04_FrogRiverOne.py

"""
Frog River One

A small frog wants to get to the other side of a river. The frog is initially located on one bank of the river (position
 0) and wants to get to the opposite bank (position X+1). Leaves fall from a tree onto the surface of the river.

You are given an array A consisting of N integers representing the falling leaves. A[K] represents the position where
one leaf falls at time K, measured in seconds.

The goal is to find the earliest time when the frog can jump to the other side of the river. The frog can cross only
when leaves appear at every position across the river from 1 to X (that is, we want to find the earliest moment when
all the positions from 1 to X are covered by leaves). You may assume that the speed of the current in the river is
negligibly small, i.e. the leaves do not change their positions once they fall in the river.

For example, you are given integer X = 5 and array A such that:
  A[0] = 1
  A[1] = 3
  A[2] = 1
  A[3] = 4
  A[4] = 2
  A[5] = 3
  A[6] = 5
  A[7] = 4

In second 6, a leaf falls into position 5. This is the earliest time when leaves appear in every position across the
river.

Write a function:

    def solution(X, A)

that, given a non-empty array A consisting of N integers and integer X, returns the earliest time when the frog can jump
to the other side of the river.

If the frog is never able to jump to the other side of the river, the function should return −1.

For example, given X = 5 and array A such that:
  A[0] = 1
  A[1] = 3
  A[2] = 1
  A[3] = 4
  A[4] = 2
  A[5] = 3
  A[6] = 5
  A[7] = 4

the function should return 6, as explained above.

Write an efficient algorithm for the following assumptions:

        N and X are integers within the range [1..100,000];
        each element of array A is an integer within the range [1..X].

Copyright 2009–2022 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure
prohibited.
"""
import unittest
import random


def solution(X, A):
    """
    :param X: [int] The frog's destination.
    :param A: [int] A sequence of leaf arrivals at these positions.
    :returns: [int]  The earliest time the frog can get to position x, or -1 if never.

    * Create a set to put all the collected leaves into.
    * Start collecting leaves and, if needed to get to our destination, put
        it into the set.  Python sets are made to prevent duplicates, so that
        makes things easy.
    * When we have enough leaves in the set to reach our destination we're done!
        (We know they are all uniquely numbered because it is set).
    """
    leaves = set()
    for count, leaf in enumerate(A):
        if leaf <= X:   # Only collect the leaves we need to get to X.
            leaves.add(leaf)
            if len(leaves) >= X:
                return count
    return -1


class TestExercise(unittest.TestCase):
    def test_example1(self):
        self.assertEqual(solution(5, [1, 3, 1, 4, 2, 3, 5, 4]), 6)

    def test_extreme_single(self):
        self.assertEqual(solution(1, [1]), 0)
        self.assertEqual(solution(0, [2]), -1)
        self.assertEqual(solution(2, [2]), -1)
        self.assertEqual(solution(2, [2, 1]), 1)
        self.assertEqual(solution(2, [1, 2]), 1)
        self.assertEqual(solution(2, [1, 1]), -1)
        self.assertEqual(solution(2, [2, 2]), -1)
        self.assertEqual(solution(3, [1, 2]), -1)

    def test_simple(self):
        self.assertEqual(solution(3, [1, 2, 3]), 2)
        self.assertEqual(solution(3, [2, 3, 1]), 2)
        self.assertEqual(solution(3, [3, 2, 1]), 2)
        self.assertEqual(solution(4, [3, 2, 1]), -1)
        self.assertEqual(solution(3, [3, 2, 2]), -1)
        self.assertEqual(solution(3, [3, 3, 3, 1, 1, 2]), 5)

    def test_extreme_max(self):
        X = 100000
        A = list(range(1, X))
        random.shuffle(A)
        A.append(X)
        self.assertEqual(solution(X, A), X - 1)

    def test_issue_1(self):
        """Someone actually bothered to report a mistake.

        https://github.com/johnmee/codility/issues/1
        """
        self.assertEqual(-1, solution(3, [3, 3, 1, 1, 4]))

    # Codility Tests
    # extreme_frog - frog never crosses
    # small_random1 - 3 random permutation X=50
    # small_random2 - 5 random permutation X=60
    # extreme_leaves = all leaves in the same place
    # medium_random = 6 and 2 random permutations, X = ~5000
    # medium_range = arithmetic sequences X = 5000
    # large_random = 10 and 100 random permutation, X = ~10000
    # large_permutation = permutation tests (2)
    # large_range = arithmetic sequences X=30000


if __name__ == "__main__":
    unittest.main()