Adding a simple cascading solution to generate a Power Set (trekhleb#975)

* Add a simple cascading version of generating a PowerSet.

* Update README.

* Update README.

* Update README.

Author: trekhleb

README.md

@@ -99,7 +99,7 @@ a set of rules that precisely define a sequence of operations.
 * **Sets**
   * `B` [Cartesian Product](src/algorithms/sets/cartesian-product) - product of multiple sets
   * `B` [Fisher–Yates Shuffle](src/algorithms/sets/fisher-yates) - random permutation of a finite sequence
-  * `A` [Power Set](src/algorithms/sets/power-set) - all subsets of a set (bitwise and backtracking solutions)
+  * `A` [Power Set](src/algorithms/sets/power-set) - all subsets of a set (bitwise, backtracking, and cascading solutions)
   * `A` [Permutations](src/algorithms/sets/permutations) (with and without repetitions)
   * `A` [Combinations](src/algorithms/sets/combinations) (with and without repetitions)
   * `A` [Longest Common Subsequence](src/algorithms/sets/longest-common-subsequence) (LCS)

src/algorithms/sets/power-set/README.md

@@ -1,7 +1,7 @@
 # Power Set
 
 Power set of a set `S` is the set of all of the subsets of `S`, including the
-empty set and `S` itself. Power set of set `S` is denoted as `P(S)`. 
+empty set and `S` itself. Power set of set `S` is denoted as `P(S)`.
 
 For example for `{x, y, z}`, the subsets
 are:
@@ -21,37 +21,37 @@ are:
 
 ![Power Set](https://www.mathsisfun.com/sets/images/power-set.svg)
 
-Here is how we may illustrate the elements of the power set of the set `{x, y, z}` ordered with respect to 
+Here is how we may illustrate the elements of the power set of the set `{x, y, z}` ordered with respect to
 inclusion:
 
 ![](https://upload.wikimedia.org/wikipedia/commons/e/ea/Hasse_diagram_of_powerset_of_3.svg)
 
 **Number of Subsets**
 
-If `S` is a finite set with `|S| = n` elements, then the number of subsets 
-of `S` is `|P(S)| = 2^n`. This fact, which is the motivation for the 
+If `S` is a finite set with `|S| = n` elements, then the number of subsets
+of `S` is `|P(S)| = 2^n`. This fact, which is the motivation for the
 notation `2^S`, may be demonstrated simply as follows:
 
-> First, order the elements of `S` in any manner. We write any subset of `S` in 
-the format `{γ1, γ2, ..., γn}` where `γi , 1 ≤ i ≤ n`, can take the value 
+> First, order the elements of `S` in any manner. We write any subset of `S` in
+the format `{γ1, γ2, ..., γn}` where `γi , 1 ≤ i ≤ n`, can take the value
 of `0` or `1`. If `γi = 1`, the `i`-th element of `S` is in the subset;
-otherwise, the `i`-th element is not in the subset. Clearly the number of 
+otherwise, the `i`-th element is not in the subset. Clearly the number of
 distinct subsets that can be constructed this way is `2^n` as `γi ∈ {0, 1}`.
 
 ## Algorithms
 
 ### Bitwise Solution
 
-Each number in binary representation in a range from `0` to `2^n` does exactly 
-what we need: it shows by its bits (`0` or `1`) whether to include related 
-element from the set or not. For example, for the set `{1, 2, 3}` the binary 
+Each number in binary representation in a range from `0` to `2^n` does exactly
+what we need: it shows by its bits (`0` or `1`) whether to include related
+element from the set or not. For example, for the set `{1, 2, 3}` the binary
 number of `0b010` would mean that we need to include only `2` to the current set.
 
 |       | `abc` | Subset        |
 | :---: | :---: | :-----------: |
 | `0`   | `000` | `{}`          |
 | `1`   | `001` | `{c}`         |
-| `2`   | `010` | `{b}`         | 
+| `2`   | `010` | `{b}`         |
 | `3`   | `011` | `{c, b}`      |
 | `4`   | `100` | `{a}`         |
 | `5`   | `101` | `{a, c}`      |
@@ -68,6 +68,44 @@ element.
 
 > See [btPowerSet.js](./btPowerSet.js) file for backtracking solution.
 
+### Cascading Solution
+
+This is, arguably, the simplest solution to generate a Power Set.
+
+We start with an empty set:
+
+```text
+powerSets = [[]]
+```
+
+Now, let's say:
+
+```text
+originalSet = [1, 2, 3]
+```
+
+Let's add the 1st element from the originalSet to all existing sets:
+
+```text
+[[]] ← 1 = [[], [1]]
+```
+
+Adding the 2nd element to all existing sets:
+
+```text
+[[], [1]] ← 2 = [[], [1], [2], [1, 2]]
+```
+
+Adding the 3nd element to all existing sets:
+
+```
+[[], [1], [2], [1, 2]] ← 3 = [[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]
+```
+
+And so on, for the rest of the elements from the `originalSet`. On every iteration the number of sets is doubled, so we'll get `2^n` sets.
+
+> See [caPowerSet.js](./caPowerSet.js) file for cascading solution.
+
 ## References
 
 * [Wikipedia](https://en.wikipedia.org/wiki/Power_set)

src/algorithms/sets/power-set/__test__/caPowerSet.test.js

@@ -0,0 +1,28 @@
+import caPowerSet from '../caPowerSet';
+
+describe('caPowerSet', () => {
+  it('should calculate power set of given set using cascading approach', () => {
+    expect(caPowerSet([1])).toEqual([
+      [],
+      [1],
+    ]);
+
+    expect(caPowerSet([1, 2])).toEqual([
+      [],
+      [1],
+      [2],
+      [1, 2],
+    ]);
+
+    expect(caPowerSet([1, 2, 3])).toEqual([
+      [],
+      [1],
+      [2],
+      [1, 2],
+      [3],
+      [1, 3],
+      [2, 3],
+      [1, 2, 3],
+    ]);
+  });
+});

src/algorithms/sets/power-set/caPowerSet.js

@@ -0,0 +1,37 @@
+/**
+ * Find power-set of a set using CASCADING approach.
+ *
+ * @param {*[]} originalSet
+ * @return {*[][]}
+ */
+export default function caPowerSet(originalSet) {
+  // Let's start with an empty set.
+  const sets = [[]];
+
+  /*
+    Now, let's say:
+    originalSet = [1, 2, 3].
+
+    Let's add the first element from the originalSet to all existing sets:
+    [[]] ← 1 = [[], [1]]
+
+    Adding the 2nd element to all existing sets:
+    [[], [1]] ← 2 = [[], [1], [2], [1, 2]]
+
+    Adding the 3nd element to all existing sets:
+    [[], [1], [2], [1, 2]] ← 3 = [[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]
+
+    And so on for the rest of the elements from originalSet.
+    On every iteration the number of sets is doubled, so we'll get 2^n sets.
+  */
+  for (let numIdx = 0; numIdx < originalSet.length; numIdx += 1) {
+    const existingSetsNum = sets.length;
+
+    for (let setIdx = 0; setIdx < existingSetsNum; setIdx += 1) {
+      const set = [...sets[setIdx], originalSet[numIdx]];
+      sets.push(set);
+    }
+  }
+
+  return sets;
+}