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Problem solution · Python

Minimum Edge Toggles on a Tree

Minimum Edge Toggles on a Tree: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Direct simulation
Source
Kamyu LeetCode Solutions
Length
89 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

Direct simulation

For Minimum Edge Toggles on a Tree, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 89 lines of Python from the credited upstream file minimum-edge-toggles-on-a-tree.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeMinimum Edge Toggles on a Tree · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n)# Space: O(n) # greedy, topological sort, bitmasksclass Solution(object):    def minimumFlips(self, n, edges, start, target):        """        :type n: int        :type edges: List[List[int]]        :type start: str        :type target: str        :rtype: List[int]        """        diff = [1 if start[u] != target[u] else 0 for u in xrange(n)]        if sum(diff)%2:            return [-1]        degree = [0]*n        adj = [0]*n        edge = [0]*n        for idx, (u, v) in enumerate(edges):            degree[u] += 1            degree[v] += 1            adj[u] ^= v            adj[v] ^= u            edge[u] ^= idx            edge[v] ^= idx        lookup = [False]*len(edges)        for u in xrange(n):            while degree[u] == 1:                v, idx = adj[u], edge[u]                degree[u] -= 1                degree[v] -= 1                adj[u] ^= v                adj[v] ^= u                edge[u] ^= idx                edge[v] ^= idx                if diff[u]:                    diff[u] ^= 1                    diff[v] ^= 1                    lookup[idx] = True                u = v        return [i for i in xrange(len(lookup)) if lookup[i]]  # Time:  O(n)# Space: O(n)# greedy, topological sortclass Solution2(object):    def minimumFlips(self, n, edges, start, target):        """        :type n: int        :type edges: List[List[int]]        :type start: str        :type target: str        :rtype: List[int]        """        def topological_sort():            lookup = [False]*len(adj)            q = [u for u in xrange(len(adj)) if degree[u] == 1]            while q:                new_q = []                for u in q:                    for v, idx in adj[u]:                        if degree[v] == 0:                            continue                        degree[u] -= 1                        degree[v] -= 1                        if degree[v] == 1:                            new_q.append(v)                        if diff[u]:                            diff[u] ^= 1                            diff[v] ^= 1                            lookup[idx] = True                q = new_q            return lookup         diff = [1 if start[u] != target[u] else 0 for u in xrange(n)]        if sum(diff)%2:            return [-1]        degree = [0]*n        adj = [[] for _ in xrange(n)]        for idx, (u, v) in enumerate(edges):            degree[u] += 1            degree[v] += 1            adj[u].append((v, idx))            adj[v].append((u, idx))        lookup = topological_sort()        return [i for i in xrange(len(lookup)) if lookup[i]]

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