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Find the minimum cost combination algorithm

Given an array N which contains at least 5 items, I want to find 2 numbers(P and Q) in which 0 < P < Q < N - 1.
Suppose we have the following array:
const N = [1, 9, 4, 5, 8];
if P = 1 , Q = 2 , the cost will be N[P] + N[Q] = N[1] + N[2] = 9 + 4 = 13
if P = 1, Q = 3 , the cost will be N[P] + N[Q] = N[1] + N[3] = 9 + 5 = 14
if P = 2, Q = 3 , the cost will be N[P] + N[Q] = N[2] + N[3] = 4 + 5 = 9
From here the combination which gives the minimum cost is P = 2 and Q = 3.
Here is the solution that I found and I am looking for your help if I can improve its time complexity:
function solution(N) {
// since 0 < P < Q < N - 1
const sliced = N.slice(1, N.length - 1);
const sorted = sliced.sort((a, b) => a - b);
// the minimum should be from the start since we have sorted the array
const P = 0;
const Q = 1;
return getCost(P, Q, sorted);
}
function getCost(P, Q, N) {
return N[P] + N[Q];
}
// output should be 9
console.log(solution([1, 9, 4, 5, 8]))
In a best-case scenario it's 0(n log(n)) because of the sort, but I am wondering if we can improve it to O(n) for example.
Thanks for your help

function twoSmallest(arr) {
let [first, second] = [arr[1], arr[2]]
for (let i = 3; i < arr.length - 1; i++) {
const el = arr[i]
if (el < first && el < second) {
[first, second] = [Math.min(first, second), el]
} else if (el < first) {
[first, second] = [second, el]
} else if (el < second) {
second = el
}
}
return first + second
}
This is an O(n) time and O(1) space solution. It also makes sure that the element with the smaller index is kept in first for the case where you need to use the indices and it is of interest for some reason.
The algorithm is clear, IMO, but the JS code is probably not the best implementation. I haven't written JS for some time.

What do you think of this solution?
function solution([_, ...n]) {
n.pop()
n.sort((a, b) => a - b);
return n[0] + n[1];
}
// output should be 9
console.log(solution([1, 9, 4, 5, 8]))
The logic is the same that you outlined - only using some other approach that JS offers.

I'm pretty sure this is O(n):
const solution = (arr) => {
// find smallest that's not at either end
let idx = 1;
let P = arr[1];
for(let i = 2; i < arr.length-1; i++) {
if(arr[i] < P) {
idx = i;
P = arr[i];
}
}
// find second smallest that's not at either end
let Q = Infinity;
for(let i = 1; i < arr.length-1; i++) {
if(i == idx) continue;
if(arr[i] < Q) Q = arr[i];
}
return P + Q;
}

Here is the fastest way to find k smallest numbers in a list with Python. The rest is trivial
fastest method of getting k smallest numbers in unsorted list of size N in python?

JavaScript, HeapSort - count swaps and comparisons

How i should count swaps and comparisons?
I'm new in programmin and algorithms. So, I got a problem to count swaps and comparisons, the problem is that i don't know how to save the value of counters in recursive function
There is code explanation
https://levelup.gitconnected.com/heapsort-for-javascript-newbies-598d25477d55
function heapify(arr, length, i) {
let largest = i
let left = i * 2 + 1
let right = left + 1
if (left < length) {
if(arr[left] > arr[largest]) {
largest = left
}
}
if (right < length) {
if(arr[right] > arr[largest]) {
largest = right
}
}
if(largest != i) {
[arr[i], arr[largest]] = [arr[largest], arr[i]]
heapify(arr, length, i)
}
return arr
}
function heapSort(arr) {
let length = arr.length
let i = Math.floor(length / 2 - 1)
let k = length - 1
while (i >= 0) {
heapify(arr, length, i)
i--
}
while (k >= 0) {
[arr[0], arr[k]] = [arr[k], arr[0]]
heapify(arr, k, 0)
k--
}
return arr
}

Since it is an inplace sorting algorithm, you don't really have to return the array. The caller already has passed the array, so they don't really need the same array reference again. You can then instead use the return value for the number of swaps.
Side note: there is a bug in your heapify function: the recursive call should get largest as last argument instead of i. I have corrected that below as well:
function heapify(arr, length, i) {
let largest = i;
let left = i * 2 + 1;
let right = left + 1;
if (left < length) {
if(arr[left] > arr[largest]) {
largest = left;
}
}
if (right < length) {
if(arr[right] > arr[largest]) {
largest = right;
}
}
if(largest != i) {
[arr[i], arr[largest]] = [arr[largest], arr[i]];
// count the above swap, and those made by the recursive calls
return 1 + heapify(arr, length, largest);
}
return 0; // Nothing was swapped here
}
function heapSort(arr) {
let length = arr.length;
let i = Math.floor(length / 2 - 1);
let k = length - 1;
let swapCount = 0; // running sum
while (i >= 0) {
swapCount += heapify(arr, length, i);
i--
}
while (k >= 0) {
[arr[0], arr[k]] = [arr[k], arr[0]];
// Count the above swap and those made by heapify:
swapCount += 1 + heapify(arr, k, 0);
k--;
}
return swapCount;
}
let arr = [4, 2, 9, 7, 1, 3, 8, 0, 5, 6];
let swapCount = heapSort(arr);
console.log("sorted:", ...arr);
console.log("swaps:", swapCount);
If you want to both count the comparisons and the swaps, then you would need to return an object/array with this pair. In that case it may be easier to pass that object as optional argument to heapify, which will then adjust the counts in that object:
function heapify(arr, length, i, counter = { comparisons: 0, swaps: 0 }) {
let largest = i;
let left = i * 2 + 1;
let right = left + 1;
if (left < length) {
counter.comparisons++; // For the next `if` condition
if(arr[left] > arr[largest]) {
largest = left;
}
}
if (right < length) {
counter.comparisons++; // For the next `if` condition
if(arr[right] > arr[largest]) {
largest = right;
}
}
if(largest != i) {
counter.swaps++;
[arr[i], arr[largest]] = [arr[largest], arr[i]];
heapify(arr, length, largest, counter);
}
}
function heapSort(arr) {
let length = arr.length;
let i = Math.floor(length / 2 - 1);
let k = length - 1;
let counter = {
comparisons: 0, // running sum
swaps: 0 // running sum
}
while (i >= 0) {
heapify(arr, length, i, counter);
i--;
}
while (k >= 0) {
counter.swaps++;
[arr[0], arr[k]] = [arr[k], arr[0]];
heapify(arr, k, 0, counter);
k--;
}
return counter;
}
let arr = [4, 2, 9, 7, 1, 3, 8, 0, 5, 6];
let counter = heapSort(arr);
console.log("sorted:", ...arr);
console.log("counters:", counter);

