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Problem solution · C++

Handling Sum Queries After Update

Handling Sum Queries After Update: a C++ solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Segment tree or range structure
Source
walkccc LeetCode Solutions
Length
85 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 Handling Sum Queries After Update, 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

  • 85 lines of C++ from the credited upstream file 2569.cpp.
  • The implementation visibly relies on sequence storage.
  • 1 loop block 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 walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeHandling Sum Queries After Update · C++C++
Use this to learn the idea, then write your own version.
class LazySegmentTree { public:  explicit LazySegmentTree(const vector<int>& nums) {    const int n = nums.size();    tree.resize(4 * n);    lazy.resize(4 * n);    build(nums, 0, 0, n - 1);  }   //            i := index of the current node  // [start, end] := range of the current node  //       [l, r] := range of the query  void updateRange(int i, int start, int end, int l, int r) {    if (lazy[i])      propogate(i, start, end);    if (start > r || end < l)      return;    if (start >= l && end <= r) {      flip(i, start, end);      return;    }    const int mid = (start + end) / 2;    updateRange(i * 2 + 1, start, mid, l, r);    updateRange(i * 2 + 2, mid + 1, end, l, r);    tree[i] = tree[2 * i + 1] + tree[2 * i + 2];  }   int getTreeSum() const {    return tree[0];  }  private:  vector<int> tree;  vector<bool> lazy;   void build(const vector<int>& nums, int i, int start, int end) {    if (start == end) {      tree[i] = nums[start];      return;    }    const int mid = (start + end) / 2;    build(nums, 2 * i + 1, start, mid);    build(nums, 2 * i + 2, mid + 1, end);    tree[i] = tree[2 * i + 1] + tree[2 * i + 2];  }   void propogate(int i, int start, int end) {    flip(i, start, end);    lazy[i] = false;  }   void flip(int i, int start, int end) {    tree[i] = (end - start + 1) - tree[i];  // Flip 0/1.    if (start < end) {      lazy[2 * i + 1] = !lazy[2 * i + 1];      lazy[2 * i + 2] = !lazy[2 * i + 2];    }  }}; class Solution { public:  vector<long long> handleQuery(vector<int>& nums1, vector<int>& nums2,                                vector<vector<int>>& queries) {    vector<long long> ans;    LazySegmentTree tree(nums1);    long sumNums2 = accumulate(nums2.begin(), nums2.end(), 0L);     for (const vector<int>& query : queries) {      const int type = query[0];      const int l = query[1];      const int r = query[2];      if (type == 1) {        tree.updateRange(0, 0, nums1.size() - 1, l, r);      } else if (type == 2) {        sumNums2 += static_cast<long>(l) * tree.getTreeSum();      } else {  // type == 3        ans.push_back(sumNums2);      }    }     return ans;  }};

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