Generate All Subsets Of A Set Problem Statement

Generate ALL possible subsets of a given set. The set is given in the form of a string s containing distinct lowercase characters 'a' - 'z'.

Example

{
"s": "xy"
}

Output:

["", "x", "y", "xy"]

Notes

• Any set is a subset of itself.
• Empty set is a subset of any set.
• Output contains ALL possible subsets of given string.
• Order of strings in the output does not matter. E.g. s = "a", arrays ["", "a"] and ["a", ""] both will be accepted.
• Order of characters in any subset must be same as in the input string. For s = "xy", array ["", "x", "y", "xy"] will be accepted, but ["", "x", "y", "yx"] will not be accepted.

Constraints:

• 0 <= length of s <= 19
• s only contains distinct lowercase English letters.

We have provided 2 solutions:

• Recursive Solution : other_solution.cpp.
• Iterative Solution : Generate All Subsets Of A Set Solution: Optimal.

Have a look at both the solutions.

optimal_solution.cpp

Both solutions are valid, but the recursive solution is slightly slower because of function calls and variable passing.

Also you should observe that the number of subsets will always be a power of 2, specifically 2ⁿ, where n represents the length of string s.

Time Complexity

O(2ⁿ * n).

As we will generate 2ⁿ strings of length O(n).

Auxiliary Space Used

O(2ⁿ * n).

As we will store 2ⁿ strings of length O(n) in the output array to be returned.

Space Complexity

O(2ⁿ * n).

As auxiliary space used is O(2ⁿ * n) and input is O(n), hence O(2ⁿ * n) + O(n) = O(2ⁿ * n).

Code For Generate All Subsets Of A Set Solution: Optimal

/*
Asymptotic complexity in terms of the length of `s` `n`:
* Time: O(2^n * n).
* Auxiliary space: O(2^n * n).
* Total space: O(2^n * n).
*/

vector<string> generate_all_subsets(string &s) {
    int n = s.length();
    vector<string> all_subsets;
    // Base case when n = 0 i.e. s = "".
    all_subsets.push_back("");
    /*
    Suppose s = "xyz".
    Now try to find pattern in following steps (think how any step is related to previous step!):
    - [""]
    - ["", "x"]
    - ["", "x", "y", "xy"]
    - ["", "x", "y", "xy", "z", "xz", "yz", "xyz"]
    Let me explain what we have done in last step.
    First take array from previous step, that is ["", "x", "y", "xy"], append 'z' to each string,
    that is ["z", "xz", "yz", "xyz"], now merge it with array in previous step!
    */
    for (int i = 1; i <= n; i++) {
        int old_len = all_subsets.size();
        for (int j = 0; j < old_len; j++) {
            /*
            Note that we are doing push_back that is adding value at the end of array, so we
            already have the old values stored in it.
            */
            all_subsets.push_back(all_subsets[j] + s[i - 1]);
        }
    }
    return all_subsets;
}

We hope that these solutions to generate all subsets of a set problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

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