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Problem solution · Python

Count the Number of Houses at a Certain Distance I

Count the Number of Houses at a Certain Distance I: a Python 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
93 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 I, 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

  • 93 lines of Python from the credited upstream file 3015.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit 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 I · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def countOfPairs(self, n: int, x: int, y: int) -> list[int]:    if x > y:      x, y = y, x     def bothInRing(ringLen: int) -> list[int]:      """      Returns the contribution from the scenario where two houses are located      in the ring.      """      res = [0] * n      for k in range(1, (ringLen - 1) // 2 + 1):        res[k - 1] += ringLen      if ringLen % 2 == 0:        res[ringLen // 2 - 1] += ringLen // 2      return res     def bothInTheSameLine(lineLen: int) -> list[int]:      """      Returns the contribution from the scenario where two houses are either      located in the left line [1, x) or the right line (y, n].      """      res = [0] * n      for k in range(1, lineLen + 1):        res[k - 1] += lineLen - k      return res     def lineToRing(lineLen: int, ringLen: int) -> list[int]:      """      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.      """      res = [0] * n      for k in range(1, lineLen + ringLen):        # 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        # )        maxInRingLen = min(k - 1, ringLen // 2)        # max(at least 0, at lest k - lineLen)        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     def lineToLine(leftLineLen: int, rightLineLen: int) -> list[int]:      """      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].      """      res = [0] * n      for k in range(leftLineLen + rightLineLen + 2):        # 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".        # )        maxInLeft = min(leftLineLen, k - 1 - (x < y))        # max(at least 1, at least k - rightLineLen - (x < y))        minInLeft = max(1, k - rightLineLen - (x < y))        if minInLeft <= maxInLeft:          res[k - 1] += maxInLeft - minInLeft + 1      return res     ringLen = y - x + 1    leftLineLen = x - 1    rightLineLen = (n - y)     ans = [0] * n    ans = list(map(operator.add, ans, bothInRing(ringLen)))    ans = list(map(operator.add, ans, bothInTheSameLine(leftLineLen)))    ans = list(map(operator.add, ans, bothInTheSameLine(rightLineLen)))    ans = list(map(operator.add, ans, lineToRing(leftLineLen, ringLen)))    ans = list(map(operator.add, ans, lineToRing(rightLineLen, ringLen)))    ans = list(map(operator.add, ans, lineToLine(leftLineLen, rightLineLen)))    return [freq * 2 for freq in ans]

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