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Problem solution · C++

Count Almost Equal Pairs II

Count Almost Equal Pairs II: 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
48 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 Almost Equal Pairs II, 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

  • 48 lines of C++ from the credited upstream file 3267.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 8 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 Almost Equal Pairs II · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  // Similar to 3265. Count Almost Equal Pairs I  int countPairs(vector<int>& nums) {    int ans = 0;    unordered_map<int, int> count;    const int maxLen = to_string(ranges::max(nums)).length();     for (const int num : nums) {      const string digits =          string(maxLen - to_string(num).length(), '0') + to_string(num);      for (const int swap : getSwaps(digits))        ans += count[swap];      ++count[num];    }     return ans;  }  private:  // Returns all possible numbers after 1 or 2 swaps.  unordered_set<int> getSwaps(const string& digits) {    const int n = digits.length();    unordered_set<int> swaps{{stoi(digits)}};     // Add all numbers after 1 swap.    for (int i = 0; i < n; ++i)      for (int j = i + 1; j < n; ++j) {        string newDigits = digits;        swap(newDigits[i], newDigits[j]);        swaps.insert(stoi(newDigits));      }     // Add all numbers after 2 swaps.    for (int i1 = 0; i1 < n; ++i1)      for (int j1 = i1 + 1; j1 < n; ++j1)        for (int i2 = 0; i2 < n; ++i2)          for (int j2 = i2 + 1; j2 < n; ++j2) {            string newDigits = digits;            swap(newDigits[i1], newDigits[j1]);            swap(newDigits[i2], newDigits[j2]);            swaps.insert(stoi(newDigits));          }     return swaps;  }};

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