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Problem solution · C++

Count Triplets with Even XOR Set Bits II

Count Triplets with Even XOR Set Bits II: 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
28 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 Triplets with Even XOR Set Bits II, 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

  • 28 lines of C++ from the credited upstream file 3215.cpp.
  • The implementation visibly relies on sequence storage.
  • 1 loop block 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 Triplets with Even XOR Set Bits II · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  // Same as 3199. Count Triplets with Even XOR Set Bits I  long long tripletCount(vector<int>& a, vector<int>& b, vector<int>& c) {    const auto [evenA, oddA] = getEvenOddBitCount(a);    const auto [evenB, oddB] = getEvenOddBitCount(b);    const auto [evenC, oddC] = getEvenOddBitCount(c);    return static_cast<long>(evenA) * oddB * oddC +           static_cast<long>(oddA) * evenB * oddC +           static_cast<long>(oddA) * oddB * evenC +           static_cast<long>(evenA) * evenB * evenC;  }  private:  // Returns the count of numbers in the `nums` arrays that have even number of  // ones and odd number of ones in their binary representation.  pair<int, int> getEvenOddBitCount(const vector<int>& nums) {    int even = 0;    int odd = 0;    for (const unsigned num : nums)      if (popcount(num) % 2 == 0)        ++even;      else        ++odd;    return {even, odd};  }};

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