I have been looking around the web for a while and I am wondering if there is a 'stable' defacto implementation of quicksort that is generally used? I can write my own but why reinvent the wheel...
Quicksort (recursive)
function quicksort(array) {
if (array.length <= 1) {
return array;
}
var pivot = array[0];
var left = [];
var right = [];
for (var i = 1; i < array.length; i++) {
array[i] < pivot ? left.push(array[i]) : right.push(array[i]);
}
return quicksort(left).concat(pivot, quicksort(right));
};
var unsorted = [23, 45, 16, 37, 3, 99, 22];
var sorted = quicksort(unsorted);
console.log('Sorted array', sorted);
You can easily "stabilize" an unstable sort using a decorate-sort-undecorate pattern
function stableSort(v, f)
{
if (f === undefined) {
f = function(a, b) {
a = ""+a; b = ""+b;
return a < b ? -1 : (a > b ? 1 : 0);
}
}
var dv = [];
for (var i=0; i<v.length; i++) {
dv[i] = [v[i], i];
}
dv.sort(function(a, b){
return f(a[0], b[0]) || (a[1] - b[1]);
});
for (var i=0; i<v.length; i++) {
v[i] = dv[i][0];
}
}
the idea is to add the index as last sorting term so that no two elements are now "the same" and if everything else is the same the original index will be the discriminating factor.
Put your objects into an array.
Call Array.sort(). It's very fast.
var array = [3,7,2,8,2,782,7,29,1,3,0,34];
array.sort();
console.log(array); // prints [0, 1, 2, 2, 29, 3, 3, 34, 7, 7, 782, 8]
Why does that print in lexicographic order? That's how array.sort() works by default, e.g. if you don't provide a comparator function. Let's fix this.
var array = [3,7,2,8,2,782,7,29,1,3,0,34];
array.sort(function (a, b)
{
return a-b;
});
console.log(array); // prints [0, 1, 2, 2, 3, 3, 7, 7, 8, 29, 34, 782]
Quick Sort (ES6)
function quickSort(arr) {
if (arr.length < 2) {
return arr;
}
const pivot = arr[Math.floor(Math.random() * arr.length)];
let left = [];
let right = [];
let equal = [];
for (let val of arr) {
if (val < pivot) {
left.push(val);
} else if (val > pivot) {
right.push(val);
} else {
equal.push(val);
}
}
return [
...quickSort(left),
...equal,
...quickSort(right)
];
}
Few notes:
A random pivot keeps the algorithm efficient even when the data is sorted.
As much as it nice to use Array.filter instead of using for of loop, like some of the answers here, it will increase time complexity (Array.reduce can be used instead though).
A Functional equivalent
In celebration of Functional Javascript, which appears to be the in thing
at the moment, especially given ES6+ wonderful syntactic sugar additions. Arrow functions and destructuring I propose a very clean, short functional equivalent of the quicksort function. I have not tested it for performance or compared it to the built-in quicksort function but it might help those who are struggling to understand the practical use of a quicksort. Given its declarative nature it is very easy to see what is happening as oppose to how it works.
Here is a JSBin version without comments https://jsbin.com/zenajud/edit?js,console
function quickSortF(arr) {
// Base case
if (!arr.length) return []
// This is a ES6 addition, it uses destructuring to pull out the
// first value and the rest, similar to how other functional languages
// such as Haskell, Scala do it. You can then use the variables as
// normal below
const [head, ...tail] = arr,
// here we are using the arrow functions, and taking full
// advantage of the concise syntax, the verbose version of
// function(e) => { return e < head } is the same thing
// so we end up with the partition part, two arrays,
// one smaller than the pivot and one bigger than the
// pivot, in this case is the head variable
left = tail.filter( e => e < head),
right = tail.filter( e => e >= head)
// this is the conquer bit of divide-and-conquer
// recursively run through each left and right array
// until we hit the if condition which returns an empty
// array. These results are all connected using concat,
// and we get our sorted array.
return quickSortF(left).concat(head, quickSortF(right))
}
const q7 = quickSortF([11,8,14,3,6,2,7])
//[2, 3, 6, 7, 8, 11, 14]
const q8 = quickSortF([11,8,14,3,6,2,1, 7])
//[1, 2, 3, 6, 7, 8, 11, 14]
const q9 = quickSortF([16,11,9,7,6,5,3, 2])
//[2, 3, 5, 6, 7, 9, 11, 16]
console.log(q7,q8,q9)
The comments should provide enough if it is already not clear what is happening. The actual code is very short without comments, and you may have noticed I am not a fan of the semicolon. :)
In this blog http://www.nczonline.net/blog/2012/11/27/computer-science-in-javascript-quicksort/ which has pointed out that
Array.sort is implemented in quicksort or merge sort internaly.
