CPMagical CP Library

Problem solution · C++

Find X Value of Array II

Find X Value of Array II: 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
103 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 Find X Value of Array II, 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

  • 103 lines of C++ from the credited upstream file 3525.cpp.
  • The implementation visibly relies on sequence storage.
  • 6 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 codeFind X Value of Array II · C++C++
Use this to learn the idea, then write your own version.
struct Node {  int remain[5] = {0};  int prod = 1;}; class SegmentTree { public:  explicit SegmentTree(const vector<int>& nums, int k)      : n(nums.size()), k(k), 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 result of the range query from nums[i..j].  Node query(int i, int j) const {    return query(0, 0, n - 1, i, j);  }  private:  const int n;        // the size of the input array  const int k;        // the modulo value  vector<Node> tree;  // the segment tree   void build(const vector<int>& nums, int cur, int left, int right) {    if (left == right) {      tree[cur].remain[nums[left]] = 1;      tree[cur].prod = nums[left];      return;    }    const int mid = (left + right) / 2;    build(nums, 2 * cur + 1, left, mid);    build(nums, 2 * cur + 2, mid + 1, right);    tree[cur] = merge(tree[2 * cur + 1], tree[2 * cur + 2]);  }   void update(int treeIndex, int lo, int hi, int i, int val) {    if (lo == hi) {      for (int j = 0; j < k; ++j)        tree[treeIndex].remain[j] = 0;      tree[treeIndex].remain[val] = 1;      tree[treeIndex].prod = 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]);  }   Node 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 Node();    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));  }   Node merge(const Node& left, const Node& right) const {    Node node;    node.prod = (left.prod * right.prod) % k;    for (int i = 0; i < k; ++i)      node.remain[i] = left.remain[i];    for (int i = 0; i < k; ++i)      node.remain[(i * left.prod) % k] += right.remain[i];    return node;  }}; class Solution { public:  vector<int> resultArray(vector<int>& nums, int k,                          vector<vector<int>>& queries) {    for (int& num : nums)      num %= k;     for (vector<int>& query : queries)      query[1] %= k;     const int n = nums.size();    vector<int> ans;    SegmentTree tree(nums, k);     for (const vector<int>& query : queries) {      const int index = query[0];      const int value = query[1];      const int start = query[2];      const int x = query[3];      tree.update(index, value);      ans.push_back(tree.query(start, n - 1).remain[x]);    }     return ans;  }};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