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6 | 6 | * @return {*[][]} - All subsets of original set.
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7 | 7 | */
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8 | 8 | function btPowerSetRecursive(originalSet, allSubsets = [[]], currentSubSet = [], startAt = 0) {
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9 |
| - // In order to avoid duplication we need to start from next element every time we're forming a |
10 |
| - // subset. If we will start from zero then we'll have duplicates like {3, 3, 3}. |
| 9 | + // Let's iterate over originalSet elements that may be added to the subset |
| 10 | + // without having duplicates. The value of startAt prevents adding the duplicates. |
11 | 11 | for (let position = startAt; position < originalSet.length; position += 1) {
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12 |
| - // Let's push current element to the subset. |
| 12 | + // Let's push current element to the subset |
13 | 13 | currentSubSet.push(originalSet[position]);
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| 14 | + |
14 | 15 | // Current subset is already valid so let's memorize it.
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| 16 | + // We do array destruction here to save the clone of the currentSubSet. |
| 17 | + // We need to save a clone since the original currentSubSet is going to be |
| 18 | + // mutated in further recursive calls. |
15 | 19 | allSubsets.push([...currentSubSet]);
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16 |
| - // Let's try to form all other subsets for the current subset. |
| 20 | + |
| 21 | + // Let's try to generate all other subsets for the current subset. |
| 22 | + // We're increasing the position by one to avoid duplicates in subset. |
17 | 23 | btPowerSetRecursive(originalSet, allSubsets, currentSubSet, position + 1);
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18 |
| - // BACKTRACK. Exclude last element from the subset and try the next one. |
| 24 | + |
| 25 | + // BACKTRACK. Exclude last element from the subset and try the next valid one. |
19 | 26 | currentSubSet.pop();
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20 | 27 | }
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21 | 28 |
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