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Problem solution · C++

Maximum Sum of Subsequence With Non-adjacent Elements

Maximum Sum of Subsequence With Non-adjacent Elements: 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
99 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 Maximum Sum of Subsequence With Non-adjacent Elements, 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

  • 99 lines of C++ from the credited upstream file 3165.cpp.
  • The implementation visibly relies on sequence storage.
  • 3 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 walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum Sum of Subsequence With Non-adjacent Elements · C++C++
Use this to learn the idea, then write your own version.
using NodeType = array<array<int, 2>, 2>; class SegmentTree { public:  explicit SegmentTree(const vector<int>& nums) : n(nums.size()), tree(4 * n) {    build(nums, 0, 0, n - 1);  }   // Updates nums[i] to val.  void update(int i, int val) {    update(0, 0, n - 1, i, val);  }   // Returns the four values of the range query from nums[i..j].  //  // The four values are:  //   1. nums[i] is not selected, nums[j] is not selected  //   2. nums[i] is not selected, nums[j] is selected  //   3. nums[i] is selected, nums[j] is not selected  //   4. nums[i] is selected, nums[j] is selected  NodeType query(int i, int j) const {    return query(0, 0, n - 1, i, j);  }  private:  static constexpr int kInf = 1'000'000'000;  static constexpr NodeType kDefaultNode = {{{-kInf, -kInf}, {-kInf, -kInf}}};  const int n;  // the size of the input array  // tree[i][l][r] := the value of the i-th node, where `l` and `r` represent if  // the leftmost or rightmost element is selected, respectively  vector<NodeType> tree;   void build(const vector<int>& nums, int treeIndex, int lo, int hi) {    if (lo == hi) {      tree[treeIndex] = {{{0, -kInf}, {-kInf, nums[lo]}}};      return;    }    const int mid = (lo + hi) / 2;    build(nums, 2 * treeIndex + 1, lo, mid);    build(nums, 2 * treeIndex + 2, mid + 1, hi);    tree[treeIndex] = merge(tree[2 * treeIndex + 1], tree[2 * treeIndex + 2]);  }   void update(int treeIndex, int lo, int hi, int i, int val) {    if (lo == hi) {      tree[treeIndex] = {{{0, -kInf}, {-kInf, val}}};      return;    }    const int mid = (lo + hi) / 2;    if (i <= mid)      update(2 * treeIndex + 1, lo, mid, i, val);    else      update(2 * treeIndex + 2, mid + 1, hi, i, val);    tree[treeIndex] = merge(tree[2 * treeIndex + 1], tree[2 * treeIndex + 2]);  }   NodeType query(int treeIndex, int lo, int hi, int i, int j) const {    if (i <= lo && hi <= j)  // [lo, hi] lies completely inside [i, j].      return tree[treeIndex];    if (j < lo || hi < i)  // [lo, hi] lies completely outside [i, j].      return kDefaultNode;    const int mid = (lo + hi) / 2;    return merge(query(2 * treeIndex + 1, lo, mid, i, j),                 query(2 * treeIndex + 2, mid + 1, hi, i, j));  }   // Merges the result of the left node and the right node.  NodeType merge(const NodeType& a, const NodeType& b) const {    NodeType node = {{{0, 0}, {0, 0}}};    for (int l = 0; l < 2; ++l)      for (int r = 0; r < 2; ++r)        node[l][r] =            max({a[l][0] + b[0][r], a[l][0] + b[1][r], a[l][1] + b[0][r]});    return node;  }}; class Solution { public:  int maximumSumSubsequence(vector<int>& nums, vector<vector<int>>& queries) {    constexpr int kMod = 1'000'000'007;    const int n = nums.size();    int ans = 0;    SegmentTree tree(nums);     for (const vector<int>& query : queries) {      const int pos = query[0];      const int x = query[1];      tree.update(pos, x);      NodeType res = tree.query(0, n - 1);      ans = (ans + static_cast<long>(                       max({res[0][0], res[0][1], res[1][0], res[1][1]}))) %            kMod;    }     return ans;  }};

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