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Problem solution · Python

CCC 2010 S3 - Firehose

CCC 2010 S3 - Firehose: a Python solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Binary search
Source
CCCSolutions
Length
81 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

Binary search

For CCC 2010 S3 - Firehose, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 81 lines of Python from the credited upstream file ccc10s3.py.
  • The implementation visibly relies on sequence storage.
  • 1 loop block detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

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 2010 S3 - Firehose · PythonPython
Use this to learn the idea, then write your own version.
## This algorithm is by Sumudu Fernando## He discovered that the other two algorithms have a flaw.# The assumption that the minimum number of hydrants# can be achieved by skipping the "largest gap"# is false. # A small testcase shows this:# # 6# 0# 240000# 292000# 448000# 500000# 740000# 2# # The largest gap here is from the last house to the first,# of length 260000.  The erroneous programs give an answer# of 146000, while a hose of length 130000 is enough# (one hydrant covers the first & last houses,# while the second can cover the other four).## this situation did NOT arise in the CCC test data,# so the other solutions are technically "correct".## BUT this is "more correct" :-)  STREET_LEN = 1000000 # Returns the minimum number of hydrants needed to service# the given houses, assuming hoses of the given length.## i is the starting house, and it can be any of the houses# 0,1,2,3... which fall within the given hose diameter## the internal j loop is tricky to understand, but it# takes care of "wrap-around" and calculates the number of# hoses needed starting at house i. def hydrants(hose, houses):    best = len(houses)    diam = hose * 2        i = 0    while i < len(houses) and houses[i] <= houses[0] + diam:        count = 1        curEnd = houses[i]        j = i + 1        while j < len(houses) and (houses[i] > (houses[j] + diam - STREET_LEN)):            if houses[j] > curEnd:                count += 1                curEnd = houses[j] + diam            j = j + 1        best = min(best, count)        i = i + 1    return best  # A standard binary search of the hose lengths# does the trick. file = open("s3.1.in",'r')h = eval(file.readline())houses = []for i in range(h):    houses.append(eval(file.readline()))houses.sort()k = eval(file.readline())  lo = -1hi = STREET_LENwhile (hi > (lo+1)):    mid = (lo + hi) // 2    if (hydrants(mid, houses) > k):        lo = mid    else:        hi = midprint hi

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