Undestanding Recursion with return values in Merge Sort

I'm trying to solve a basic problem in Hackerearth Given an array A of size N, you need to find the number of ordered pairs (i,j) such that i < j and A[i] > A[j].
I was able to find out an idea actually implemented it by having a global variable. But having a global value is not a good practice, hence I tried to pass it as parameter and I couldn't able to solve it. Am stuck with keeping an already returned value and adding an updated value to it.
// let ans = 0;
let mergeSort = (left, right, arr, ans) => {
let i = 0,
j = 0,
k = 0;
let leftLen = left.length,
rightLen = right.length;
while (i < leftLen && j < rightLen) {
if (left[i] <= right[j]) {
arr[k] = left[i];
i++;
} else {
arr[k] = right[j];
ans += leftLen - i;
j++;
}
k++;
}
while (i < leftLen) {
arr[k] = left[i];
i++;
k++;
}
while (j < rightLen) {
arr[k] = right[j];
j++;
k++;
}
return { arr, ans };
};
let divideArray = (arr, ans) => {
if (arr.length < 2) return { arr, ans };
let N = arr.length;
let mid = Math.round(N / 2);
let left = arr.slice(0, mid);
let right = arr.slice(mid, N);
ans = ans;
divideArray(left, ans);
divideArray(right, ans);
let blabla = mergeSort(left, right, arr, ans);
return blabla;
};
let merge = (arr, ans) => {
let res = divideArray(arr, ans);
return res;
};
function main(input) {
let arr = [1, 4, 3, 2, 5];
let ans = 0;
let output = merge(arr, ans);
console.log('Final', output);
}
main();
In mergeSort function When the input of the left array is [1,4] and the right array is [3] the ans will be updated as 1, also when the left array is [1,3,4] and right is [2,5] the ans will be updated as 2. I would like to add both this ans values and return it as 3. But somewhere am making a mistake while returning. Any help will be appreciated.
JsFiddle
EDIT:
Please note that am trying to achieve it via MergeSort and recursion i know in lot of other ways i can solve this problem for instance i have clearly mentioned i had solved it by having a global variable,which is not a good practice so please provide me a solution only via recursion and merge sort

There is no need to pass an inversion count to divideArray(), it only needs to return a sub_count = left_count + right_count + merged_count. The sub_counts originate from each instance of merging and will be accumulated as recursion returns sub-counts back up the call chain to produce a total_count.
Example of an optimized top down merge sort updated to return an inversion count. A helper function (mergesort()) does a one time allocation of an auxiliary array (aux). To avoid unneeded copying of data, two mutually recursive functions are used, sortarrtoarr() sorts data from arr[] back to arr[], while sortarrtoaux() sorts data from arr[] to aux[]. Each of the mutually recursive functions calls the other in order to change the direction of merge based on the level of recursion.
function merge(arr, aux, bgn, mid, end) {
var i = bgn;
var j = mid;
var k = bgn;
var cnt = 0; // count of inversions
while(true){
if(arr[i] <= arr[j]){ // if left element <= right element
aux[k++] = arr[i++]; // copy left element
if(i < mid) // if not end of left run
continue; // continue back to while
do // else copy rest of right run
aux[k++] = arr[j++]; // and break
while(j < end);
break;
} else { // else left element > right element
cnt += mid - i; // arr.slice(i,mid) is > arr[j]
aux[k++] = arr[j++]; // copy right element
if(j < end) // if not end of right run
continue; // continue back to while
do // else copy rest of left run
aux[k++] = arr[i++]; // and break
while(i < mid);
break;
}
}
return cnt; // return inversion count
}
// sort from arr[] to aux[]
function sortarrtoaux(arr, aux, bgn, end) {
if ((end-bgn) < 2){ // if only 1 element
aux[bgn] = arr[bgn]; // copy it to aux
return 0; // return inversion count == 0
}
var cnt = 0; // inversion count = 0
var mid = Math.floor(bgn + (end - bgn) / 2);
cnt += sortarrtoarr(arr, aux, bgn, mid); // sort left arr back to arr
cnt += sortarrtoarr(arr, aux, mid, end); // sort right arr back to arr
cnt += merge(arr, aux, bgn, mid, end); // merge arr to aux
return cnt; // return inversion count
}
// sort from arr[] back to arr[]
function sortarrtoarr(arr, aux, bgn, end) {
if ((end-bgn) < 2) // if only 1 element
return 0; // return inversion count == 0
var cnt = 0; // inversion count = 0
var mid = Math.floor(bgn + (end - bgn) / 2);
cnt += sortarrtoaux(arr, aux, bgn, mid); // sort left arr to aux
cnt += sortarrtoaux(arr, aux, mid, end); // sort right arr to aux
cnt += merge(aux, arr, bgn, mid, end); // merge aux to arr
return cnt; // return inversion count
}
// entry function for mergesort
function mergesort(arr) {
if(arr.length < 2) // if less than 2 elements
return 0; // return inversion count == 0
var aux = new Array(arr.length); // allocate aux[] and start merge sort
return sortarrtoarr(arr, aux, 0, arr.length);
}
var arr = [8, 6, 7, 5, 3, 0, 9];
var cnt = mergesort(arr);
console.log(cnt);
for (i = 1; i < arr.length; i++) {
if(arr[i-1] > arr[i]){
console.log('error');
break;
}
}

