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Problem solution · Python

Minimum Moves to Balance Circular Array II

Minimum Moves to Balance Circular Array II: a Python solution using heap or priority queue. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Heap or priority queue
Source
Kamyu LeetCode Solutions
Length
138 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

Heap or priority queue

For Minimum Moves to Balance Circular Array II, the implementation repeatedly takes the currently best candidate from a heap while inserting newly available choices.

  1. Define the priority key and whether the smallest or largest item should lead.
  2. Push each candidate when it becomes eligible.
  3. Discard stale entries when necessary and process the best live candidate.

Code notes

  • 138 lines of Python from the credited upstream file minimum-moves-to-balance-circular-array-ii.py.
  • The implementation visibly relies on sequence storage, work queue.
  • No explicit loop blocks detected.

Complexity

Count heap pushes and pops; each normally contributes a logarithmic factor in the heap size.

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 Moves to Balance Circular Array II · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n^2 * logn)# Space: O(n) import heapq  # l1 isotonic regression, heap, greedyclass Solution(object):    def minMoves(self, balance):        """        :type balance: List[int]        :rtype: int        """        def clamp(x, l, r):            return min(max(x, l), r)            def cost(i):            result = prefix = 0            max_heap = []            for j in xrange(len(balance)):                prefix += balance[(i+j)%len(balance)]                c = clamp(prefix, 0, total)                result += abs(prefix-c)                # heap-based l1 isotonic regression, reference: https://codeforces.com/contest/13/problem/C                heapq.heappush(max_heap, -c)                if -max_heap[0] > c:                    result += -heapq.heappop(max_heap)-c                    heapq.heappush(max_heap, -c)            return result         total = sum(balance)        return min(cost(i) for i in xrange(len(balance))) if total >= 0 else -1  # Time:  O(F * E * logV) = O(n^2 * logn), V = O(n), E = O(n), and each augmentation saturates a supply or demand edge, so there are only O(n) augmentations# Space: O(V + E) = O(n)# min-cost max-flow, ssp, dijkstra's algorithm, johnson potentialimport heapq  # Time: O(F * E * logV)# Space: O(V + E)# Template: https://github.com/kth-competitive-programming/kactl/blob/main/content/graph/MinCostMaxFlow.hINF = float("inf")class Edge(object):    def __init__(self, from_node, to, rev, cap, cost, flow):        self.from_node = from_node        self.to = to        self.rev = rev        self.cap = cap        self.cost = cost        self.flow = flow  class MCMF(object):    def __init__(self, n):        self.N = n        self.ed = [[] for _ in xrange(n)]        self.seen = [0]*n        self.dist = [INF]*n        self.pi = [0]*n        self.par = [None]*n     def addEdge(self, from_node, to, cap, cost):        if from_node == to:            return        self.ed[from_node].append(Edge(from_node, to, len(self.ed[to]), cap, cost, 0))        self.ed[to].append(Edge(to, from_node, len(self.ed[from_node])-1, 0, -cost, 0))     def path(self, s):        self.seen = [0]*self.N        self.dist = [INF]*self.N        self.par = [None]*self.N        self.dist[s] = 0        q = [(0, s)]        while q:            d, u = heapq.heappop(q)            if d != self.dist[u]:                continue            self.seen[u] = 1            for edge in self.ed[u]:                if edge.cap-edge.flow <= 0:                    continue                val = d+self.pi[u]-self.pi[edge.to]+edge.cost                if val < self.dist[edge.to]:                    self.dist[edge.to] = val                    self.par[edge.to] = edge                    heapq.heappush(q, (val, edge.to))        for i in xrange(self.N):            if self.dist[i] != INF:                self.pi[i] += self.dist[i]     def maxflow(self, s, t):        total_flow = total_cost = 0        while True:            self.path(s)            if not self.seen[t]:                break            flow = INF            edge = self.par[t]            while edge:                flow = min(flow, edge.cap-edge.flow)                edge = self.par[edge.from_node]            total_flow += flow            edge = self.par[t]            while edge:                edge.flow += flow                self.ed[edge.to][edge.rev].flow -= flow                edge = self.par[edge.from_node]        for edges in self.ed:            for edge in edges:                total_cost += edge.cost*edge.flow        return total_flow, total_cost//2  class Solution2(object):    def minMoves(self, balance):        """        :type balance: List[int]        :rtype: int        """        if sum(balance) < 0:            return -1        source, sink = len(balance), len(balance)+1        mcmf = MCMF(len(balance)+2)        for i in xrange(len(balance)):            mcmf.addEdge(i, (i+1)%len(balance), INF, 1)            mcmf.addEdge((i+1)%len(balance), i, INF, 1)        demand = 0        for i in xrange(len(balance)):            if balance[i] > 0:                mcmf.addEdge(source, i, balance[i], 0)            elif balance[i] < 0:                mcmf.addEdge(i, sink, -balance[i], 0)                demand += -balance[i]        flow, cost = mcmf.maxflow(source, sink)        return cost if flow == demand else -1

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