Quicksort is generally considered to be efficient and fast and so is
used by V8 as the implementation for Array.prototype.sort() on arrays
with more than 23 items. For less than 23 items, V8 uses insertion
sort[2]. Merge sort is a competitor of quicksort as it is also
efficient and fast but has the added benefit of being stable. This is
why Mozilla and Safari use it for their implementation of
Array.prototype.sort().
and when using Array.sort,you should return -1 0 1 instead of true or false in Chrome.
arr.sort(function(a,b){
return a<b;
});
// maybe--> [21, 0, 3, 11, 4, 5, 6, 7, 8, 9, 10, 1, 2, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22]
arr.sort(function(a,b){
return a > b ? -1 : a < b ? 1 : 0;
});
// --> [22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
var array = [8, 2, 5, 7, 4, 3, 12, 6, 19, 11, 10, 13, 9];
quickSort(array, 0, array.length -1);
document.write(array);
function quickSort(arr, left, right)
{
var i = left;
var j = right;
var tmp;
pivotidx = (left + right) / 2;
var pivot = parseInt(arr[pivotidx.toFixed()]);
/* partition */
while (i <= j)
{
while (parseInt(arr[i]) < pivot)
i++;
while (parseInt(arr[j]) > pivot)
j--;
if (i <= j)
{
tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
i++;
j--;
}
}
/* recursion */
if (left < j)
quickSort(arr, left, j);
if (i < right)
quickSort(arr, i, right);
return arr;
}
Using ES6 rest, spread:
smaller = (a, list) => list.filter(x => x <= a)
larger = (a, list) => list.filter(x => x > a)
qsort = ([x, ...list]) => (!isNaN(x))
? [...qsort(smaller(x, list)), x, ...qsort(larger(x, list))]
: []
This algorithm work almost as fast as the default implementation
of Array.prototype.sort in chrome.
function quickSort(t){
_quickSort(t,0,t.length-1,0,t.length-1);
}
function _quickSort(t, s, e, sp, ep){
if( s>=e ) return;
while( sp<ep && t[sp]<t[e] ) sp++;
if( sp==e )
_quickSort(t,s,e-1,s,e-1);
else{
while(t[ep]>=t[e] && sp<ep ) ep--;
if( sp==ep ){
var temp = t[sp];
t[sp] = t[e];
t[e] = temp;
if( s!=sp ){
_quickSort(t,s,sp-1,s,sp-1);
}
_quickSort(t,sp+1,e,sp+1,e);
}else{
var temp = t[sp];
t[sp] = t[ep];
t[ep] = temp;
_quickSort(t,s,e,sp+1,ep);
}
}
}
quickSort time (ms): 738
javaScriptSort time (ms): 603
var m = randTxT(5000,500,-1000,1000);
VS(m);
function VS(M){
var t;
t = Date.now();
for(var i=0;i<M.length;i++){
quickSort(M[i].slice());
}console.log("quickSort time (ms): "+(Date.now()-t));
t = Date.now();
for(var i=0;i<M.length;i++){
M[i].slice().sort(compare);
}console.log("javaScriptSort time (ms): "+(Date.now()-t));
}
function compare(a, b) {
if( a<b )
return -1;
if( a==b )
return 0;
return 1;
}
function randT(n,min,max){
var res = [], i=0;
while( i<n ){
res.push( Math.floor(Math.random()*(max-min+1)+min) );
i++;
}
return res;
}
function randTxT(n,m,min,max){
var res = [], i=0;
while( i<n ){
res.push( randT(m,min,max) );
i++;
}
return res;
}
Yet another quick sort demonstration, which takes middle of the array as pivot for no specific reason.
const QuickSort = function (A, start, end) {
//
if (start >= end) {
return;
}
// return index of the pivot
var pIndex = Partition(A, start, end);
// partition left side
QuickSort(A, start, pIndex - 1);
// partition right side
QuickSort(A, pIndex + 1, end);
}
const Partition = function (A, start, end) {
if (A.length > 1 == false) {
return 0;
}
let pivotIndex = Math.ceil((start + end) / 2);
let pivotValue = A[pivotIndex];
for (var i = 0; i < A.length; i++) {
var leftValue = A[i];
//
if (i < pivotIndex) {
if (leftValue > pivotValue) {
A[pivotIndex] = leftValue;
A[i] = pivotValue;
pivotIndex = i;
}
}
else if (i > pivotIndex) {
if (leftValue < pivotValue) {
A[pivotIndex] = leftValue;
A[i] = pivotValue;
pivotIndex = i;
}
}
}
return pivotIndex;
}
const QuickSortTest = function () {
const arrTest = [3, 5, 6, 22, 7, 1, 8, 9];
QuickSort(arrTest, 0, arrTest.length - 1);
console.log("arrTest", arrTest);
}
//
QuickSortTest();
I really thought about this question. So first I found the normal search mode and wrote.