Scott's answer offers a functional approach. Generators, function* below, offer another capable and flexible way of encoding this kind of problem -
const descendingPairs = function* (a = [])
{ for (const i of range(0, a.length)) // for all (0 <= i < a.length)
for (const j of range(0, a.length)) // for all (0 <= i < a.length)
if (i < j) // such that i < j
if (a[i] > a[j]) // and a[i] > a[j]
yield [ a[i], a[j] ] // output descending pair
}
We can optimise this by using i as the input for the j range's start -
const descendingPairs = function* (a = [])
{ for (const i of range(0, a.length)) // for all (0 < i < a.length)
for (const j of range(i + 1, a.length)) // for all (i < j < a.length)
if (a[i] > a[j]) // such that a[i] > a[j]
yield [ a[i], a[j] ] // output descending pair
}
range is nicely-encoded using a generator as well -
const range = function* (start = 0, stop = 0)
{ for (let x = start; x < stop; x++)
yield x
}
We can output the results of each descending pair -
const input =
[ 1, 4, 3, 2, 5 ]
for (const pair of descendingPairs(input))
console.log(pair)
// [ 4, 3 ]
// [ 4, 2 ]
// [ 3, 2 ]
Or we can collect all pairs into an array -
Array.from(descendingPairs(input))
// => [ [ 4, 3 ], [ 4, 2 ], [ 3, 2 ] ]
Or we can simply count them -
Array.from(descendingPairs(input)).length
// => 3
Expand the snippet below to verify the results in your own browser -
const range = function* (start = 0, stop = 0)
{ for (let x = start; x < stop; x++)
yield x
}
const descendingPairs = function* (a = [])
{ for (const i of range(0, a.length))
for (const j of range(i, a.length))
if (a[i] > a[j])
yield [ a[i], a[j] ]
}
const input =
[ 1, 4, 3, 2, 5 ]
console.log(Array.from(descendingPairs(input)))
// [ [ 4, 3 ], [ 4, 2 ], [ 3, 2 ] ]
console.log(Array.from(descendingPairs(input)).length)
// 3

I'm having a hard time figuring why you are writing this with code that's all about a mergesort. It seems to me that all you need to do is to generate the index pairs where j > i (a fairly easy task) and then count those for which A[i] > A[j]. Recursion is a fine way (though by no means the only easy way) to create those index pairs. The rest is a filter/length combination or a reduce.
Here's one variation:
const countDescendingPairs = (xs) =>
xs .map ((x, i) => xs .slice (i + 1) .filter(y => x > y) .length)
.reduce ((a, b) => a + b, 0)
console .log (
countDescendingPairs ([8, 6, 7, 5, 3, 0, 9])
)
But there are many simple alternatives.
And if you wanted to retrieve those pairs, it's a straightforward modification:
const descendingPairs = (xs) =>
xs .flatMap (
(x, i) =>
xs .slice (i + 1)
.filter (y => x > y)
.map ((y) => ({x, y}))
)
console .log (
descendingPairs ([8, 6, 7, 5, 3, 0, 9])
)
Updated to add flatMap and to remove the incorrect indices from the second version. (You can't filter, then expect the old index to still work!)

https://codepen.io/Santhoshsanz/pen/dybedgm?editors=1112
// let ans = 0;
let mergeSort = (left, right, arr, ans) => {
// console.log(left)
// console.log("*****")
// console.log(right)
// console.log("*****£££")
let i = 0,
j = 0,
k = 0;
let leftLen = left.length,
rightLen = right.length;
while (i < leftLen && j < rightLen) {
if (left[i] <= right[j]) {
arr[k] = left[i];
i++;
} else {
arr[k] = right[j];
ans += leftLen - i;
j++;
}
k++;
}
while (i < leftLen) {
arr[k] = left[i];
i++;
k++;
}
while (j < rightLen) {
arr[k] = right[j];
j++;
k++;
}
return { arr, ans };
};
let divideArray = (arr, ans) => {
if (arr.length < 2) return { arr, ans };
let N = arr.length;
let mid = Math.round(N / 2);
let left = arr.slice(0, mid);
let right = arr.slice(mid, N);
ans = ans;
let lans=divideArray(left, ans).ans;
let bAns=divideArray(right, ans).ans;
// console.log(bAns)
let blabla= mergeSort(left, right, arr, lans+bAns);
return blabla
};
let merge = (arr, ans) => {
let res=0+ divideArray(arr, ans).ans;
// console.log("asdad")
// console.log(res)
return res;
};
function main(input) {
let arr = [1,4,3,2,5];
let ans = 0;
let output = merge(arr, ans);
console.log('Final', output);
}
main();
I have persisted the count val inside your divide array and used it to merge the 2 counts from the split array i.e left and right direction split

