Searching for a very efficient solution for permutations on array elements in JavaScript

permutations-in-js-exploration.js

let listOfArticles = ["HLP63P2I","HLP63P2I","R9T7FVD9","PLI82PFU","M62JFW8Q","W6FXWZHC","URVX9S22","UVF2BX3S","HH3R72BH","EUKZS3QR","PKRP3W4J","56YI28BH","WWZF8Z2V","YKEBAEX7","SB66FSJQ","HHJ739BB","VV26P2K9","ZZIDA78D","NWHASTVQ","RBLMDELC","BUEWZ258","3CQWK77M","WZBL4WV3","ZMPW6UQE","3APZQIYN","3XR95DME","C238ZP3G","PUIK8SYL","PQ7GP3G9","8EGSYTF4","SLEFYRWL","KVBLVJUU","YAPQI5C8","46NMCHCI","MBR34K6L","8K5HT8XM","WGLQYDYA","6MIN5N8G","9K6BU7U4","VWBWABD9","RWHZFBTC","GUBE5PM2","UCZZ7MFD","9T29K75C","J39EPBQU","44AFAI9W","JN8EEGB2"];
console.log(listOfArticles.length);

/* == The functional programming solution == https://stackoverflow.com/questions/43241174/javascript-generating-all-combinations-of-elements-in-a-single-array-in-pairs == */
let firstApproach = listOfArticles.flatMap(
    (v, i) => listOfArticles.slice(i+1).map( w => v + ' ' + w )
);
// console.log(firstApproach);
console.log(firstApproach.length);

/* === Second criptic solution (not suitable) === */
const combinations = (arr) => [...Array(2 ** arr.length - 1).keys()].map((n) => ((n + 1) >>> 0).toString(2).split("").reverse().map((n, i) => (+n ? arr[i] : false)).filter(Boolean)).sort((a, b) => (a.length > b.length ? 1 : -1));
console.log(combinations(["apple", "banana", "lemon", "mango"]));

/* === Third solution === */ 
var results = [];
for (var i = 0; i < listOfArticles.length - 1; i++) {
    // This is where you'll capture that last value
    for (var j = i + 1; j < listOfArticles.length; j++) {
      results.push(listOfArticles[i] + ' ' + listOfArticles[j]);
    }
  }

console.log(results);


function* generateDistinctPairs(array) {
    for (let i = 0; i < array.length; i++) {
      for (let j = i + 1; j < array.length; j++) {
        yield [array[i], array[j]];
      }
    }
  };

console.log(generateDistinctPairs(listOfArticles))

// Another solution
const permutator = (inputArr) => {
  let result = [];

  const permute = (arr, m = []) => {
    if (arr.length === 0) {
      result.push(m)
    } else {
      for (let i = 0; i < arr.length; i++) {
        let curr = arr.slice();
        let next = curr.splice(i, 1);
        permute(curr.slice(), m.concat(next))
     }
   }
 }

 permute(inputArr)

 return result;
}
// source https://stackoverflow.com/questions/9960908/permutations-in-javascript/30013816#30013816

// This looks very promissing
function* permute(permutation) {
  var length = permutation.length,
      c = Array(length).fill(0),
      i = 1, k, p;

  yield permutation.slice();
  while (i < length) {
    if (c[i] < i) {
      k = i % 2 && c[i];
      p = permutation[i];
      permutation[i] = permutation[k];
      permutation[k] = p;
      ++c[i];
      i = 1;
      yield permutation.slice();
    } else {
      c[i] = 0;
      ++i;
    }
  }
}

// Memory efficient iteration through permutations:
for (var permutation of permute([1, 2, 3])) console.log(permutation);

// Simple array conversion:
var permutations = [...permute([1, 2, 3])];
// investigate!!!

/*
Documentation
https://stackoverflow.com/questions/43241174/javascript-generating-all-combinations-of-elements-in-a-single-array-in-pairs
Cool package: https://www.npmjs.com/package/js-combinatorics
Super cool proj: https://github.com/foo123/Abacus Explore!!!
*/

For benchmarking: https://benchmarkjs.com/