数组最小乘积子集的 JavaScript 程序

[2], [3], [-1], [4], [-2], [2, 3], [2, -1], [2, 4], [2, -2], [3, -1], [3, 4], [3, -2], [-1, 4], [-1, -2], [4, -2], [2, 3, -1], [2, 3, 4], [2, 3, -2], [2, -1, 4], [2, -1, -2], [2, 4, -2], [3, -1, 4], [3, -1, -2], [3, 4, -2], [-1, 4, -2], and [2, 3, -1, 4, -2]. 

function minProductSubset(a, n) {
   if (n === 1) {
      return [a[0], a[0]];
   }
   let negmax = Number.NEGATIVE_INFINITY;
   let posmin = Number.POSITIVE_INFINITY;
   let count_neg = 0, count_zero = 0;
   let subsets = [[]];  
   for (let i = 0; i < n; i++) {
      if (a[i] === 0) {
         count_zero++;
         continue;
      }
      if (a[i] < 0) {
         count_neg++;
         negmax = Math.max(negmax, a[i]);
      }
      if (a[i] > 0 && a[i] < posmin) {
         posmin = a[i];
      }
      const subsetsLength = subsets.length;
      for(let j = 0; j < subsetsLength; j++){
         const subset = [...subsets[j], a[i]];
         subsets.push(subset);
      }
   }
   if (count_zero === n || (count_neg === 0 && count_zero > 0)) {
      return [0, 0];
   }
   if (count_neg === 0) {
      return [posmin, posmin];
   }
   const negativeSubsets = subsets.filter(subset => subset.reduce((acc, cur) => acc * cur, 1) < 0);
   let minSubset = negativeSubsets[0];
   let minProduct = minSubset.reduce((acc, cur) => acc * cur, 1);
   for (let i = 1; i < negativeSubsets.length; i++) {
      const product = negativeSubsets[i].reduce((acc, cur) => acc * cur, 1);
      if (product < minProduct) {
         minSubset = negativeSubsets[i];
         minProduct = product;
      }
   }  
   return [minSubset, minProduct];
}
let a = [-1, -1, -2, 4, 3];
let n = 5;
const [minSubset, minProduct] = minProductSubset(a, n);
console.log(`The minimum subset is [ ${minSubset.join(', ')} ] and the minimum product is ${minProduct}.`);