Day 10

Author: tuukkasarvi

data/input_10.txt

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+138
+3
+108
+64
+92
+112
+44
+53
+27
+20
+23
+77
+119
+62
+121
+11
+2
+37
+148
+34
+83
+24
+10
+79
+96
+98
+127
+7
+115
+19
+16
+78
+133
+61
+82
+91
+145
+39
+33
+13
+97
+55
+141
+1
+134
+40
+71
+54
+103
+101
+26
+47
+90
+72
+126
+124
+110
+131
+58
+12
+142
+105
+63
+75
+50
+95
+69
+25
+68
+144
+86
+132
+89
+128
+135
+65
+125
+76
+116
+32
+18
+6
+38
+109
+111
+30
+70
+143
+104
+102
+120
+31
+41
+17

day_10.py

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+"""
+Each of your joltage adapters is rated for a specific output joltage (your puzzle input). Any given adapter
+can take an input 1, 2, or 3 jolts lower than its rating and still produce its rated output joltage.
+
+In addition, your device has a built-in joltage adapter rated for 3 jolts higher than the highest-rated adapter
+in your bag. (If your adapter list were 3, 9, and 6, your device's built-in adapter would be rated for 12 jolts.)
+
+Treat the charging outlet near your seat as having an effective joltage rating of 0.
+
+Since you have some time to kill, you might as well test all of your adapters. Wouldn't want to get to your resort
+and realize you can't even charge your device!
+
+If you use every adapter in your bag at once, what is the distribution of joltage differences between the charging
+outlet, the adapters, and your device?
+
+1.
+Find a chain that uses all of your adapters to connect the charging outlet to your device's built-in adapter and count
+the joltage differences between the charging outlet, the adapters, and your device. What is the number of 1-jolt
+differences multiplied by the number of 3-jolt differences?
+
+2.
+What is the total number of distinct ways you can arrange the adapters to connect the charging outlet to your device?
+
+"""
+import numpy as np
+
+
+filepath = "data/input_10.txt"
+
+
+def part1():
+    with open(filepath) as fp:
+        lines = [line.strip("\n") for line in fp]
+    joltages = np.array([int(item) for item in lines])
+    joltages.sort()
+    joltages = np.concatenate(([0], joltages, [joltages.max() + 3]))
+    diffs = joltages[1:] - joltages[:(len(joltages)-1)]
+    unique_elements, counts_elements = np.unique(diffs, return_counts=True)
+    print(unique_elements)
+    answer = counts_elements[0] * counts_elements[1]
+    print("Answer %d." % answer)
+
+
+def get_number_of_paths(ones_group):
+    """ For a set of ones [1]_k (k ones) the number of paths f([1]_k) =
+    f([1]_(k-1)) + f([1]_(k-2)) + f([1]_(k-3)).
+    """
+    assert set(ones_group) == {1} or len(ones_group) == 0
+    if len(ones_group) == 0:
+        return 1
+    elif len(ones_group) == 1:
+        return 1
+    elif len(ones_group) == 2:
+        return get_number_of_paths(ones_group[1:]) + get_number_of_paths(ones_group[2:])
+    else:  # len >= 3
+        return get_number_of_paths(ones_group[1:]) + get_number_of_paths(ones_group[2:]) + \
+               get_number_of_paths(ones_group[3:])
+
+
+def part2():
+    with open(filepath) as fp:
+        lines = [line.strip("\n") for line in fp]
+    joltages = np.array([int(item) for item in lines])
+    joltages.sort()
+    joltages = np.concatenate(([0], joltages, [joltages.max() + 3]))
+    diffs = joltages[1:] - joltages[:(len(joltages)-1)]
+    # in diffs there are only 1s and 3s
+    # paths can be formed by combining different 1s. Each subset of 1s between 3s is an independent set of paths
+    # total number of paths found by multiplying number of paths for each subset of ones: n1 * n2 * n3 ... nk
+    diffs_string = "".join([str(i) for i in diffs])
+    ones_groups = diffs_string.split("3")
+    ones_groups = [list(item) for item in ones_groups]
+    ones_groups = [[int(i) for i in item] for item in ones_groups]
+    numbers_of_paths = [get_number_of_paths(og) for og in ones_groups]
+    total_number_paths = np.prod(np.array(numbers_of_paths))
+    print("Answer %d" % total_number_paths)
+
+
+if __name__ == '__main__':
+    part1()
+    print("*****")
+    part2()