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Problem solution · Python

Count Pairs Of Nodes

Count Pairs Of Nodes: 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
43 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 Count Pairs Of Nodes, 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

  • 43 lines of Python from the credited upstream file 1782.py.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 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 codeCount Pairs Of Nodes · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def countPairs(      self,      n: int,      edges: list[list[int]],      queries: list[int],  ) -> list[int]:    ans = [0] * len(queries)     # count[i] := the number of edges of node i    count = [0] * (n + 1)     # shared[i][j] := the number of edges incident to i or j, where i < j    shared = [collections.Counter() for _ in range(n + 1)]     for u, v in edges:      count[u] += 1      count[v] += 1      shared[min(u, v)][max(u, v)] += 1     sortedCount = sorted(count)     for k, query in enumerate(queries):      i = 1      j = n      while i < j:        if sortedCount[i] + sortedCount[j] > query:          # sortedCount[i] + sortedCount[j] > query          # sortedCount[i + 1] + sortedCount[j] > query          # ...          # sortedCount[j - 1] + sortedCount[j] > query          # So, there are (j - 1) - i + 1 = j - i pairs > query          ans[k] += j - i          j -= 1        else:          i += 1      for i in range(1, n + 1):        for j, sh in shared[i].items():          if count[i] + count[j] > query and count[i] + count[j] - sh <= query:            ans[k] -= 1     return ans

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