let QuickSort = (arr, low, high) => {
if (low < high) {
p = Partition(arr, low, high);
QuickSort(arr, low, p - 1);
QuickSort(arr, p + 1, high);
}
return arr.A;
}
let Partition = (arr, low, high) => {
let pivot = arr.A[high];
let i = low;
for (let j = low; j <= high; j++) {
if (arr.A[j] < pivot) {
[arr.A[i], arr.A[j]] = [arr.A[j], arr.A[i]];
i++;
}
}
[arr.A[i], arr.A[high]] = [arr.A[high], arr.A[i]];
return i;
}
let arr = { A/* POINTER */: [33, 22, 88, 23, 45, 0, 44, 11] };
let res = QuickSort(arr, 0, arr.A.length - 1);
console.log(res);
Result is [0, 11, 22, 23, 33, 44, 45, 88]
But its not stable; so I checked the other answers and the Idea of #6502 was interesting to me that "two items do not have to be the same" to be distinguishable.
Well, I have a solution in my mind, but it is not optimal. We can keep the indexes of the items in a separate array. Memory consumption will almost double in this idea.
arr.A => Array of numbers
arr.I => Indexes related to each item of A
influencer => This should be a very very small number; I want to use this as a factor to be able to distinguish between similar items.
So we can change the partition like this:
let Partition = (arr, low, high) => {
let pivot = arr.A[high];
let index = arr.I[high];
let i = low;
for (let j = low; j <= high; j++) {
if (arr.A[j] + (arr.I[j] * influencer) < pivot + (index * influencer)) {
[arr.A[i], arr.A[j]] = [arr.A[j], arr.A[i]];
[arr.I[i], arr.I[j]] = [arr.I[j], arr.I[i]];
i++;
}
}
[arr.A[i], arr.A[high]] = [arr.A[high], arr.A[i]];
[arr.I[i], arr.I[high]] = [arr.I[high], arr.I[i]];
return i;
}
let influencer = 0.0000001;
let arr = {
I/* INDEXES */: [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
A/* POINTER */: [33, 22, 88, 33, 23, 45, 33, 89, 44, 11]
};
let res = QuickSort(arr, 0, arr.A.length - 1);
console.log(res);
Result:
I: [19, 11, 14, 10, 13, 16, 18, 15, 12, 17],
A: [11, 22, 23, 33, 33, 33, 44, 45, 88, 89]
More compact and easy to understand quicksort implementation
const quicksort = arr =>
arr.length <= 1
? arr
: [
...quicksort(arr.slice(1).filter((el) => el < arr[0])),
arr[0],
...quicksort(arr.slice(1).filter((el) => el >= arr[0])),
];
try my solution
const quickSort = (arr) => {
// base case
if(arr.length < 2) return arr;
// recurisve case
// pick a random pivot
let pivotIndex = Math.floor(Math.random() * arr.length);
let pivot = arr[pivotIndex];
let left = [];
let right = [];
// make array of the elements less than pivot and greater than it
for(let i = 0; i < arr.length; i++) {
if(i === pivotIndex) {
continue;
}
if(arr[i] < pivot) {
left.push(arr[i])
} else {
right.push(arr[i])
}
}
// call the recursive case again
return quickSort(left).concat([pivot], quickSort(right));
}
when testing this
quickSort([7, 5, 3, 2, 8, 1, 5]) // returns [[1, 2, 3, 5, 5, 7, 8]]
This is it !!!
