I have a slightly different approach which I believe is O(n^2) time complexity, I imagine the space complexity is the same O(n^3) but correct me if I'm wrong.

class Solution {
  findSubarrays(arr, target) {
    let result = [];
    let l = 0;
    let r = 0;
    let runningSum = 1;

    while (r <= arr.length) {
      runningSum *= arr[r];
      
      if (runningSum < target) {
        result.push(arr.slice(l, r + 1));
        r++
      } else {
        l += 1
        runningSum = 1;
        r = l
      }
    }

    return result;
  }
}