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Problem solution · Python

Maximum Good Subtree Score

Maximum Good Subtree Score: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
111 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Maximum Good Subtree Score, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 111 lines of Python from the credited upstream file maximum-good-subtree-score.py.
  • The implementation visibly relies on sequence storage, hash lookup, cached states.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum Good Subtree Score · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n * (2^10)^2)# Space: O(2^10) import collections  # bitmasks, iterative dfs, tree dpclass Solution(object):    def goodSubtreeSum(self, vals, par):        """        :type vals: List[int]        :type par: List[int]        :rtype: int        """        MOD = 10**9+7        def get_mask(x):            mask = 0            while x:                x, d = divmod(x, 10)                if mask&(1<<d):                    return -1                mask |= 1<<d            return mask         def iter_dfs():            result = 0            ret = collections.defaultdict(int)            stk = [(1, (0, ret))]            while stk:                step, args = stk.pop()                if step == 1:                    u, ret = args                    ret[0] = 0                    mask = get_mask(vals[u])                    if mask != -1:                        ret[mask] = vals[u]                    stk.append((4, (u, ret)))                    stk.append((2, (u, 0, ret)))                elif step == 2:                    u, i, ret = args                    if i == len(adj[u]):                        continue                    v = adj[u][i]                    stk.append((2, (u, i+1, ret)))                    new_ret = collections.defaultdict(int)                    stk.append((3, (new_ret, ret)))                    stk.append((1, (v, new_ret)))                elif step == 3:                    new_ret, ret = args                    for m1, v1 in ret.items():                        for m2, v2 in new_ret.iteritems():                            if m1&m2:                                continue                            ret[m1|m2] =  max(ret[m1|m2], v1+v2)                elif step == 4:                    u, ret = args                    result = (result+max(ret.itervalues()))%MOD            return result         adj = [[] for _ in xrange(len(vals))]        for u in xrange(1, len(par)):            adj[par[u]].append(u)        return iter_dfs()  # Time:  O(n * (2^10)^2)# Space: O(2^10)import collections  # bitmasks, dfs, tree dpclass Solution2(object):    def goodSubtreeSum(self, vals, par):        """        :type vals: List[int]        :type par: List[int]        :rtype: int        """        MOD = 10**9+7        def get_mask(x):            mask = 0            while x:                x, d = divmod(x, 10)                if mask&(1<<d):                    return -1                mask |= 1<<d            return mask         def dfs(u):            dp = collections.defaultdict(int)            dp[0] = 0            mask = get_mask(vals[u])            if mask != -1:                dp[mask] = vals[u]            for v in adj[u]:                new_dp = dfs(v)                for m1, v1 in dp.items():                    for m2, v2 in new_dp.iteritems():                        if m1&m2:                            continue                        dp[m1|m2] =  max(dp[m1|m2], v1+v2)            result[0] = (result[0]+max(dp.itervalues()))%MOD            return dp         adj = [[] for _ in xrange(len(vals))]        for u in xrange(1, len(par)):            adj[par[u]].append(u)        result = [0]        dfs(0)        return result[0]

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