Added new coding challenge

Author: seba3c

src/trip_stops.py

@@ -0,0 +1,90 @@
+# -*- coding: utf-8 -*-
+import random
+import math
+import itertools
+
+from concurrent.futures.thread import ThreadPoolExecutor
+
+"""
+Return the number of possible combinations of three kind of stops during a trip, before
+a previous combination must be repeated.
+
+Each list represent the distance from the trip origin where the stop is placed.
+Each list can be disordered.
+
+Example:
+
+    A = [29, 50]
+    B = [61, 37]
+    C = [37, 70]
+
+(29, 37, 70), (29, 61, 70), (50, 61, 70)
+
+Should return 3.
+"""
+
+
+def solution_v1(A, B, C):
+
+    def countt(it):
+        itt = filter(lambda t: t[0] < t[1] and t[1] < t[2], it[0])
+        return sum(1 for _ in itt)
+
+    N = len(A)
+    NN = N * N * N
+
+    prod = itertools.product(A, B, C)
+    t = ThreadPoolExecutor(4)
+    chunk = math.ceil(NN / 4)
+
+    f1 = t.submit(countt, (itertools.islice(prod, chunk),))
+    f2 = t.submit(countt, (itertools.islice(prod, chunk),))
+    f3 = t.submit(countt, (itertools.islice(prod, chunk),))
+    f4 = t.submit(countt, (itertools.islice(prod, chunk),))
+    count = f1.result() + f2.result() + f3.result() + f4.result()
+    return count if count < 1000000000 else -1
+
+
+if __name__ == '__main__':
+
+    A = [29, 50]
+    B = [61, 37]
+    C = [37, 70]
+
+    ans = solution_v1(A, B, C)
+    assert(ans == 3)
+
+    A = [5]
+    B = [5]
+    C = [5]
+
+    ans = solution_v1(A, B, C)
+    assert(ans == 0)
+
+    A = [5]
+    B = [6]
+    C = [7]
+
+    ans = solution_v1(A, B, C)
+    assert(ans == 1)
+
+    A = [5, 6]
+    B = [5, 7]
+    C = [5, 8]
+
+    ans = solution_v1(A, B, C)
+    assert(ans == 2)
+
+    N = 100
+    A = [random.randint(1, 500) for x in range(N)]
+    B = [random.randint(501, 1000) for x in range(N)]
+    C = [random.randint(1001, 1500) for x in range(N)]
+    ans = solution_v1(A, B, C)
+    assert(ans == N * N * N)
+
+    N = 100
+    A = [random.randint(1, 500) for x in range(N)]
+    B = [random.randint(501, 1000) for x in range(N)]
+    C = [x for x in range(N - 1)] + [1001]
+    ans = solution_v1(A, B, C)
+    assert(ans == N * N * 1)