function typeCheck(a, b){
if(typeof a === typeof b){
return true;
}else{
return false;
}
}
function qSort(arr){
if(arr.length === 0){
return [];
}
var leftArr = [];
var rightArr = [];
var pivot = arr[0];
for(var i = 1; i < arr.length; i++){
if(typeCheck(arr[i], parseInt(0))){
if(arr[i] < pivot){
leftArr.push(arr[i]);
}else { rightArr.push(arr[i]) }
}else{
throw new Error("All must be integers");
}
}
return qSort(leftArr).concat(pivot, qSort(rightArr));
}
var test = [];
for(var i = 0; i < 10; i++){
test[i] = Math.floor(Math.random() * 100 + 2);
}
console.log(test);
console.log(qSort(test));
Slim version:
function swap(arr,a,b){
let temp = arr[a]
arr[a] = arr[b]
arr[b] = temp
return 1
}
function qS(arr, first, last){
if(first > last) return
let p = first
for(let i = p; i < last; i++)
if(arr[i] < arr[last])
p += swap(arr, i, p)
swap(arr, p, last)
qS(arr, first, p - 1)
qS(arr, p + 1, last)
}
Tested with random values Arrays, and seems to be always faster than Array.sort()
quickSort = (array, left, right) => {
if (left >= right) {
return;
}
const pivot = array[Math.trunc((left + right) / 2)];
const index = partition(array, left, right, pivot);
quickSort(array, left, index - 1);
quickSort(array, index, right);
}
partition = (array, left, right, pivot) => {
while (left <= right) {
while (array[left] < pivot) {
left++;
}
while (array[right] > pivot) {
right--;
}
if (left <= right) {
swap(array, left, right);
left++;
right--;
}
}
return left;
}
swap = (array, left, right) => {
let temp = array[left];
array[left] = array[right];
array[right] = temp;
}
let array = [1, 5, 2, 3, 5, 766, 64, 7678, 21, 567];
quickSort(array, 0, array.length - 1);
console.log('final Array: ', array);
A Fastest implementation
const quickSort = array =>
(function qsort(arr, start, end) {
if (start >= end) return arr;
let swapPos = start;
for (let i = start; i <= end; i++) {
if (arr[i] <= arr[end]) {
[arr[swapPos], arr[i]] = [arr[i], arr[swapPos]];
swapPos++;
}
}
qsort(arr, start, --swapPos - 1);
qsort(arr, swapPos + 1, end);
return arr;
})(Array.from(array), 0, array.length - 1);
Quicksort using ES6, filter and spread operation.
We establish a base case that 0 or 1 elements in an array are already sorted. Then we establish an inductive case that if quicksort works for 0 or 1 elements, it can work for an array of size 2. We then divide and conquer until and recursively call our function until we reach our base case in the call stack to get our desired result.
O(n log n)
const quick_sort = array => {
if (array.length < 2) return array; // base case: arrays with 0 or 1 elements are already "sorted"
const pivot = array[0]; // recursive case;
const slicedArr = array.slice(1);
const left = slicedArr.filter(val => val <= pivot); // sub array of all elements less than pivot
const right = slicedArr.filter(val => val > pivot); // sub array of all elements greater than pivot
return [...quick_sort(left), pivot, ...quick_sort(right)];
}
const quicksort = (arr)=>{
const length = Math.ceil(arr.length/2);
const pivot = arr[length];
let newcondition=false;
for(i=0;i<length;i++){
if(arr[i]>arr[i+1]){
[arr[i], arr[i+1]] = [arr[i+1], arr[i]]
newcondition =true
}
}
for(i=arr.length;i>length-1;i--){
if(arr[i]>arr[i+1]){
[arr[i], arr[i+1]] = [arr[i+1], arr[i]]
newcondition =true
}
}
return newcondition? quicksort(arr) :arr
}
const t1 = performance.now();
const t2 = performance.now();
console.log(t2-t1);
console.log(quicksort([3, 2, 4, 9, 1, 0, 8, 7]));
const quicksort = (arr)=>{
if (arr.length < 2) return arr;
const pivot = arr[0];
const left = [];
const right = [];
arr.shift();
arr.forEach(number => {
(number<pivot) ? left.push(number) : right.push(number);
});
return ([...quicksort(left), pivot, ...quicksort(right)]);
}
console.log(quicksort([6, 23, 29, 4, 12, 3, 0, 97]));
How about this non-mutating functional QuickSort:
const quicksort = (arr, comp, iArr = arr) => {
if (arr.length < 2) {
return arr;
}
const isInitial = arr.length === iArr.length;
const arrIndexes = isInitial ? Object.keys(arr) : arr;
const compF = typeof comp === 'function'
? comp : (left, right) => left < right ? -1 : right < left ? 1 : 0;
const [pivotIndex, ...indexesSansPivot] = arrIndexes;
const indexSortReducer = isLeftOfPivot => [
(acc, index) => isLeftOfPivot === (compF(iArr[index], iArr[pivotIndex]) === -1)
? acc.concat(index) : acc,
[]
];
const ret = quicksort(indexesSansPivot.reduce(...indexSortReducer(true)), compF, iArr)
.concat(pivotIndex)
.concat(quicksort(indexesSansPivot.reduce(...indexSortReducer(false)), compF, iArr));
return isInitial ? ret.reduce((acc, index) => acc.concat([arr[index]]), []) : ret;
};
As a bonus, it supports optional comparing function which enables sorting of array of objects per property/properties, and doesn't get slower if dealing with larger values/objects.
First quick sorts original array keys, then returns sorted copy of original array.
Related
I am trying to make a program that returns ["1-5"] if I give [1,2,3,4,5].
