huntober

Author: IntelliJinceTech

.idea/.gitignore

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+# You will be given an array of numbers in which two numbers occur once and the rest occur only twice. Your task will be to return the sum of the numbers that occur only once.
+
+# For example, repeats([4,5,7,5,4,8]) = 15 because only the numbers 7 and 8 occur once, and their sum is 15. Every other number occurs twice.
+
+# repeats([4,5,7,5,4,8]),15)
+# repeats([9, 10, 19, 13, 19, 13]),19)
+# repeats([16, 0, 11, 4, 8, 16, 0, 11]),12)
+# repeats([5, 17, 18, 11, 13, 18, 11, 13]),22)
+# repeats([5, 10, 19, 13, 10, 13]),24)
+
+# input array/list of numbers
+# positive and negative?, characters?
+# numbers can have 1 + occurrence
+# one number output
+
+# initiate table
+# iterate over input array
+# if number doesn't exist in table, add
+# if number does exist in table + 1
+# add all key values
+
+def repeats(input):
+    # unique_num = list(filter(lambda n: input.count(n)==1, input))
+    # return sum(unique_num)
+    unique_num = list(filter(lambda n: input.count(n) == 1, input))
+    return sum(unique_num)
+print(repeats([4,5,7,5,4,8]),15)

.idea/codeWars.iml

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+# Return the number of unique arrays that can be formed by picking exactly one element from each subarray.
+
+# For example: solve([[1,2],[4],[5,6]]) = 4, because it results in only 4 possibilites. They are [1,4,5],[1,4,6],[2,4,5],[2,4,6].
+
+# Make sure that you don't count duplicates; for example solve([[1,2],[4,4],[5,6,6]]) = 4, since the extra outcomes are just duplicates.
+
+# solve([[1,2],[4],[5,6]]),4)
+# solve([[1,2],[4,4],[5,6,6]]),4)
+# solve([[1,2],[3,4],[5,6]]),8)
+# solve([[1,2,3],[3,4,6,6,7],[8,9,10,12,5,6]]),72)
+
+
+# array of arrays of only numbers
+# number outcome
+
+# iterate over array input
+# convert each nested array to a set
+# multiply length
+
+
+def solve(arr):
+    prod = 1
+    for num in arr:
+        prod *= len(set(num))
+    return prod
+    
+
+
+
+# print(solve([[1,2],[4],[5,6]]),4)
+print(solve([[1,2],[4,4],[5,6,6]]),4)
+print(solve([[1,2],[3,4],[5,6]]),8)
+print(solve([[1,2,3],[3,4,6,6,7],[8,9,10,12,5,6]]),72)