Add backtracking solution for finding the power-set of a set.

README.md

@@ -77,7 +77,7 @@ 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
+  * `A` [Power Set](src/algorithms/sets/power-set) - all subsets of a set (bitwise and backtracking 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)
@@ -190,6 +190,7 @@ if it satisfies all conditions, and only then continue generating subsequent sol
 different path of finding a solution. Normally the DFS traversal of state-space is being used.
   * `B` [Jump Game](src/algorithms/uncategorized/jump-game)
   * `B` [Unique Paths](src/algorithms/uncategorized/unique-paths)
+  * `B` [Power Set](src/algorithms/sets/power-set) - all subsets of a set
   * `A` [Hamiltonian Cycle](src/algorithms/graph/hamiltonian-cycle) - Visit every vertex exactly once
   * `A` [N-Queens Problem](src/algorithms/uncategorized/n-queens)
   * `A` [Knight's Tour](src/algorithms/uncategorized/knight-tour)

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

@@ -1,11 +1,62 @@
 # Power Set
 
-Power set of a set A is the set of all of the subsets of A.
+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)`. 
 
-Eg. for `{x, y, z}`, the subsets are : `{{}, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}}`
+For example for `{x, y, z}`, the subsets
+are:
+
+```text
+{
+  {}, // (also denoted empty set ∅ or the null set)
+  {x},
+  {y},
+  {z},
+  {x, y},
+  {x, z},
+  {y, z},
+  {x, y, z}
+}
+```
 
 ![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 
+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 
+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 
+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 
+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 
+number of `0b010` would mean that we need to include only `2` to the current set.
+
+> See [bwPowerSet.js](./bwPowerSet.js) file for bitwise solution.
+
+### Backtracking Solution
+
+In backtracking approach we're constantly trying to add next element of the set
+to the subset, memorizing it and then removing it and try the same with the next
+element.
+
+> See [btPowerSet.js](./btPowerSet.js) file for backtracking solution.
+
 ## References
 
 * [Wikipedia](https://en.wikipedia.org/wiki/Power_set)

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

@@ -0,0 +1,21 @@
+import btPowerSet from '../btPowerSet';
+
+describe('btPowerSet', () => {
+  it('should calculate power set of given set using backtracking approach', () => {
+    expect(btPowerSet([1])).toEqual([
+      [],
+      [1],
+    ]);
+
+    expect(btPowerSet([1, 2, 3])).toEqual([
+      [],
+      [1],
+      [1, 2],
+      [1, 2, 3],
+      [1, 3],
+      [2],
+      [2, 3],
+      [3],
+    ]);
+  });
+});

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

@@ -0,0 +1,21 @@
+import bwPowerSet from '../bwPowerSet';
+
+describe('bwPowerSet', () => {
+  it('should calculate power set of given set using bitwise approach', () => {
+    expect(bwPowerSet([1])).toEqual([
+      [],
+      [1],
+    ]);
+
+    expect(bwPowerSet([1, 2, 3])).toEqual([
+      [],
+      [1],
+      [2],
+      [1, 2],
+      [3],
+      [1, 3],
+      [2, 3],
+      [1, 2, 3],
+    ]);
+  });
+});

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

@@ -1,24 +0,0 @@
-import powerSet from '../powerSet';
-
-describe('powerSet', () => {
-  it('should calculate power set of given set', () => {
-    const powerSets1 = powerSet([1]);
-    const powerSets2 = powerSet([1, 2, 3]);
-
-    expect(powerSets1).toEqual([
-      [],
-      [1],
-    ]);
-
-    expect(powerSets2).toEqual([
-      [],
-      [1],
-      [2],
-      [1, 2],
-      [3],
-      [1, 3],
-      [2, 3],
-      [1, 2, 3],
-    ]);
-  });
-});

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

@@ -0,0 +1,34 @@
+/**
+ * @param {*[]} originalSet - Original set of elements we're forming power-set of.
+ * @param {*[][]} allSubsets - All subsets that have been formed so far.
+ * @param {*[]} currentSubSet - Current subset that we're forming at the moment.
+ * @param {number} startAt - The position of in original set we're starting to form current subset.
+ * @return {*[][]} - All subsets of original set.
+ */
+function btPowerSetRecursive(originalSet, allSubsets = [[]], currentSubSet = [], startAt = 0) {
+  // In order to avoid duplication we need to start from next element every time we're forming a
+  // subset. If we will start from zero then we'll have duplicates like {3, 3, 3}.
+  for (let position = startAt; position < originalSet.length; position += 1) {
+    // Let's push current element to the subset.
+    currentSubSet.push(originalSet[position]);
+    // Current subset is already valid so let's memorize it.
+    allSubsets.push([...currentSubSet]);
+    // Let's try to form all other subsets for the current subset.
+    btPowerSetRecursive(originalSet, allSubsets, currentSubSet, position + 1);
+    // BACKTRACK. Exclude last element from the subset and try the next one.
+    currentSubSet.pop();
+  }
+
+  // Return all subsets of a set.
+  return allSubsets;
+}
+
+/**
+ * Find power-set of a set using BACKTRACKING approach.
+ *
+ * @param {*[]} originalSet
+ * @return {*[][]}
+ */
+export default function btPowerSet(originalSet) {
+  return btPowerSetRecursive(originalSet);
+}

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

@@ -1,3 +1,9 @@
+/**
+ * Find power-set of a set using BITWISE approach.
+ *
+ * @param {*[]} originalSet
+ * @return {*[][]}
+ */
 export default function powerSet(originalSet) {
   const subSets = [];
 
@@ -7,8 +13,8 @@ export default function powerSet(originalSet) {
   const numberOfCombinations = 2 ** originalSet.length;
 
   // Each number in binary representation in a range from 0 to 2^n does exactly what we need:
-  // it shoes 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 010 would mean that we need to
+  // 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.
   for (let combinationIndex = 0; combinationIndex < numberOfCombinations; combinationIndex += 1) {
     const subSet = [];