I have made it but I can't filter it. So I want a code that will filter my output code. Or any code that is better than mine.
let array = [1,2,3,5,6,7,8,10,11, 34, 56,57,];
let x = [];
for(let i = 0; i < array.length; i++){
for(let j = 0; j < array.length; j++){
if(array[i] + j == array[j]){
x.push(array[i] + "-" + array[j]);
}
if(array[j] > array[i] + j && array[j + 1]){
let y = array.slice(j, array.length)
array = y;
i, j = 0;
}
if(array[i] - array[i + 1] != -1 && array[i + 1] - array[i] != 1 && array[i + 1] != undefined){
x.push(array[i]);
}
}
}
console.log(x);
The phrasing of the question makes this somewhat difficult to answer, but based on your code snippet I can gather that you are either:
Attempting to find the range of the entire array OR
Attempting to find contiguous ranges within the array
Based on these interpretations, you could answer this question as follows:
function detectRange(a) {
// clone a
const b = [...a]
// remove first value
const min = max = b.splice(0, 1)[0]
// compute range
const range = b.reduce(({min, max}, i) => {
if(i < min) min = i
if(i > max) max = i
return { min, max }
}, {min, max})
return range
}
function detectRanges(a) {
// clone a
const b = [...a]
// remove first value
const min = max = b.splice(0, 1)[0]
// init ranges array
const ranges = [ ]
// compute ranges
const range = b.reduce(({min, max}, i) => {
if(i === max + 1) {
return {min , max: i}
} else {
ranges.push({min, max})
return {min: i, max: i}
}
}, {min, max})
// push the remaining range onto the array
ranges.push(range)
return ranges
}
function printRange(r) {
console.log(`["${r.min}-${r.max}"]`)
}
function printRanges(r) {
r.forEach(i => {
printRange(i)
})
}
// detect and print range of whole array
printRange(detectRange([1, 2, 3, 5, 6, 7, 8, 10, 11, 34, 56, 57]))
// detect and print only contiguous ranges within array
printRanges(detectRanges([1, 2, 3, 5, 6, 7, 8, 10, 11, 34, 56, 57]))
If you assume that the list is sorted, we only need to traverse the list sequentially. There's no need to have double-nested loops. If you maintain sufficient states, you can determine whether you are in a group and you merely manage the start versus the last element in the group.
To simplify things I made use of ES6 string interpolation ${start}-${last}.
let array = [1,2,3,5,6,7,8,10,11, 34, 56,57];
let result = [ ];
let hasStart = false;
let start = 0;
let last = 0;
for (let num of array) {
if (!hasStart) {
hasStart = true;
last = start = num;
continue;
}
if (num === last + 1) {
last = num;
continue;
}
result.push( start === last ? start : `${start}-${last}` );
last = start = num;
}
if (hasStart) {
result.push( start === last ? start : `${start}-${last}` );
}
console.log(result);
Input: [1,2,3,4,5]
Output: ["1-5"]
So I assume you want to get string in format:
["smallestelement-largestelement"]
var input1 = [1,2,3,4,5]
console.log( "["+'"'+Math.min(...input1)+"-"+Math.max(...input1)+'"'+"]")
If what you want is string in format:
["firstelement-lastelement"]
var input1 = [1,2,3,4,5]
console.log( "["+'"'+input1[0]+"-"+input1.pop()+'"'+"]")
If you have an integer array, and if you want to output the range, you could natively sort() it (you can also provide rules for sorting) and use shift() for the first element and slice(-1) for the last:
let arr = [4,1,5,3].sort();
console.log(arr.shift()+'-'+arr.slice(-1));
As said in the comments, you should clarify if you wish "1-57" for the snippet array, or describe your use case more broadly.
const array = [1, 2, 3, 5, 6, 7, 8, 10, 11, 34, 56, 57];
let s = null;
const result = array.sort((a, b) => a - b).reduce((p, c, i, arr) => {
if (!s) s = c;
if (c + 1 !== arr[i + 1]) {
p.push(s === c ? s : `${s}-${c}`);
s = null;
}
return p
}, [])
console.log(result);
Added maximum number according to the input length should be returned.
For example, if the length is 2 then the max among arr[0] + arr[1], arr[1] + arr[2], arr[2] + arr[3] should be returned.
Input is array and length.
I solved this in a real job interview but I think there will be a way not to use nested loop.
const add = (arr, len) => {
let rtnVal = 0
for (let i = len - 1; i < arr.length; i++) {
let temp_idx = i;
let temp_sum = 0;
for (let j = 0; j < len; j++) {
temp_sum = (temp_sum || 0) + arr[temp_idx]
temp_idx -= 1
}
if (temp_sum > rtnVal) {
rtnVal = temp_sum
}
}
return rtnVal
}
console.log(add([1, 2, 3, 4, 5, 6, 7, 8, 9], 4))
I expect the output 30
// enhanced nested loop
const add = (arr, len) => {
let rtnVal = 0;
for(let i=len-1;i<arr.length;i++){
let sum = 0
for(let j=i;j>=i-len+1;j--){
sum += arr[j]
}
if (sum > rtnVal) rtnVal = sum
}
return rtnVal
}
console.log(add([9, 9, 9, 4, 5, 6, 7, 8, 9], 3))
Use a moving window. Add up len numbers starting at the beginning. Then continue through the array adding the next number and subtracting the trailing number.
