def combinationSum(self, nums, target): def backtrack(start, combination, current_sum): if current_sum == target: result.append(list(combination)) return elif current_sum > target: return for i in range(start, len(nums)): combination.append(nums[i]) backtrack(i, combination, current_sum + nums[i]) combination.pop() result = [] backtrack(0, [], 0) return result What is the time and space complexity of this algorithm? Give in terms of Big O notation
def combinationSum(self, nums, target): def backtrack(start, combination, current_sum): if current_sum == target: result.append(list(combination)) return elif current_sum > target: return for i in range(start, len(nums)): combination.append(nums[i]) backtrack(i, combination, current_sum + nums[i]) combination.pop() result = [] backtrack(0, [], 0) return result What is the time and space complexity of this algorithm? Give in terms of Big O notation
Related questions
Question
def combinationSum(self, nums, target):
def backtrack(start, combination, current_sum):
if current_sum == target:
result.append(list(combination))
return
elif current_sum > target:
return
for i in range(start, len(nums)):
combination.append(nums[i])
backtrack(i, combination, current_sum + nums[i])
combination.pop()
result = []
backtrack(0, [], 0)
return result
What is the time and space complexity of this algorithm? Give in terms of Big O notation
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