CODEWITHSUNDEEP
pythonalgorithmtime-complexitycode-complexity
Jake
Jake

Reputation: 16837

Leetcode 78 - Generate power set time complexity

Leetcode 78 question potential solution:

class Solution(object):
    def __init__(self):
        self.map = {}
    
    def helper(self, count, nums, vals, result):       
        
        if count == 0:
            result += [vals]        
            
        for i in range(len(nums)):
            self.helper(count - 1, nums[i+1:], vals + [nums[i]], result)
        
    
    def subsets(self, nums):
        result = []
        result.append([])
        
        for count in range(1,len(nums)+1):
            self.helper(count, nums, [], result)   
            
        return result

For the above solution, will the time complexity be O(2^n) or will it be O(n * 2^n) ?

Upvotes: 1

Views: 243

Answers (1)

dreamcrash
dreamcrash

Reputation: 51463

One can find out the complexity by looking for different N how many times is the helper function called, code-wise would look like the following:

class Solution(object):

    def __init__(self):
        self.map = {}
        self.counter = 0

    def helper(self, count, nums, vals, result):
        self.counter += 1
        if count == 0:
            result += [vals]

        for i in range(len(nums)):
            self.helper(count - 1, nums[i + 1:], vals + [nums[i]], result)

    def subsets(self, nums):
        result = [[]]

        for count in range(1, len(nums) + 1):
            self.helper(count, nums, [], result)

        return self.counter

So for:

N Time the helper function gets called
1 2
2 8
3 24
4 64
5 160
6 384
7 896
8 2048
9 4608
10 10240
... ....
N O(n * 2^n)

The helper function has a complexity of O(2^n) and since you called for each element of the list nums:

for count in range(1,len(nums)+1):
    self.helper(count, nums, [], result)

the overall time complexity is O(n * 2^n)

Upvotes: 1

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