const add = (arr, len) => {
return arr.reduce((a,v,i,arr) => {
a.running += v;
if(i >= len) {
a.running -= arr[i-len];
}
if(i+1 >= len && a.largest < a.running) {
a.largest = a.running;
}
return a;
}, {
largest: Number.NEGATIVE_INFINITY,
running: 0
}).largest;
}
console.log(add([1,2,3,4,5,6,7,8,9],4)); // 30
console.log(add([-1,-1,-1,-1],2)); // -2
console.log(add([1],1)); // 1
Assuming its sorted like your example. You can use negative slice to select from end and then reduce the array.
const add = (arr, len) => arr.slice(len > arr.len ? arr.len : -len).reduce((total, num) => total + num)
console.log(add([1, 2, 3, 4, 5, 6, 7, 8, 9], 4))
Why array can't be displayed immediately after swap function call in JS? Why result of the function call is undefined?
const swap = (arr) => {
for (let i = 0; i < arr.length - 1; i++) {
var tmp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = tmp;
}
}
r = [...Array(10)].map( (_, i) => (i + 1) * 3);
console.log(r); // --> Array [ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 ]
//swap(r);
// why array can't be displayed immediately after swap function call?
console.log( swap(r) ); // --> displays undefined but the swap function worked out
console.log(r); // --> Array [ 6, 9, 12, 15, 18, 21, 24, 27, 30, 3 ]
I think it's because the swap function returns nothing but I'm not sure. But
swap(r);
console.log(r);
works fine.
You don't need to return anything and leave your function swap as it. The passed array is a reference of r = [...Array(10)].map((_, i) => (i + 1) * 3) so, every modification will modify that array after the function swap ends.
Just print the array console.log(r);.
const swap = (arr) => {
for (let i = 0; i < arr.length - 1; i++) {
var tmp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = tmp;
}
}
r = [...Array(10)].map((_, i) => (i + 1) * 3);
console.log(r); // --> Array [ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 ]
swap(r);
// why array can't be displayed immediately after swap function call?
// console.log(swap(r)); // --> displays undefined but the swap function worked out
console.log(r); // --> Array [ 6, 9, 12, 15, 18, 21, 24, 27, 30, 3 ]
.as-console-wrapper { max-height: 100% !important; top: 0; }
Yes you are correct, it's because you are trying to output the result of the swap function call but you are not returning anything specific. So the default result is undefined.
If you want to output the array after the swap, then return that in your function like so:
const swap = (arr) => {
for (let i = 0; i < arr.length - 1; i++) {
var tmp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = tmp;
}
return arr;
}
r = [...Array(10)].map((_, i) => (i + 1) * 3);
console.log(swap(r));
Ultimately it is up to you to decide how you want the function to work. As you have identified there is no need to return anything at all as long as you don't try and use the result of the function call.
With regards to why it doesn't automatically return the array, well you have to remember that a function can do lots of different things to many objects. There would be no way to know which object or value should automatically be returned, so it doesn't make sense for it to try and do anything magic like that. If you want something returned, then make sure you return it.
Yes you are right; you should use return arr; but I think what you want is rotating the array.
arr = new Array(10).fill().map( (_, i) => (i + 1) * 3);
console.log(arr);
res = [...arr.slice(1),arr[0]]; // prettier but can be dramatically slow on large arrays
// or for even faster one :
rotate = (arr) => !(arr.push(arr.shift())) || arr;
// or
/*const rotate = (arr) => {
var first = arr[0];
for(var c = 0, l = arr.length - 1; c < l;)
arr[c] = arr[++c];
arr[arr.length-1] = first;
return arr;
}*/
console.log(res);
console.log(rotate(arr));
I am trying to come up with a function that will take an array of integers and output the 2nd highest number and the 2nd smallest number. The function will take into account floats, duplicates, and negative numbers.
