CPMagical CP Library

Problem solution · C++

Count Pairs With XOR in a Range

Count Pairs With XOR in a Range: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
55 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

Direct simulation

For Count Pairs With XOR in a Range, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 55 lines of C++ from the credited upstream file 1803.cpp.
  • The implementation visibly relies on sequence storage.
  • 3 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeCount Pairs With XOR in a Range · C++C++
Use this to learn the idea, then write your own version.
struct TrieNode {  vector<shared_ptr<TrieNode>> children;  int count = 0;  TrieNode() : children(2) {}}; class Solution { public:  int countPairs(vector<int>& nums, int low, int high) {    int ans = 0;     for (const int num : nums) {      ans += getCount(num, high + 1) - getCount(num, low);      insert(num);    }     return ans;  }  private:  static constexpr int kHeight = 14;  shared_ptr<TrieNode> root = make_shared<TrieNode>();   void insert(int num) {    shared_ptr<TrieNode> node = root;    for (int i = kHeight; i >= 0; --i) {      const int bit = num >> i & 1;      if (node->children[bit] == nullptr)        node->children[bit] = make_shared<TrieNode>();      node = node->children[bit];      ++node->count;    }  }   // Returns the number of numbers < limit.  int getCount(int num, int limit) {    int count = 0;    shared_ptr<TrieNode> node = root;    for (int i = kHeight; i >= 0; --i) {      const int bit = num >> i & 1;      const int bitLimit = limit >> i & 1;      if (bitLimit == 1) {        if (node->children[bit] != nullptr)          count += node->children[bit]->count;        node = node->children[bit ^ 1];      } else {        node = node->children[bit];      }      if (node == nullptr)        break;    }    return count;  }};

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 ↗