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Problem solution · C++

Minimum Operations to Equalize Subarrays

Minimum Operations to Equalize Subarrays: a C++ solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Segment tree or range structure
Source
Kamyu LeetCode Solutions
Length
118 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

Segment tree or range structure

For Minimum Operations to Equalize Subarrays, the implementation stores interval information in a range-query data structure so updates and queries avoid rescanning the full input.

  1. Choose the aggregate stored for each interval or prefix.
  2. Build or initialize the structure from the input.
  3. Apply updates and combine the affected nodes to answer each query.

Code notes

  • 118 lines of C++ from the credited upstream file minimum-operations-to-equalize-subarrays.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 7 loop blocks detected.

Complexity

Count the build once, then multiply the logarithmic update or query path by the number of operations.

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 Operations to Equalize Subarrays · C++C++
Use this to learn the idea, then write your own version.
// Time:  O((n + q) * logn)// Space: O(nlogn) // prefix sum, persistent segment tree, binary searchclass PersistentSegmentTree {public:    PersistentSegmentTree(const auto& vals)       : sorted_unique_vals_(vals) {         sort(begin(sorted_unique_vals_), end(sorted_unique_vals_));        sorted_unique_vals_.erase(unique(begin(sorted_unique_vals_), end(sorted_unique_vals_)), end(sorted_unique_vals_));        n_ = size(sorted_unique_vals_);        val_to_idx_.reserve(n_);        for (int i = 0; i < n_; ++i) {            val_to_idx_[sorted_unique_vals_[i]] = i;        }        build(vals);    }     int64_t query(int l, int r) {        int a = roots_[l];        int b = roots_[r + 1];        int64_t left_cnt = 0, left_total = 0;        auto med_cnt = ((r - l + 1) / 2) + 1;        int left = 0, right = n_ - 1;        while (left < right) {            const auto& mid = left + (right - left) / 2;            const auto& cnt = nodes_[nodes_[b].left].cnt - nodes_[nodes_[a].left].cnt;            if (med_cnt <= cnt) {                a = nodes_[a].left;                b = nodes_[b].left;                right = mid;            } else {                left_cnt += cnt;                left_total += nodes_[nodes_[b].left].total - nodes_[nodes_[a].left].total;                med_cnt -= cnt;                a = nodes_[a].right;                b = nodes_[b].right;                left = mid + 1;            }        }        return (sorted_unique_vals_[left] * left_cnt - left_total) +               (((nodes_[roots_[r + 1]].total - nodes_[roots_[l]].total - left_total)) - sorted_unique_vals_[left] * ((r - l + 1) - left_cnt));    } private:    inline int new_node() {        nodes_.emplace_back();        return size(nodes_) - 1;    }     inline int copy_node(int idx) {        nodes_.emplace_back(nodes_[idx]);        return size(nodes_) - 1;    }     void build(const auto& vals) {        int root = new_node();        roots_.emplace_back(root);        for (const auto& x : vals) {            root = copy_node(root);            roots_.emplace_back(root);            int curr = root;            int left = 0, right = n_ - 1;            const auto& i = val_to_idx_[x];            while (left < right) {                ++nodes_[curr].cnt;                nodes_[curr].total += x;                const auto& mid = left + (right - left) / 2;                if (i <= mid) {                    curr = nodes_[curr].left = copy_node(nodes_[curr].left);                    right = mid;                } else {                    curr = nodes_[curr].right = copy_node(nodes_[curr].right);                    left = mid + 1;                }            }            ++nodes_[curr].cnt;            nodes_[curr].total += x;        }    }     struct Node {        int left = 0;        int right = 0;        int cnt = 0;        int64_t total = 0;    };     int n_;    vector<int64_t> sorted_unique_vals_;    unordered_map<int64_t, int> val_to_idx_;    vector<Node> nodes_;    vector<int> roots_;}; class Solution {public:    vector<long long> minOperations(vector<int>& nums, int k,  vector<vector<int>>& queries) {        vector<int> prefix(size(nums) + 1);        for (int i = 0; i < size(nums); ++i) {            prefix[i + 1] = prefix[i] + (i - 1 >= 0 && nums[i] % k != nums[i - 1] % k ? 1 : 0);        }        vector<int64_t> vals(size(nums));        for (int i = 0; i < size(vals); ++i) {            vals[i] = nums[i] / k;        }        PersistentSegmentTree pst(vals);        vector<long long> result;        result.reserve(size(queries));        for (const auto& q : queries) {            const auto& s = q[0], &t = q[1];            result.emplace_back(prefix[t + 1] - prefix[s + 1] == 0 ? pst.query(s, t) : -1);        }        return result;    }};

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