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Problem solution · Python

Path Existence Queries in a Graph II

Path Existence Queries in a Graph II: a Python solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sorting and greedy selection
Source
walkccc LeetCode Solutions
Length
53 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

Sorting and greedy selection

For Path Existence Queries in a Graph II, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 53 lines of Python from the credited upstream file 3534.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

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 codePath Existence Queries in a Graph II · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def pathExistenceQueries(      self,      n: int,      nums: list[int],      maxDiff: int,      queries: list[list[int]],  ) -> list[int]:    sortedNumAndIndexes = sorted((num, i) for i, num in enumerate(nums))    sortedNums = [num for num, _ in sortedNumAndIndexes]    indexMap = {originalIndex: sortedIndex for sortedIndex,                (_, originalIndex) in enumerate(sortedNumAndIndexes)}    maxLevel = n.bit_length() + 1    # jump[i][j] is the index of the j-th ancestor of i    jump = [[0] * maxLevel for _ in range(n)]     right = 0    for i in range(n):      while right + 1 < n and sortedNums[right + 1] - sortedNums[i] <= maxDiff:        right += 1      jump[i][0] = right     for level in range(1, maxLevel):      for i in range(n):        prevJump = jump[i][level - 1]        jump[i][level] = jump[prevJump][level - 1]     def minJumps(start: int, end: int, level: int) -> int:      """      Returns the minimum number of jumps from `start` to `end` using binary      lifting.      """      if start == end:        return 0      if jump[start][0] >= end:        return 1      if jump[start][level] < end:        return math.inf      for j in range(level, -1, -1):        if jump[start][j] < end:          break      return (1 << j) + minJumps(jump[start][j], end, j)     def minDist(u: int, v: int) -> int:      uIndex = indexMap[u]      vIndex = indexMap[v]      start = min(uIndex, vIndex)      end = max(uIndex, vIndex)      res = minJumps(start, end, maxLevel - 1)      return res if res < math.inf else -1     return [minDist(u, v) for u, v in queries]

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