you would get with minimist.
const splitParameters = (args) => {
const split = (arg) => (arg.includes("=") ? arg.split("=") : [arg]);
return args.reduce((params, arg) => [...params, ...split(arg)], []);
};
const createPairs = (args) =>
Array.from({ length: args.length / 2 }, (_, i) =>
args.slice(i * 2, i * 2 + 2)
);
const createParameters = (pairs) =>
pairs.reduce(
(flags, value) => ({
...flags,
...{ [value[0].replace("--", "")]: value[1] }
}),
{}
);
const getCliParameters = (args) => {
const pairs = createPairs(splitParameters(args));
const paramaters = createParameters(pairs);
console.log(paramaters);
return paramaters;
};
//const argsFromNodeCli = process.argv.slice(2); // For node
const testArgs = [
"--url",
"https://www.google.com",
"--phrases=hello,hi,bye,ok"
];
const output = getCliParameters(testArgs);
document.body.innerText = JSON.stringify(output);
Here is another concise but still efficient solution using modern JavaScript (arrow function, Array.prototype.at):
splitPairs = arr =>
arr.reduce((pairs, n, i) =>
(i % 2 ? pairs.at(-1).push(n)
: pairs.push([n]),
pairs), []);
It is (memory-)efficient because it just creates one array for the result and one array for each pair and then modifies them. The case where there is an odd number of elements is handled naturally.
When minified, it is also really concise code:
splitPairs = a=>a.reduce((p,n,i)=>(i%2?p.at(-1)[1]=n:p.push([n]),p),[]);
Using ES6 features:
const arr = [2, 3, 4, 5, 6, 4, 3, 5, 5]
const result = arr.slice(arr.length/2).map((_,i)=>arr.slice(i*=2,i+2))
console.log(result)
Here is another generic solution that uses a generator function.
/**
* Returns a `Generator` of all unique pairs of elements from the given `iterable`.
* #param iterable The collection of which to find all unique element pairs.
*/
function* pairs(iterable) {
const seenItems = new Set();
for (const currentItem of iterable) {
if (!seenItems.has(currentItem)) {
for (const seenItem of seenItems) {
yield [seenItem, currentItem];
}
seenItems.add(currentItem);
}
}
}
const numbers = [1, 2, 3, 2];
const pairsOfNumbers = pairs(numbers);
console.log(Array.from(pairsOfNumbers));
// [[1,2],[1,3],[2,3]]
What I like about this approach is that it will not consume the next item from the input until it actually needs it. This is especially handy if you feed it a generator as input, since it will respect its lazy execution.