Recursive algorithm fails to complete tests in allotted time

I was doing a test that required an algorithm for Binary Tomography. A set of 38 test values are supplied that test correctness, but there is also a time limit of 1 CPU sec to complete all the tests. The problem is as follows:
Output “Yes” if there exists an m-by-n matrix A, with each element either being 0 or 1, such that
Otherwise output “No”.
For each test, 2 arrays are provided:
r (the sum of each row in the matrix)
c (the sum of each column in the matrix)
In the equation:
m is the length of the r array, where 1 <= m
n is the length of the c array, where n <= 1000
ri is an element of r, where 0 <= ri <= n
cj is an element of c, where 0 <= cj <= m
A "Yes" example
m = 3;
n = 4;
r = [2, 3, 2];
c = [1, 1, 3, 2];
A "No" example
m = 3;
n = 3;
r = [0, 0, 3];
c = [0, 0, 3];
I have a solution that appears to give correct answers, however it only manages 12 / 38 tests before the 1 second of CPU time is exceeded.
I originally wrote the code in ES5 and then went back and converted to to ES3 to try and get more performance out of it. (originally managed 9 tests as ES5). There doesn't seem a great deal left that I can do to the current algorithm to improve the performance (unless I am mistaken). This leads me to believe that my algorithm is at fault an that there must be a faster algorithm for doing this. I did a ton of reading trying to find one and ended up with a headache :)
So I'm turning to the community to see if anyone can suggest a faster algorithm than I am currently using.
'use strict';
const ZEROS = (function (seed) {
let string = seed;
for (let i = 0; i < 19; i += 1) {
string += seed;
}
return string;
}('00000000000000000000000000000000000000000000000000'));
const ZEROSLEN = ZEROS.length;
const permutate = function (n, ri) {
const result = [];
const memoize = {};
let count = 0;
do {
const bin = count.toString(2);
if (ZEROSLEN + bin.length > ZEROSLEN + n) {
break;
}
if (!memoize[bin] && (bin.split('1').length - 1) === ri) {
const string = (ZEROS + bin).slice(-n);
const sLen = string.length;
const perm = new Array(sLen);
for (let i = sLen - 1; i >= 0; i -= 1) {
perm[i] = +string[i];
}
memoize[bin] = result.push(perm);
}
count += 1;
} while (count);
return result;
};
const getMatrixSum = function (n, matrix) {
const mLength = matrix.length;
const rows = new Array(mLength);
const a = new Array(n);
const last = mLength - 1;
for (let x = n - 1; x >= 0; x -= 1) {
for (let y = last; y >= 0; y -= 1) {
rows[y] = matrix[y][x];
}
let sum = 0;
for (let i = rows.length - 1; i >= 0; i -= 1) {
sum += rows[i];
}
a[x] = sum;
}
return a;
};
const isEqual = function (a, b) {
const length = a.length;
if (length !== b.length) {
return false;
}
for (let i = length - 1; i >= 0; i -= 1) {
if (a[i] !== b[i]) {
return false;
}
}
return true;
};
const addRow = function (i, prev, r, c, result) {
if (result) {
return result;
}
const n = c.length;
const ri = r[i];
if (ri < 0 || ri > n) {
throw new RangeError('ri out of range');
}
const p = permutate(n, ri);
const m = r.length;
const rsLast = m - 1;
const nextI = i + 1;
for (let x = p.length - 1; x >= 0; x -= 1) {
const permutation = p[x];
const next = prev.slice();
next.push(permutation);
const sums = getMatrixSum(n, next);
if (i < rsLast) {
let memo = 0;
for (let j = sums.length - 1; j >= 0; j -= 1) {
if (sums[j] > c[j]) {
memo += 1;
}
}
if (!memo && addRow(nextI, next, r, c, result)) {
return true;
}
} else if (isEqual(sums, c)) {
return true;
}
}
return false;
};
const isSolvable = function (r, c) {
const m = r.length;
const n = c.length;
if (m < 1 || n > 1000) {
throw new Error('Bad data');
}
for (let j = n; j >= 0; j -= 1) {
const cj = c[j];
if (cj < 0 || cj > m) {
throw new RangeError('cj out of range');
}
}
return addRow(0, [], r, c, false) ? 'Yes' : 'No';
};
console.log(isSolvable([2, 3, 2], [1, 1, 3, 2]));
console.log(isSolvable([0, 0, 3], [0, 0, 3]));
It may be worth noting that the tests are being run on SpiderMonkey version JavaScript-C24.2.0
Refs:
https://en.wikipedia.org/wiki/Discrete_tomography
https://open.kattis.com/problems/tomography

