LC-2440 Hard LeetCode

2440. Create Components With Same Value

Array Math Tree Depth-First Search Enumeration

Read the full problem statement on LeetCode.

Difficulty: hard Acceptance: 52% Topics: Array, Math, Tree, Depth-First Search, Enumeration

Reading material

Arrays & Hashing 12 min Math & Geometry 12 min Trees & Traversals 18 min

Reference solution (spoiler · python)

# Time:  O(n * sqrt(n))
# Space: O(n)

# bfs, greedy
class Solution(object):
    def componentValue(self, nums, edges):
        """
        :type nums: List[int]
        :type edges: List[List[int]]
        :rtype: int
        """
        def bfs(target):
            total = nums[:]
            lookup = [len(adj[u]) for u in xrange(len(adj))]
            q = [u for u in xrange(len(adj)) if lookup[u] == 1]
            while q:
                new_q = []
                for u in q:
                    if total[u] > target:
                        return False
                    if total[u] == target:
                        total[u] = 0
                    for v in adj[u]:
                        total[v] += total[u]
                        lookup[v] -= 1
                        if lookup[v] == 1:
                            new_q.append(v)
                q = new_q
            return True

        result = 0
        adj = [[] for _ in xrange(len(nums))]
        for u, v in edges:
            adj[u].append(v)
            adj[v].append(u)
        total = sum(nums)
        for cnt in reversed(xrange(2, len(nums)+1)):
            if total%cnt == 0 and bfs(total//cnt):
                return cnt-1
        return 0


# Time:  O(n * sqrt(n))
# Space: O(n)
# iterative dfs, greedy
class Solution2(object):
    def componentValue(self, nums, edges):
        """
        :type nums: List[int]
        :type edges: List[List[int]]
        :rtype: int
        """
        def iter_dfs(target):
            total = nums[:]
            stk = [(1, (0, -1))]
            while stk:
                step, (u, p) = stk.pop()
                if step == 1:
                    stk.append((2, (u, p)))
                    for v in adj[u]:
                        if v == p:
                            continue
                        stk.append((1, (v, u)))
                elif step == 2:
                    for v in adj[u]:
                        if v == p:
                            continue
                        total[u] += total[v]
                    if total[u] == target:
                        total[u] = 0
            return total[0]

        result = 0
        adj = [[] for _ in xrange(len(nums))]
        for u, v in edges:
            adj[u].append(v)
            adj[v].append(u)
        total = sum(nums)
        for cnt in reversed(xrange(2, len(nums)+1)):
            if total%cnt == 0 and iter_dfs(total//cnt) == 0:
                return cnt-1
        return 0


# Time:  O(n * sqrt(n))
# Space: O(n)
# dfs, greedy
class Solution3(object):
    def componentValue(self, nums, edges):
        """
        :type nums: List[int]
        :type edges: List[List[int]]
        :rtype: int
        """
        def dfs(u, p, target):
            total = nums[u]
            for v in adj[u]:
                if v == p:
                    continue
                total += dfs(v, u, target)
            return total if total != target else 0

        result = 0
        adj = [[] for _ in xrange(len(nums))]
        for u, v in edges:
            adj[u].append(v)
            adj[v].append(u)
        total = sum(nums)
        for cnt in reversed(xrange(2, len(nums)+1)):
            if total%cnt == 0 and dfs(0, -1, total//cnt) == 0:
                return cnt-1
        return 0

Solution from kamyu104/LeetCode-Solutions · MIT