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Problem solution · C++

Minimum Moves to Balance Circular Array II

Minimum Moves to Balance Circular Array II: a C++ 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
142 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

  • 142 lines of C++ from the credited upstream file minimum-moves-to-balance-circular-array-ii.cpp.
  • The implementation visibly relies on sequence storage, work queue.
  • 12 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 · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n^2 * logn)// Space: O(n) // l1 isotonic regression, heap, greedyclass Solution {public:    long long minMoves(vector<int>& balance) {        const auto& total = accumulate(cbegin(balance), cend(balance), 0LL);        if (total < 0) {            return -1;        }         const auto& cost = [&](int i) {            long long result = 0, prefix = 0;            priority_queue<int64_t> max_heap;            for (int j = 0; j < size(balance); ++j) {                prefix += balance[(i + j) % size(balance)];                const auto c = clamp(prefix, 0LL, total);                result += llabs(prefix - c);                // heap-based l1 isotonic regression                // reference: https://codeforces.com/contest/13/problem/C                max_heap.emplace(c);                if (max_heap.top() > c) {                    result += max_heap.top() - c;                    max_heap.pop();                    max_heap.emplace(c);                }            }            return result;        };         long long result = numeric_limits<long long>::max();        for (int i = 0; i < size(balance); ++i) {            result = min(result, cost(i));        }        return result;    }}; // 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 potential#include <bits/stdc++.h> // Time: O(F * E * logV// Space: O(V + E)// Template: https://github.com/kth-competitive-programming/kactl/blob/main/content/graph/MinCostMaxFlow.hconst long long INF = numeric_limits<long long>::max(); struct MCMF {    struct edge {        int from, to, rev;        long long cap, cost, flow;    };     int N;    vector<vector<edge>> ed;    vector<int> seen;    vector<long long> dist, pi;    vector<edge*> par;     MCMF(int N) : N(N), ed(N), seen(N), dist(N), pi(N), par(N) {}     void addEdge(int from, int to, long long cap, long long cost) {        if (from == to) return;        ed[from].push_back({from, to, (int)ed[to].size(), cap, cost, 0});        ed[to].push_back({to, from, (int)ed[from].size() - 1, 0, -cost, 0});    }     void path(int s) {        fill(seen.begin(), seen.end(), 0);        fill(dist.begin(), dist.end(), INF);        fill(par.begin(), par.end(), nullptr);        dist[s] = 0;        using State = pair<long long, int>;        priority_queue<State, vector<State>, greater<State>> q;        q.push({0, s});        while (!q.empty()) {            auto [d, u] = q.top();            q.pop();            if (d != dist[u]) continue;            seen[u] = 1;            for (edge& e : ed[u]) {                if (e.cap - e.flow <= 0) continue;                long long val = d + pi[u] - pi[e.to] + e.cost;                if (val < dist[e.to]) {                    dist[e.to] = val;                    par[e.to] = &e;                    q.push({val, e.to});                }            }        }        for (int i = 0; i < N; ++i)            if (dist[i] != INF)                pi[i] += dist[i];    }     pair<long long, long long> maxflow(int s, int t) {        long long totflow = 0, totcost = 0;        while (path(s), seen[t]) {            long long fl = INF;            for (edge* x = par[t]; x; x = par[x->from])                fl = min(fl, x->cap - x->flow);            totflow += fl;            for (edge* x = par[t]; x; x = par[x->from]) {                x->flow += fl;                ed[x->to][x->rev].flow -= fl;            }        }        for (int i = 0; i < N; ++i)            for (edge& e : ed[i])                totcost += e.cost * e.flow;        return {totflow, totcost / 2};    }}; class Solution2 {public:    long long minMoves(vector<int>& balance) {        if (accumulate(cbegin(balance), cend(balance), 0LL) < 0) {            return -1;        }        int S = size(balance), T = size(balance) + 1;        MCMF mcmf(size(balance) + 2);        for (int i = 0; i < size(balance); ++i) {            mcmf.addEdge(i, (i + 1) % size(balance), INF, 1);            mcmf.addEdge((i + 1) % size(balance), i, INF, 1);        }        int64_t demand = 0;        for (int i = 0; i < size(balance); ++i) {            if (balance[i] > 0) {                mcmf.addEdge(S, i, balance[i], 0);            } else if (balance[i] < 0) {                mcmf.addEdge(i, T, -balance[i], 0);                demand += -balance[i];            }        }        const auto& [flow, cost] = mcmf.maxflow(S, T);        return flow == demand ? cost : -1;    }};

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