Since permutations yield to brute force, they should be the last resort when developing algorithms similar to this one. Most of the time they are not needed.
As i have commented above, I have a feeling that one strategy could be first sorting the r and c arrays descending and start with the bigger ones. I haven't had time to implemented a JS code to work out this, so I haven't had a chance to test thoroughly. Please have a look and if you discover a flaw please mention.
In the below visual representation of the algorithm we try r = [1,3,1,3] and c = [3,2,1,2]. X denotes an occupied cell and a red dot denotes an untouchable cell while the empty ones are obviously the free cells. So in the real algorithm to represent a cell we need a data type like {value: false, avail: false} for a red dot while {value: false, avail: true} would mean a free space. Or to save space and speed you may use a data type like 0b00 for red dot, 0b01 for free space and 0b1X for occupied (X here means don't care) cells.
Note: It's worth mentioning Step 3 where we process c[0]. After we insert the three Xs we have to check the rows occupied by the Xs to update the status of the empty cells in those rows. In this case for r[2], all empty cells become untouchable.
Edit:
Well.. OK since we don't need to construct the solution in a 2D array like structure but only need an answer on wheather the supplied data is meaningful or not, I have come up with another and simpler idea which is essentially based on the above approach. I really don't think it can get any faster than this. It solves a 999 by 1000 board in like 50ms.
Lets get into it.
The input is r = [2, 3, 2]; c = [1, 1, 3, 2]; However one important condition here is both c and r arrays should sum up to the same number. We can simply check this at the beginning of our code or leave it, go through the following steps and if they pass check only if c is full of 0s. The following code prefers the latter approach.
Sort r descending so; r = [3, 2, 2]; c = [1, 1, 3, 2];
Try reducing r[0] (3 in the first case) many non-zero elements of c by 1. Now c becomes [0, 0, 2, 2]. If it fails then try no more and return false.
Now that we have finished with row r[0], recursivelly call function with r = [2, 2]; c = [0, 0, 2, 2]; while r.length is bigger than 0 and the bool argument b is true. Next call will be r = [2]; c = [0, 0, 1, 1]; and finally r = []; c = [0, 0, 0, 0];
If finally a recursive call with empty r is invoked then check b is true and all items of c are 0. (b && cs.every(n => !n)).
I believe this is just fine but as i don't have your test cases it's for you to try. I am sure it will pass the time test though. Here is the code in it's simplest. Here i am testing rs = [7,3,5,4,6,2,8] and cs = [7,1,6,3,4,5,2,7]. It looks like;
71634527
7 x xxxxxx
3 x x x
5 x x xx x
4 x x x x
6 x xxxx x
2 x x
8 xxxxxxxx
function nonogram(rs,cs){
function runner(rs,cs, b = true){//console.log(rs,cs,b)
return b && rs.length ? runner(rs.slice(1), // rows argument
cs.map(e => rs[0] ? e ? (b = !--rs[0], e-1) // cols argument
: e
: e),
b) // bool argument
: b && cs.every(n => !n);
}
return runner(rs.sort((a,b) => b-a), cs);
}
var rs = [7,3,5,4,6,2,8],
cs = [7,1,6,3,4,5,2,7],
result;
console.time("test");
result = nonogram(rs,cs);
console.timeEnd("test");
console.log(result);

I didn't have this ready for my test, but I found a far more efficient algorithm after the event.
'use strict';
const sortNumber = function (a, b) {
return b - a;
};
const isSolvable = function (r, c) {
const m = r.length;
const n = c.length;
if (m < 1 || n > 1000) {
throw new Error('Bad data');
}
for (let j = n; j >= 0; j -= 1) {
const cj = c[j];
if (cj < 0 || cj > m) {
throw new RangeError('cj out of range');
}
}
while (r.length) {
c.sort(sortNumber);
const ri = r.pop();
if (ri < 0 || ri > n) {
throw new RangeError('ri out of range');
}
if (ri) {
if (!c[ri - 1]) {
return 'No';
}
for (let j = ri - 1; j >= 0; j -= 1) {
c[j] -= 1;
}
}
}
for (let j = n - 1; j >= 0; j -= 1) {
if (c[j]) {
return 'No';
}
}
return 'Yes';
};
console.log(isSolvable([2, 3, 2], [1, 1, 3, 2]));
console.log(isSolvable([0, 0, 3], [0, 0, 3]));

Partial sort in JavaScript

Is there any built-in JavaScript function to do a partial sort? If not, what is a good way to implement it?
Given an unsorted array of N elements, I would like to find K elements that are minimal with respect to some weighting function. K is much smaller than N, so it would be inefficient to sort the whole array and take the first K elements.
I would be happy even if there was something non-standard, browser-dependent. I could still fallback to the custom JavaScript implementation.
PS: This is my current custom implementation (without taking a weighting function into account, just sorting the elements as they are for simplicity):
function bisect(items, x, lo, hi) {
var mid;
if (typeof(lo) == 'undefined') lo = 0;
if (typeof(hi) == 'undefined') hi = items.length;
while (lo < hi) {
mid = Math.floor((lo + hi) / 2);
if (x < items[mid]) hi = mid;
else lo = mid + 1;
}
return lo;
}
function insort(items, x) {
items.splice(bisect(items, x), 0, x);
}
function partialSort(items, k) {
var smallest = [];
for (var i = 0, len = items.length; i < len; ++i) {
var item = items[i];
if (smallest.length < k || item < smallest[smallest.length - 1]) {
insort(smallest, item);
if (smallest.length > k)
smallest.splice(k, 1);
}
}
return smallest;
}
console.log(partialSort([5, 4, 3, 2, 1, 6, 7, 8, 1, 9], 3));
The algorithm walks through the given array one single time, keeping track of a sorted list of the k smallest items so far, using binary search to insert new elements.
Please post alternative solutions if you think they might be faster or more elegant. Timings are very welcome.