Two functions pass all the tests below except for test2.
var test1 = [7, 7, 12, 98, 106]
answer1 = {2nd Min: 12, 2nd Max: 98}
var test2 = [5, 23, -112, 6, 70, 70, -112]
answer2 = {2nd Min: 5, 2nd Max: 23}
var test3 = [-22, 22]
answer3 = {2nd Min: 22, 2nd Max: -22}
var test4 = [10, 89, 3]
answer = {2nd Min: 10, 2nd Max: 10}
var test5 = [10.4, -12.09, .75, 22]
answer3 = {2nd Min: 0.75, 2nd Max: 10.4}
/*/ \ \
---SOLUTION 1---
\ \ /*/
function secondGreatLow1(arr) {
//make copy of array because it will be spliced in the following functions
var arrCopy = arr.slice();
//push returned values of each function into this the answer array
var answer = []
answer.push(secondMin(arrCopy));
answer.push(secondMax(arrCopy));
return answer;
};
//helper function 1
var secondMin = function (arr){
var arrCopy = arr.slice();
//check length of array
if (arr.length == 2) {
return arr[1];
} else {
var min = Math.min.apply(null, arrCopy);
arrCopy.splice(arrCopy.indexOf(min), 1);
//check for duplicates
for (var i = 0; i < arrCopy.length; i++) {
if (arrCopy.indexOf(min) === -1) {
//.apply is used for arrays
return Math.min.apply(null, arrCopy);
} else {
arrCopy.splice(arrCopy.indexOf(min), 1);
return Math.min.apply(null, arrCopy);
}
};
}
};
//helper function 2
var secondMax = function (arr){
var arrCopy = arr.slice();
if (arr.length == 2) {
return arr[0];
} else {
var max = Math.max.apply(null, arrCopy);
arrCopy.splice(arrCopy.indexOf(max), 1);
//check for duplicates
for (var i = 0; i < arrCopy.length; i++) {
if (arrCopy.indexOf(max) === -1) {
return Math.max.apply(null, arrCopy);
} else {
arrCopy.splice(arrCopy.indexOf(max), 1);
return Math.min.apply(null, arrCopy);
}
};
}
};
/*/ \ \
---SOLUTION 2---
\ \ /*/
function secondGreatLow2 (numbers) {
var arr = withoutDuplicates(numbers);
arr.sort(function(a,b) { return a-b; });
return arr[1] + ' ' + arr[arr.length-2];
};
// helpers
var withoutDuplicates = function(arr) {
var out = [];
for(var i=0; i<arr.length; i++) {
if(i === 0 || arr[i] !== arr[i-1]) {
out.push(arr[i]);
}
}
return out;
};
In your second solution, your withoutDuplicates function appears to operate on the assumption that the list is sorted (checking for duplicates by comparing an element to the previous element); however, in secondGreatLow2, you call withoutDuplicates without performing some sort of sorting.
If you changed the order of those two lines, solution #2 looks valid assuming you don't have any floating point mismatches, ie 3.9999999999997 != 3.99999999998
Not performance efficient for large arrays, but concise:
var a = [5, 23, -112, 6, 70, 70, -112], b = [];
// remove duplicates
b = a.filter(function (item, pos) {
return a.indexOf(item) == pos;
});
// sort
b.sort(function (a, b) {
return a > b;
});
console.log(b[1]); // 2nd min:12
console.log(b[b.length-2]); // 2nd max:12
function get_seconds(a) {
var b = uniq(a); // remove duplicates and sort
var l = b.length;
return [b[1], b[l-2]];
}
Check full tests below:
var tests = {
"case1": {
"input": [7, 7, 12, 98, 106],
"output": [12, 98]
},
"case2": {
"input": [5, 23, -112, 6, 70, 70, -112],
"output": [5, 23]
},
"case3": {
"input": [-22, 22],
"output": [22, -22]
},
"case4": {
"input": [10, 89, 3],
"output": [10, 10]
},
"case5": {
"input": [10.4, -12.09, .75, 22],
"output": [0.75, 10.4]
}
};
function do_tests() {
var res, logs = '',
j_exp, j_res;
$.each(tests, function(c, io) {
j_exp = JSON.stringify(io.output);
res = get_seconds(io.input);
j_res = JSON.stringify(res);
if (j_res == j_exp) {
logs += '<div class="success">' + c + ' passed.</div>';
} else {
logs += '<div class="failed">' + c + ' failed. Expected: ' + j_exp + ', Got: ' + j_res + '</div>';
}
});
$("#log").html(logs);
}
function get_seconds(a) {
var b = uniq(a);
console.log(b, a);
var l = b.length;
return [b[1], b[l - 2]];
}
function uniq(a) {
return a.sort(sortNumber).filter(function(item, pos, ary) {
return !pos || item != ary[pos - 1];
})
}
function sortNumber(a, b) {
return a - b;
}
div#log {
font-family: monospace;
}
div#log > div {
padding: 4px;
margin: 2px;
}
.success {
color: green;
}
.failed {
color: red;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<button onclick="do_tests();">Run tests</button>
<div id="log"></div>
This would also work and it appears do be faster (no sorting). What do you guys think?
