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Problem solution · C++

Count Good Triplets in an Array

Count Good Triplets in an Array: a C++ solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
67 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

Hash-based lookup

For Count Good Triplets in an Array, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 67 lines of C++ from the credited upstream file 2179.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 7 loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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 Good Triplets in an Array · C++C++
Use this to learn the idea, then write your own version.
class FenwickTree { public:  FenwickTree(int n) : sums(n + 1) {}   void add(int i, int delta) {    while (i < sums.size()) {      sums[i] += delta;      i += lowbit(i);    }  }   int get(int i) const {    int sum = 0;    while (i > 0) {      sum += sums[i];      i -= lowbit(i);    }    return sum;  }  private:  vector<int> sums;   static inline int lowbit(int i) {    return i & -i;  }}; class Solution { public:  long long goodTriplets(vector<int>& nums1, vector<int>& nums2) {    const int n = nums1.size();    long ans = 0;    unordered_map<int, int> numToIndex;    vector<int> arr;    // leftSmaller[i] := the number of arr[j] < arr[i], where 0 <= j < i    vector<int> leftSmaller(n);    // rightLarger[i] := the number of arr[j] > arr[i], where i < j < n    vector<int> rightLarger(n);    FenwickTree tree1(n);  // Calculates `leftSmaller`.    FenwickTree tree2(n);  // Calculates `rightLarger`.     for (int i = 0; i < n; ++i)      numToIndex[nums1[i]] = i;     // Remap each number in `nums2` to the according index in `nums1` as `arr`.    // So the problem is to find the number of increasing tripets in `arr`.    for (const int num : nums2)      arr.push_back(numToIndex[num]);     for (int i = 0; i < n; ++i) {      leftSmaller[i] = tree1.get(arr[i]);      tree1.add(arr[i] + 1, 1);    }     for (int i = n - 1; i >= 0; --i) {      rightLarger[i] = tree2.get(n) - tree2.get(arr[i]);      tree2.add(arr[i] + 1, 1);    }     for (int i = 0; i < n; ++i)      ans += static_cast<long>(leftSmaller[i]) * rightLarger[i];     return ans;  }};

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