No. There's only the full array sort, so you will need to use your own implementation.
Little improvement on your code (I had thought of exactly the same algorithm :-)):
function partialSort(items, k) {
var smallest = items.slice(0, k).sort(),
max = smallest[k-1];
for (var i = k, len = items.length; i < len; ++i) {
var item = items[i];
if (item < max) {
insort(smallest, item);
smallest.length = k;
max = smallest[k-1];
}
}
return smallest;
}
(Even seems to be a little faster, I guess due to caching the max variable)

For relatively small k it can be worth it to implement a Max Heap (by lack of a native one in JavaScript):
Create a Max Heap of the first k values
For each remaining value:
If it is less than the root of the heap, replace the root with this value. Otherwise ignore the value. Note that the heap never changes in size.
Finally sort the heap and return it.
This is in fact an improvement on another idea using a Min Heap, but that one needs to heapify the whole array, and so will not run as fast. After heapifying the whole array, you just extract k times a value from that heap, and return those values.
I have added both solutions to Bergi's jsperf.com performance tests (copied to jsbench.me). For that particular test (5000 array values, k = 10), the Max Heap solution is faster. But this advantage will shrink as k is increased.
Here is the code for the Max Heap solution:
// A few Heap-functions that operate on an array
function maxSiftDown(arr, i=0, value=arr[i]) {
if (i >= arr.length) return;
while (true) {
var j = i*2+1;
if (j+1 < arr.length && arr[j] < arr[j+1]) j++;
if (j >= arr.length || value >= arr[j]) break;
arr[i] = arr[j];
i = j;
}
arr[i] = value;
}
function maxHeapify(arr) {
for (var i = arr.length>>1; i--; ) maxSiftDown(arr, i);
return arr;
}
// The main algorithm
function partialSortWithMaxHeap(items, k) {
var heap = maxHeapify(items.slice(0, k));
for (var i = k, len = items.length; i < len; ++i) {
var item = items[i];
if (item < heap[0]) maxSiftDown(heap, 0, item);
}
return heap.sort((a,b) => a-b);
}
// Sample data & call
var arr = Array.from({length:5000}, () => Math.floor(Math.random() * 1e5));
console.log(partialSortWithMaxHeap(arr, 10));

There's no native partial sort function. The closest thing to what you want is Array.filter.
function isSmallEnough(element, index, array) {
return (element <= 10);
}
var filtered = [12, 5, 8, 130, 44].filter(isSmallEnough);
// filtered is [5, 8]
The example was borrowed (and slightly modified) from the above link.

I made a version than works with objects, like Array.sort(f):
function partialSort(items, k,f) {
function bisect(items, x, lo, hi) {
var mid;
if (typeof(lo) == 'undefined') lo = 0;
if (typeof(hi) == 'undefined') hi = items.length;
while (lo < hi) {
mid = Math.floor((lo + hi) / 2);
if (0>f(x,items[mid])) hi = mid;
else lo = mid + 1;
}
return lo;
}
function insort(items, x) {
items.splice(bisect(items, x), 0, x);
}
var smallest = items.slice(0, k).sort(f),
max = smallest[k-1];
for (var i = k, len = items.length; i < len; ++i) {
var item = items[i];
if (0>f(item,max)) {
insort(smallest, item);
smallest.length = k;
max = smallest[k-1];
}
}
return smallest;
}
// [ { e: 1 }, { e: 1 }, { e: 2 } ]
console.log(partialSort([{e:4},{e:6},{e:1},{e:8},{e:3},{e:1},{e:6},{e:2}],3,(a,b)=>a.e-b.e))
console.log()