var secondMax = function(arr){
return Math.min(...arr.reduce((acc, val) =>
(val > Math.min(...acc)) ? [Math.max(...acc), val] : acc
, [-Infinity, -Infinity]))
}
var secondMin = function(arr){
return Math.max(...arr.reduce((acc, val) =>
(val < Math.max(...acc)) ? [Math.min(...acc), val] : acc
, [Infinity, Infinity]))
}
var unique = function(arr) {
return arr.filter((value, index, arr) =>
arr.indexOf(value) === index
)
}
var arr = [7, 7, 12, 98, 106, 106];
console.log(secondMin(unique(arr)));// => 12
console.log(secondMax(unique(arr)));// => 98
console.log(secondMin(arr));// => 7
console.log(secondMax(arr));// => 106
Glad you have found the solution.I am posting this for those who are searching for a simpler solution.This would run on O(n) time.
var fl = arr[0],sl = arr[0];
var fs = Infinity,ss = Infinity;
//find the largest and smallest.
for(var i = 0;i < arr.length;i ++) {
if(arr[i] > fl) fl = arr[i];
if(arr[i] < fs) fs = arr[i];
}
//let us assume smallest is second largest and vice versa.
sl = fs;
ss = fl;
//find second smallest and largest.
for(var i = 0;i < arr.length;i ++) {
if(arr[i] < fl && arr[i] > sl) sl = arr[i];
if(arr[i] > fs && arr[i] < ss) ss = arr[i];
}
console.log("first and second largest : ",fl,sl);
console.log("first and second smallest : ",fs,ss);
This is an easy and simple method for a second highest number but the problem is we could not get a number if there have duplicate exists in the array.
function getSecondLargest(nums) {
if(nums.length<2){
return nums;
}
var first=0;
var second=0;
for(var i=0; i<nums.length;i++)
{
if(nums[i]>first)
{
second = first;
first = nums[i]
}
else(nums[i]>second && nums[i]!=first)
{
second = nums[i];
}
}
return second;
}
If array contains duplicates like nums = [2, 3, 6, 6, 5], the below solution works,
let sortedArray = new Set(nums.sort((a, b) => b - a ));
console.log('second smallest', [...sortedArray][sortedArray.size - 2])
console.log('second largest',[...sortedArray][1]);
C program for arrays to find second largest and second smallest in a array.
#include<stdio.h>
void main()
{
int arr[20];
int small,big,big2,small2;
int i,num;
printf("Enter Number of elements:");
scanf("%d",&num);
printf("Enter the elements: \n");
for(i=0;i<num;i++)
scanf("%d",&arr[i]);//Entering elements
big = arr[0];//For big in array
small = arr[0];// For small in array
for(i=0;i<num;i++)
{
if(arr[i]>big)
{
big2 = big;
big = arr[i];
}
if(arr[i]<small)
{
small2 = small;
small = arr[i];
}
}
printf("The biggest is %d\n",big);
printf("The smallest is %d\n",small);
printf("The second biggest is %d\n",big2);
printf("The second smallest is %d\n",small2);
}
#tej
let ar = [1,4,76,80,90,3,6,82];
var big = 0,sbig=0;
for(let i=0; i<ar.length;i++){
if(big < ar[i]) {
sbig = big;
big = ar[i]
continue;
}
if(sbig < ar[i]) {
sbig = ar[i];
}
}
console.log(sbig,big);
Considering a sorted array arr
1- Simple case:
var arr = [3, 6, 12, 18];
indexesAround(6)
//> [1, 1]
indexesAround(7)
//> [1, 2]
2- More complex case:
var arr = [3, 3, 6, 6, 18, 18];
indexesAround(6)
//> [2, 3]
indexesAround(7)
//> [3, 4]
How would you implement(or pseudo code) a such indexesAround(value) function ?
--
Here is what I have for now, but I think this could be enhanced:
function indexesAround(val) {
var lower = 0;
var upper = lower;
var el;
for (var i = 0, len = arr.length; i < len; i++) {
el = arr[i];
if (el > val) {break;}
if (arr[lower] < el) {lower = upper = i;}
if (arr[upper] <= el) {upper = i;}
}
return [lower, upper];
}
Considering the array is sorted:
function indexesAround(arr, val) {
if (!~arr.indexOf(val)) return false; // not found
var start = arr.indexOf(val);
var end = (arr.length - 1) - arr.reverse().indexOf(val);
arr.reverse(); // restore original order
return [start, end];
}
This solution covers every possibility and works exactly to OP's specifications. Run it on jsfiddle.
Code
function indexesAround(target,array) {
var start;
var len = array.length;
for(i = 0; i < len; i++) {
if (array[i] == target && !start) { start = i; }
if (array[i] > target) {
if(i == 0) { return [ 0, 0 ]; } // Target lower than array range
if(!start) { return [ i-1, i ]; } // Target inside array range but not found
return [ start, i-1 ]; // Target found
}
}
if(start) { return [ len-1, len-1 ]; } // Target higher than array range
return [ start, len-1 ]; // Target found and extends until end of array
}