CPMagical CP Library

Problem solution · C++

Count the Number of Houses at a Certain Distance II

Count the Number of Houses at a Certain Distance II: a C++ solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
107 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

Sliding window or two pointers

For Count the Number of Houses at a Certain Distance II, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 107 lines of C++ from the credited upstream file 3017.cpp.
  • The implementation visibly relies on sequence storage.
  • 5 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure 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 codeCount the Number of Houses at a Certain Distance II · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  // Same as 3015. Count the Number of Houses at a Certain Distance I  vector<long long> countOfPairs(int n, int x, int y) {    if (x > y)      swap(x, y);     const int ringLen = y - x + 1;    const int leftLineLen = x - 1;    const int rightLineLen = n - y;     vector<long long> ans(n);    ans = addVectors(ans, bothInRing(n, ringLen));    ans = addVectors(ans, bothInTheSameLine(n, leftLineLen));    ans = addVectors(ans, bothInTheSameLine(n, rightLineLen));    ans = addVectors(ans, lineToRing(n, leftLineLen, ringLen));    ans = addVectors(ans, lineToRing(n, rightLineLen, ringLen));    ans = addVectors(ans, lineToLine(n, x, y, leftLineLen, rightLineLen));    for (long long& freq : ans)      freq *= 2;    return ans;  }  private:  // Returns the contribution from the scenario where two houses are located in  // the ring.  vector<long long> bothInRing(int n, int ringLen) {    vector<long long> res(n);    for (int k = 1; k <= (ringLen - 1) / 2; ++k)      res[k - 1] += ringLen;    if (ringLen % 2 == 0)      res[ringLen / 2 - 1] += ringLen / 2;    return res;  }   // Returns the contribution from the scenario where two houses are either  // located in the left line [1, x) or the right line (y, n].  vector<long long> bothInTheSameLine(int n, int lineLen) {    vector<long long> res(n);    for (int k = 1; k <= lineLen; ++k)      res[k - 1] += lineLen - k;    return res;  }   // Returns the contribution from the scenario where one house is either  // located in the left line [1, x) or the right line (y, n] and the other  // house is located in the cycle.  vector<long long> lineToRing(int n, int lineLen, int ringLen) {    vector<long long> res(n);    for (int k = 1; k <= lineLen + ringLen; ++k) {      // min(      //   at most k - 1 since we need to give 1 to the line,      //   at most ringLen / 2 since for length > ringLen / 2, it can always be      //     calculated as ringLen - ringLen / 2      // )      const int maxInRingLen = min(k - 1, ringLen / 2);      // max(at least 0, at lest k - lineLen)      const int minInRingLen = max(0, k - lineLen);      if (minInRingLen <= maxInRingLen) {        // Each ring length contributes 2 to the count due to the split of        // paths when entering the ring: One path traverses the upper half of        // the ring, and the other traverses the lower half.        // This is illustrated as follows:        //   Path 1: ... -- x -- (upper half of the ring)        //   Path 2: ... -- x -- (lower half of the ring)        res[k - 1] += (maxInRingLen - minInRingLen + 1) * 2;        if (minInRingLen == 0)          // Subtract 1 since there's no split.          res[k - 1] -= 1;        if (maxInRingLen * 2 == ringLen)          // Subtract 1 since the following case only contribute one:          //   ... -- x -- (upper half of the ring) -- middle point          //   ... -- x -- (upper half of the ring) -- middle point          res[k - 1] -= 1;      }    }    return res;  }   // Returns the contribution from the scenario where one house is in the left  // line [1, x) and the other house is in the right line (y, n].  vector<long long> lineToLine(int n, int x, int y, int leftLineLen,                               int rightLineLen) {    vector<long long> res(n);    for (int k = 1; k <= leftLineLen + rightLineLen + 2; ++k) {      // min(      //   at most leftLineLen,      //   at most k - 1 - (x < y) since we need to give 1 to the right line      //     and if x < y we need to give another 1 to "x - y".      // )      const int maxInLeft = min(leftLineLen, k - 1 - (x < y));      // max(at least 1, at least k - rightLineLen - (x < y))      const int minInLeft = max(1, k - rightLineLen - (x < y));      if (minInLeft <= maxInLeft)        res[k - 1] += maxInLeft - minInLeft + 1;    }    return res;  }   vector<long long> addVectors(const vector<long long>& a,                               const vector<long long>& b) {    vector<long long> res(a.size());    transform(a.begin(), a.end(), b.begin(), res.begin(), plus<int>());    return res;  };};

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