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Problem solution · Python

CCC 2020 S1 - Surmising a Sprinter's Speed

CCC 2020 S1 - Surmising a Sprinter's Speed: a Python solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Sorting and greedy selection
Source
CCCSolutions
Length
34 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 CCC 2020 S1 - Surmising a Sprinter's Speed, 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

  • 34 lines of Python from the credited upstream file ccc20s1.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 CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2020 S1 - Surmising a Sprinter's Speed · PythonPython
Use this to learn the idea, then write your own version.
# By Oscar Zhou, Abbey Park High School# This solution still requires commented explanations. Please feel free to add them! import sysinput = sys.stdin.readline N = int(input()) obs1 = {} for _ in range(N):    input1 = input().split()    time = int(input1[0])    pos = int(input1[1])    obs1[time] = pos times = list(obs1.keys())times.sort()obs = {} for i in range(N):    obs[times[i]] = obs1[times[i]] max = 0 for i in range(N-1):    tdiff = times[i+1]-times[i]    key1 = times[i]    key2 = times[i+1]    pdiff = obs[key2] - obs[key1]    speed = abs(pdiff/tdiff)    if speed > max:        max = speedprint(max)

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