In the code block below, the nisetamafibo function keeps an array of the smallest items found so far. The array is sorted and truncated to length K after a certain number of new items have been added to the array, where the number is taken from the Fibonacci sequence so that it is first 1, next 1, then 2, then 3, then 5, and so on. The nisetamadouble method doubles the interval after which the array is sorted instead, so that it is first 1, then 2, then 4, and so on. (I also tried the approach that I sorted the array each time after a fixed number of new items like 10 had been added, but it was slower. And I also tried to initialize the array at the start of the function so that I took in a fixed number of the first items of the input and sorted them, but I found that initializing the array with 1 or 0 items was the fastest, so I removed the initialization step.)
The nisetamainsertion function uses insertion sort to sort the items. It's very slow at high K-values because insertion sort has quadratic time complexity, but it's fast at K-values of around 10 to 100 or lower, because insertion sort is fast for short arrays. The nisetamachoose method chooses nisetamainsertion for K-values of 100 or less but nisetamafibo otherwise. (In the Java JDK, the file DualPivotQuicksort.java uses insertion sort instead of quicksort for arrays with less than 47 items. A presentation about sorting algorithms in R said that "fastest for < 30 items is insert sort".)
I also tried to implement the quickselect algorithm with and without recursion. The version that didn't use recursion was a bit faster, but both versions were still slow compared to other methods especially in cases where N was high and K was low.
On another Stack Exchange site, someone came up with new variants of the Floyd-Rivest algorithm which were faster than the regular Floyd-Rivest algorithm in C: https://softwareengineering.stackexchange.com/questions/284767/kth-selection-routine-floyd-algorithm-489. I tried to implement the variant called select7MO3 in JavaScript, but it ended up being one of the slowest options in my benchmark.
function nisetamafibo(a,k=1){
let found=[],len=a.length,unsorted=0,biggestfound=Infinity,nextsort=1,prevsort=1,oldsort
for(let i=0;i<len;i++){
if(a[i]<biggestfound||i<k){
found.push(a[i])
if(++unsorted==nextsort){
found.sort((l,r)=>l<r?-1:l>r?1:0)
found=found.slice(0,k)
biggestfound=found[found.length-1]
oldsort=nextsort;nextsort+=prevsort;prevsort=oldsort
unsorted=0
}
}
}
found.sort((l,r)=>l<r?-1:l>r?1:0)
return found.slice(0,k)
}
function nisetamadouble(a,k=1){
let found=[],len=a.length,unsorted=0,biggestfound=Infinity,nextsort=1
for(let i=0;i<len;i++){
if(a[i]<biggestfound||i<k){
found.push(a[i])
if(++unsorted==nextsort){
found.sort((l,r)=>l<r?-1:l>r?1:0)
found=found.slice(0,k)
biggestfound=found[found.length-1]
nextsort*=2
unsorted=0
}
}
}
found.sort((l,r)=>l<r?-1:l>r?1:0)
return found.slice(0,k)
}
function nisetamainsertion(a,k=1){
let found=a.slice(0,k),l=a.length
found.sort((l,r)=>l<r?-1:l>r?1:0)
let biggestfound=found[k-1]
for(let i=0;i<l;i++){
let v=a[i]
if(v<biggestfound){
let insertat=k-1
for(let j=0;j<k-1;j++)if(v<found[j]||j==i){insertat=j;break}
for(let j=k-1;j>insertat;j--)found[j]=found[j-1]
found[insertat]=v
biggestfound=found[k-1]
}
}
return found
}
function nisetamachoose(a,k=1){
return k<=100?nisetamainsertion(a,k):nisetamafibo(a,k)
}
function quickselect(a,k,l,r){
l=l||0
r=r||a.length-1
while(true){
let pivot=a[r],pos=l
for(let i=l;i<=r;i++)if(a[i]<pivot){let temp=a[i];a[i]=a[pos];a[pos++]=temp}
let temp=a[r];a[r]=a[pos];a[pos]=temp
if(pos==k)break
pos<k?l=pos+1:r=pos-1
}
}
function quickselectrecursive(a,k,l,r){
l=l||0
r=r||a.length-1
let pivot=a[r],pos=l
for(let i=l;i<=r;i++)if(a[i]<pivot){let temp=a[i];a[i]=a[pos];a[pos++]=temp}
let temp=a[r];a[r]=a[pos];a[pos]=temp
if(pos<k)quickselectrecursive(a,pos+1,r,k)
if(pos>k)quickselectrecursive(a,l,pos-1,k)
}
function sortslice(a,k){
a.sort((l,r)=>l<r?-1:l>r?1:0)
return a.slice(0,k)
}
// https://softwareengineering.stackexchange.com/questions/284767/kth-selection-routine-floyd-algorithm-489
function select7MO3(a,k){
let l=0,i,r=a.length-1,rr=r,ll=l
while(r>l){
if(a[k]<a[l]){let t=a[l];a[l]=a[k];a[k]=t}
if(a[r]<a[l]){let t=a[l];a[l]=a[r];a[r]=t}
if(a[r]<a[k]){let t=a[k];a[k]=a[r];a[r]=t}
if((r-l)>k){
let n=r-l+1
i=k-l+1
let s=(2*n/3)
let div=i-n
let sd=(n*s*(n-s)/n)*(div<0?-1:div>0?1:0)
ll=Math.max(l,k-i*s/n+sd)
rr=Math.min(r,k+(n-i)*s/n+sd)
}
let pivot=a[k]
i=l
let j=r
let t=a[l];a[l]=a[k];a[k]=t
if(a[r]>pivot){t=a[r];a[r]=a[l];a[l]=t}
while(i<j){
let t=a[i];a[i]=a[j];a[j]=t
i++
j--
while(a[i]<pivot)i++
while(a[j]>pivot)j--
}
if(a[l]==pivot){i--;let t=a[l];a[l]=a[j];a[j]=t}
else{j++;let t=a[j];a[j]=a[r];a[r]=t}
if(j<=k)l=j+1
else if(k<=j)r=j-1
}
