3d printing python solution uploaded - Apprentice 2023a

solutions/3d-printing/README.md

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+# 3D Printing
+
+You are part of the executive committee of the Database Design Day festivities. You are in charge of promotions and want to print three D's to create a logo of the contest. You can choose any color you want to print them, but all three have to be printed in the same color.
+
+You were given three printers and will use each one to print one of the D's. All printers use ink from 4 individual cartridges of different colors (cyan, magenta, yellow, and black) to form any color. For these printers, a color is uniquely defined by 4 non-negative integers c, m, y, and k, which indicate the number of ink units of cyan, magenta, yellow, and black ink (respectively) needed to make the color.
+
+The total amount of ink needed to print a single D is exactly 1000000 units. For example, printing a D in pure yellow would use 1000000 units of yellow ink and 0 from all others. Printing a D in the Code Jam red uses 0 units of cyan ink, 500000 units of magenta ink, 450000 units of yellow ink, and 50000 units of black ink.
+
+To print a color, a printer must have at least the required amount of ink for each of its 4 color cartridges. Given the number of units of ink each printer has in each cartridge, output any color, defined as 4 non-negative integers that add up to 106, such that all three printers have enough ink to print it.
+
+More details: https://codingcompetitions.withgoogle.com/codejam/round/0000000000876ff1/0000000000a4672b

solutions/3d-printing/python/approach1.py

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+"""3d printing solution
+
+This is a solution for the Google Code Jam '3d printing' problem
+The first line of the input gives the number of test cases, T.
+T test cases follow. Each test case consists of 3 lines.
+The i-th line of a test case contains 4 integers
+Ci, Mi, Yi, and Ki, representing the number of ink units
+in the i-th printer's cartridge for the colors
+cyan, magenta, yellow, and black, respectively.
+More information:
+codingcompetitions.withgoogle.com/
+    codejam/round/0000000000876ff1/0000000000a4672b
+"""
+
+def solve(printers):
+    """Finds a 4-digit color solution
+
+    Finds a solution consisting of 4 numbers
+    representing cyan, yellow, magenta, and black ink units
+
+    Args:
+        printers: list of (list of int)
+            inner lists represent ink units available in a printer,
+            outer list represents the printers
+            [[c, y, m, b], [c, y, m, b], ...]
+
+    Returns:
+        A string containing 4 space separated numbers
+        representing a color made up of
+        cyan, yellow, magenta and black ink units
+    """
+
+    # Initialize the "available ink units" per ink type
+    #                    C        Y        M        B
+    maxInkPerColor = [1000000, 1000000, 1000000, 1000000]
+
+    # finds the min value per ink type in the printers
+    for printer in printers:
+        for i in range(0, 4):
+            maxInkPerColor[i] = min(maxInkPerColor[i], printer[i])
+
+    inkUnits = sum(maxInkPerColor)
+
+    if (inkUnits < 1000000):
+        # no combination can be used in EVERY printer
+        # to print with 1000000 units
+        return "IMPOSSIBLE"
+
+    for i in range(0, 4):
+        inkUnits = sum(maxInkPerColor)
+        diff = inkUnits - 1000000
+        if diff == 0:
+            # return solution in string format
+            return " ".join(str(item) for item in maxInkPerColor)
+        # if current sum is not 1000000 we need to reduce ink units
+        if maxInkPerColor[i] >= diff:
+            # just need to reduce the diff
+            # and at next iteration diff value will be 0
+            maxInkPerColor[i] -= diff
+        else:
+            maxInkPerColor[i] = 0
+
+t = int(input()) # read number of test cases
+for i in range(1, t+1):
+    printer1 = [int(s) for s in input().split(' ')]
+    printer2 = [int(s) for s in input().split(' ')]
+    printer3 = [int(s) for s in input().split(' ')]
+    print(f"Case #{i}: {solve([printer1, printer2, printer3])}")
+