Both are almost same codes for House Robber III in leetcode but one is time limit exceed code whereas other is perfect solution to the question

Question

The correct code is

class Solution:
    def rob(self, root: Optional[TreeNode]) -> int:
        def dfs(root):
            if not root:
                return [0,0]
            leftpair=dfs(root.left)
            rightpair=dfs(root.right)
            withroot=root.val+leftpair[1]+rightpair[1]
            withoutroot=max(leftpair)+max(rightpair)
            return [withroot,withoutroot]
                
        
        return max(dfs(root))

The code which is leading to time limit exceeded error in leetcode is

class Solution:
    def rob(self, root: Optional[TreeNode]) -> int:
        def dfs(root):
            if not root:
                return [0,0]
            
            left=root.val+dfs(root.left)[1]+dfs(root.right)[1]
            right=max(dfs(root.left))+max(dfs(root.right))
            return [left,right]
                
        return max(dfs(root))

I don’t think there is any difference in time complexities of above two code. Can someone help me with this ??

Asked By: Anusha Yerram

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Source

Answer 1

The first version of dfs calls itself recursively twice, saving the results and reusing them. Here are the two recursive calls:

leftpair=dfs(root.left)
rightpair=dfs(root.right)

The second version of dfs calls itself recursively four times. Here are the four recursive calls, embedded in expressions:

left=root.val+dfs(root.left)[1]+dfs(root.right)[1]
right=max(dfs(root.left))+max(dfs(root.right))

As you can see, rather than reusing the results of the recursive calls, it instead repeats the calls (twice with root.left and twice with root.right). Understandably, this results in a significantly longer tuntime.

Answered By: Tom Karzes

Both are almost same codes for House Robber III in leetcode but one is time limit exceed code whereas other is perfect solution to the question

Question:

Answers: