juan's coding challenge prompts added; solutions to follow

Coding_Challenges/Shortest_Lock_Combos/README.md

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+# Strange Sequence
+
+## Prompt 1
+
+Write a function that determines the minimum number of rotations needed to unlock a a basic number combination lock
+
+A rotation is defined as each time you increment/decrement a number by one.  
+For example, to get from 2 to 4, you would perform 2 rotations:
+2 -> 3, and 3 -> 4
+
+Furthermore, the lock can wrap around the numbers, so that it would only take 3 rotations to get from 9 to 2
+9 -> 0, 0 -> 1, and 1 ->2
+
+Your function takes two arguments:
+1. the current state of the lock, which is shown as an array of single digits representing each bit of the lock
+2. An array showing the unlocking position
+Both arrays will made up of single digits and they will be the same length, though they can be of any size
+
+Example:
+current lock state: [0, 0, 0]
+unlocking combination: [3, 1, 8]
+
+first: 0 -> 1 -> 2 -> 3   ____ 3 rotations
+Second: 0 -> 2 ___ 1 rotation
+third: 0 -> 9 -> 8 ____ 2 rotations
+Total: 6 rotations
+
+
+## Prompt 2
+
+Update your answer to the previous prompt to account for locks with arbitrary characters instead of numbers 0-9.
+They will be accepted as a third character set array, and will contain at least two strings.  
+The character set is ordered and will contain only unique values. 
+Similarly, you are able to wrap around the lock from one end of the array to the other. 
+Your first two array arguments will only contains values that appear in the character set of locks
+
+Example: 
+starting point: ['*', '#', '+']
+unlocking combination: ['+', 'i', '{\}']
+Character set: ['!', '+', '*', ')', 'i', '=', '{\}', '#']
+
+The result would be the following:
+'*' -> '+ ____ 1 rotation
+'#' -> '{\}' -> '=' -> 'i' ____ 3 rotations
+'+' -> '!' -> '#' -> '{\}' ___ 3 rotations
+Total: 7 rotations

Coding_Challenges/Strange_Sequence/README.md

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+# Strange Sequence
+
+## Prompt
+
+Imagine a sequence of numbers x0, x1, x2,..., xn where a given element is equal to the sum of the squared digits of the previous element. 
+
+The sequence ends once you reach an element that has already been in the sequence appears again.  
+
+For example, for x0 = 2 our answer would be 10 after the following sequence:
+
+x0 = 2
+x1 = 2^2 = 4
+x2 = 4^2 = 16
+x3 = 1^2 + 6^2 = 37
+x4 = 3^2 + 7^2 = 58
+x5 = 5^2 + 8^2 = 89
+x6 = 8^2 + 9^2 = 145
+x7 = 1^2 + 4^2 + 5^2 = 42
+x8 = 4^2 + 2^2 = 20
+x9 = 2^2 + 0^2 = 4
+Since we have already seen 4 in the sequence, we return the length of the sequence
+
+Write a function that accepts one argument (the first element)and returns the length of sequence