3203. Find Minimum Diameter After Merging Two Trees

# Time:  O(n + m)
# Space: O(n + m)

# iterative dfs, tree diameter
class Solution(object):
    def minimumDiameterAfterMerge(self, edges1, edges2):
        """
        :type edges1: List[List[int]]
        :type edges2: List[List[int]]
        :rtype: int
        """
        def ceil_divide(a, b):
            return (a+b-1)//2
    
        def tree_diameter(edges):
            def iter_dfs():
                result = 0
                stk = [(1, (0, -1, [0]))]
                while stk:
                    step, args = stk.pop()
                    if step == 1:
                        u, p, ret = args
                        for v in reversed(adj[u]):
                            if v == p:
                                continue
                            ret2 = [0]
                            stk.append((2, (ret2, ret)))
                            stk.append((1, (v, u, ret2)))
                    elif step == 2:
                        ret2, ret = args
                        result = max(result, ret[0]+(ret2[0]+1))
                        ret[0] = max(ret[0], ret2[0]+1)
                return result
            
            adj = [[] for _ in xrange(len(edges)+1)]
            for u, v in edges:
                adj[u].append(v)
                adj[v].append(u)
            return iter_dfs()
        
        d1 = tree_diameter(edges1)
        d2 = tree_diameter(edges2)
        return max(ceil_divide(d1, 2)+1+ceil_divide(d2, 2), d1, d2)
        

# Time:  O(n + m)
# Space: O(n + m)
# dfs, tree diameter
class Solution2(object):
    def minimumDiameterAfterMerge(self, edges1, edges2):
        """
        :type edges1: List[List[int]]
        :type edges2: List[List[int]]
        :rtype: int
        """
        def ceil_divide(a, b):
            return (a+b-1)//2
    
        def tree_diameter(edges):
            def dfs(u, p):
                mx = 0
                for v in adj[u]:
                    if v == p:
                        continue
                    curr = dfs(v, u)
                    result[0] = max(result[0], mx+(curr+1))
                    mx = max(mx, curr+1)
                return mx
            
            adj = [[] for _ in xrange(len(edges)+1)]
            for u, v in edges:
                adj[u].append(v)
                adj[v].append(u)
            result = [0]
            dfs(0, -1)
            return result[0]
        
        d1 = tree_diameter(edges1)
        d2 = tree_diameter(edges2)
        return max(ceil_divide(d1, 2)+1+ceil_divide(d2, 2), d1, d2)


# Time:  O(n + m)
# Space: O(n + m)
# bfs, tree dp, tree diameter
class Solution3(object):
    def minimumDiameterAfterMerge(self, edges1, edges2):
        """
        :type edges1: List[List[int]]
        :type edges2: List[List[int]]
        :rtype: int
        """
        def ceil_divide(a, b):
            return (a+b-1)//2
    
        def tree_diameter(edges):
            def bfs():
                result = 0
                dp = [0]*len(adj)
                degree = map(len, adj)
                q = [u for u in xrange(len(degree)) if degree[u] == 1]
                while q:
                    new_q = []
                    for u in q:
                        if degree[u] == 0:
                            continue
                        degree[u] -= 1
                        for v in adj[u]:
                            if degree[v] == 0:
                                continue
                            result = max(result, dp[v]+(dp[u]+1))
                            dp[v] = max(dp[v], (dp[u]+1))
                            degree[v] -= 1
                            if degree[v] == 1:
                                new_q.append(v)
                    q = new_q
                return result
            
            adj = [[] for _ in xrange(len(edges)+1)]
            for u, v in edges:
                adj[u].append(v)
                adj[v].append(u)
            return bfs()

        d1 = tree_diameter(edges1)
        d2 = tree_diameter(edges2)
        return max(ceil_divide(d1, 2)+1+ceil_divide(d2, 2), d1, d2)


# Time:  O(n + m)
# Space: O(n + m)
# bfs, tree diameter
class Solution4(object):
    def minimumDiameterAfterMerge(self, edges1, edges2):
        """
        :type edges1: List[List[int]]
        :type edges2: List[List[int]]
        :rtype: int
        """
        def ceil_divide(a, b):
            return (a+b-1)//2
    
        def tree_diameter(edges):
            def bfs(root):
                d = new_root = -1
                lookup = [False]*len(adj)
                lookup[root] = True
                q = [root]
                while q:
                    d, new_root = d+1, q[0]
                    new_q = []
                    for u in q:
                        for v in adj[u]:
                            if lookup[v]:
                                continue
                            lookup[v] = True
                            new_q.append(v)
                    q = new_q
                return d, new_root
            
            adj = [[] for _ in xrange(len(edges)+1)]
            for u, v in edges:
                adj[u].append(v)
                adj[v].append(u)
            _, root = bfs(0)
            d, _ = bfs(root)
            return d
        
        d1 = tree_diameter(edges1)
        d2 = tree_diameter(edges2)
        return max(ceil_divide(d1, 2)+1+ceil_divide(d2, 2), d1, d2)

Solution from kamyu104/LeetCode-Solutions · MIT