#solution where you just take subset I and replace last return statement 
class Solution:
    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:

   
        subset = [] 
        result = [] 
        
        #i is index of subset 
        def dfs(i): 
            if i >= len(nums): 
                result.append(subset.copy())
                return 
            
            #decision to include nums[i]
            subset.append(nums[i])
            #this recursive call will be given filled subset 
            dfs(i + 1)
            
            #decision NOT to include nums[i] 
            subset.pop()
            #this recursive call will be given empty subset 
            dfs(i + 1)
        
        dfs(0) 
        
        #need to remove duplicate sublists within result 
        return ([list(i) for i in {*[tuple(sorted(i)) for i in result]}])  
  

#solution to prevent duplicates in the first place
class Solution:
    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:

   
        result = []
		nums.sort() 

		def backtrack(i, subset): 
        	if i == len(nums): 
            	res.append(subset.copy())
				return 
			
			#all subsets that include nums[i] 
            subset.append(nums[i]) 
			backtrack(i + 1, subset) 
			#remove number we just added 
            subset.pop()
            
            #all subsets that don't include nums[i]
        	while i + 1 < len(nums) and nums[i] == nums[i + 1]: 
            	i += 1 
			backtrack(i + 1, subset)
		
		backtrack(0, [])
		
		return result 
        
        
        
        
#solution for array with unique integer array. only last return line is dfferent 
class Solution:
    def subsets(self, nums: List[int]) -> List[List[int]]:
        subset = [] 
        result = [] 
        
        #i is index of subset 
        def dfs(i): 
            if i >= len(nums): 
                result.append(subset.copy())
                return 
            
            #decision to include nums[i]
            subset.append(nums[i])
            #this recursive call will be given filled subset 
            dfs(i + 1)
            
            #decision NOT to include nums[i] 
            subset.pop()
            #this recursive call will be given empty subset 
            dfs(i + 1)
        
        dfs(0) 
        
        return result 
        
        
        

def solution(n, k):
    return climb(n, k, [])
    
        
        
def climb(n, k, jumps):
    if n == 0:
        return [jumps]
    
    out = []
    
    for i in range(1, k+1):
        if i > n:
            continue
            
        temp = jumps + [i] 
        out += climb(n-i, k, temp)
        
    return out