let out=a.slice(0,k)
out.sort((l,r)=>l<r?-1:l>r?1:0)
return out
}
// OP and Bergi
function bisect(items, x, lo, hi) {
var mid;
if (typeof(lo) == 'undefined') lo = 0;
if (typeof(hi) == 'undefined') hi = items.length;
while (lo < hi) {
mid = Math.floor((lo + hi) / 2);
if (x < items[mid]) hi = mid;
else lo = mid + 1;
}
return lo;
}
function insort(items, x) {
items.splice(bisect(items, x), 0, x);
}
function OP(items, k) {
var smallest = [];
for (var i = 0, len = items.length; i < len; ++i) {
var item = items[i];
if (smallest.length < k || item < smallest[smallest.length - 1]) {
insort(smallest, item);
if (smallest.length > k)
smallest.splice(k, 1);
}
}
return smallest;
}
function OP_Bergi(items, k) {
var smallest = items.slice(0, k).sort(),
max = smallest[k-1];
for (var i = k, len = items.length; i < len; ++i) {
var item = items[i];
if (item < max) {
insort(smallest, item);
smallest.length = k;
max = smallest[k-1];
}
}
return smallest;
}
// trincot
function maxSiftDown(arr, i=0, value=arr[i]) {
if (i >= arr.length) return;
while (true) {
var j = i*2+1;
if (j+1 < arr.length && arr[j] < arr[j+1]) j++;
if (j >= arr.length || value >= arr[j]) break;
arr[i] = arr[j];
i = j;
}
arr[i] = value;
}
function maxHeapify(arr) {
for (var i = arr.length>>1; i--; ) maxSiftDown(arr, i);
return arr;
}
function trincot_MaxHeap(items, k) {
var heap = maxHeapify(items.slice(0, k));
for (var i = k, len = items.length; i < len; ++i) {
var item = items[i];
if (item < heap[0]) maxSiftDown(heap, 0, item);
}
return heap.sort((a,b) => a-b);
}
// DiazJara
function DiazJara(items, k,f) {
function bisect(items, x, lo, hi) {
var mid;
if (typeof(lo) == 'undefined') lo = 0;
if (typeof(hi) == 'undefined') hi = items.length;
while (lo < hi) {
mid = Math.floor((lo + hi) / 2);
if (0>f(x,items[mid])) hi = mid;
else lo = mid + 1;
}
return lo;
}
function insort(items, x) {
items.splice(bisect(items, x), 0, x);
}
var smallest = items.slice(0, k).sort(f),
max = smallest[k-1];
for (var i = k, len = items.length; i < len; ++i) {
var item = items[i];
if (0>f(item,max)) {
insort(smallest, item);
smallest.length = k;
max = smallest[k-1];
}
}
return smallest;
}
// benchmark
for(let nk of'31 33 40 42 44 51 53 55 60 62 64 66 71 73 75'.split(' ')){
let n=parseInt(nk[0]),k0=parseInt(nk[1]),k=10**k0
let opt=[
'OP(a,k)',
'OP_Bergi(a,k)',
'trincot_MaxHeap(a,k)',
'DiazJara(a,k,(l,r)=>l-r)',
'DiazJara(a,k,(l,r)=>l<r?-1:l>r?1:0)',
'nisetamafibo(a,k)',
'nisetamadouble(a,k)',
// 'nisetamainsertion(a,k)', // this would've taken too long to run at K=1e6
'nisetamachoose(a,k)',
'quickselect(a,k);a=a.slice(0,k);a.sort((l,r)=>l<r?-1:l>r?1:0)',
'quickselectrecursive(a,k);a=a.slice(0,k);a.sort((l,r)=>l<r?-1:l>r?1:0)',
'select7MO3(a,k);a=a.slice(0,k);a.sort((l,r)=>l<r?-1:l>r?1:0)',
'sortslice(a,k)'
]
let ord=Array.from({length:100},()=>Array(opt.length).fill().map((_,i)=>i)).flat()
ord.sort(()=>Math.random()-.5)
for(let x of ord){
let o=opt[x]
let a=Array.from({length:10**n},()=>Math.random())
let t1=process.hrtime.bigint()
eval(o)
let t2=process.hrtime.bigint()-t1
console.log(n+' '+k0+' '+o+' '+t2)
}
}
This shows the median time of a hundred runs in ms and the average rank of each method (where for example 7/4 means that N was 1e7 and K was 1e4):
For most combinations of N and K, Bergi's modified version of the OP's code was actually slower than the OP's code, even though the OP's code was extremely slow in the case where N and K were both 1e6.
(l,r)=>l<r?-1:l>r?1:0 is faster than (l,r)=>l-r as you can see by comparing the two versions of Díaz-Jara's method above.
Here's also versions of my nisetamadouble and nisetamainsertion methods which return the indexes of the smallest items in addition to the values:
let a=Array.from({length:1e5},()=>Math.random())
let k=10
let l=a.length
let biggestfound=Infinity,foundind=[],foundval=[]
for(let i=0;i<l;i++){
let v=a[i]
if(i<k||v<biggestfound){
let insertat=k-1
for(let j=0;j<k-1;j++)if(v<foundval[j]||j==i){insertat=j;break}
for(let j=k-1;j>insertat;j--){foundind[j]=foundind[j-1];foundval[j]=foundval[j-1]}
foundind[insertat]=i
foundval[insertat]=v
biggestfound=foundval[k-1]
}
}
console.log(foundind)
console.log(foundval)
function nisetama(a,k=1){
let found=[],len=a.length,unsorted=0,biggestfound=Infinity,nextsort=1
for(let i=0;i<len;i++){
if(a[i]<biggestfound||i<k){
found.push(a[i])
if(++unsorted==nextsort){
found.sort((l,r)=>l<r?-1:l>r?1:0)
found=found.slice(0,k)
biggestfound=found[found.length-1]
nextsort*=2
unsorted=0
}
}
}
found.sort((l,r)=>l<r?-1:l>r?1:0)
return found.slice(0,k)
}
let a2=a
nisetama(a2,k)
biggestfound=a2[k-1],foundind=[]
for(let i=0;i<l;i++)if(a[i]<=biggestfound)foundind.push(i)
foundind.sort((l,r)=>a[l]<a[r]?-1:a[l]>a[r]?1:0)
foundind=foundind.slice(0,k)
console.log(foundind)
console.log(foundval)