First pass solutions completed

eating_cookies/eating_cookies.py

@@ -1,14 +1,30 @@
-'''
+"""
 Input: an integer
 Returns: an integer
-'''
+"""
+
+
 def eating_cookies(n):
-    # Your code here
+    """
+    Cookie Monster can eat either 1, 2, or 3 cookies at a time
+    Implement a function eating_cookies that counts the number of possible ways Cookie Monster
+    can eat all of the cookies in the jar.
+    """
+    # Plan: Use Recursion until you hit the base cases
+    if n == 3:
+        return 4
+    if n == 2:
+        return 2
+    if n == 1:
+        return 1
+    if n <= 0:
+        return 1
+    if n > 3:
+        return eating_cookies(n - 3) + eating_cookies(n - 2) + eating_cookies(n - 1)
 
-    pass
 
 if __name__ == "__main__":
     # Use the main function here to test out your implementation
-    num_cookies = 5
+    num_cookies = 4
 
     print(f"There are {eating_cookies(num_cookies)} ways for Cookie Monster to each {num_cookies} cookies")

moving_zeroes/moving_zeroes.py

@@ -1,11 +1,19 @@
-'''
-Input: a List of integers
-Returns: a List of integers
-'''
 def moving_zeroes(arr):
-    # Your code here
+    """
+    Write a function that takes an array of integers and moves each non-zero integer
+     to the left side of the array, then returns the altered array.
+     The order of the non-zero integers does not matter in the mutated array.
+     Input: a List of integers
+    Returns: a List of integers
+    """
+    # so could append to the end if it equals zero
+    for i in arr:
+        if i == 0:
+            # remove so that we don't have leftover zeroes
+            arr.remove(i)
+            arr.append(i)
 
-    pass
+    return arr
 
 
 if __name__ == '__main__':

product_of_all_other_numbers/product_of_all_other_numbers.py

@@ -1,11 +1,27 @@
-'''
-Input: a List of integers
-Returns: a List of integers
-'''
+# https://discuss.codecademy.com/t/challenge-product-of-everything-else/176317
 def product_of_all_other_numbers(arr):
-    # Your code here
-
-    pass
+    """
+    Write a function that receives an array of integers and returns an array
+    consisting of the product of all numbers in the array except the number at that index.
+    :param arr: a list of integers
+    :return: a list of integers
+    """
+    # so someway to exclude the index while we are doing multiplying
+    # results to keep the result of all the products except index
+    result = 1
+    # product_list to put each product for each value
+    product_list = []
+    # loop through the arr
+    for i in range(len(arr)):
+        # access each number in the arr
+        for n in arr:
+            if n != arr[i]:
+                # if n doesn't equal the current number for i then multiple it
+                result *= n
+        product_list.append(result)
+        # restart
+        result = 1
+    return product_list
 
 
 if __name__ == '__main__':

single_number/single_number.py

@@ -1,15 +1,18 @@
-'''
-Input: a List of integers where every int except one shows up twice
-Returns: an integer
-'''
 def single_number(arr):
-    # Your code here
-
-    pass
-
+    """
+    Given a non-empty array of integers where every element appears twice except for one. Find that single number.
+    Input: a List of integers where every int except one shows up twice
+    Returns: an integer
+    """
+    # isn't there a way to take the item in the list then iterate over it until it finds it another pair?
+    # iterate
+    for i in arr:
+        # if the number of values is 1 return the single value
+        if arr.count(i) == 1:
+            return i
 
 if __name__ == '__main__':
     # Use the main function to test your implementation
     arr = [1, 1, 4, 4, 5, 5, 3, 3, 9, 0, 0]
 
-    print(f"The odd-number-out is {single_number(arr)}")
+    print(f"The odd-number-out is {single_number(arr)}")

sliding_window_max/sliding_window_max.py

@@ -1,11 +1,47 @@
-'''
-Input: a List of integers as well as an integer `k` representing the size of the sliding window
-Returns: a List of integers
-'''
+from _collections import deque
 def sliding_window_max(nums, k):
-    # Your code here
+    """
+    Given an array of integers, there is a sliding window of size k which is moving from the left side of the array
+    to the right, one element at a time. You can only interact with the k numbers in the window. Return an array
+    consisting of the maximum value of each window of elements.
+    Input: a List of integers as well as an integer `k` representing the size of the sliding window
+    Returns: a List of integers
+    """
+    # result for max inside the windows
+    # array that will hold the list of maximums
+    # divide as you iterate over the arr to get a new subarray each time
+    # window - nums[n:k + n]
+
+    max_vals = [0 for _ in range(len(nums) - k + 1)]
+    for i in range(len(max_vals)):
+        current_elem = nums[i]
+        for j in range(1, k):
+            if nums[i +j] > current_elem:
+                current_elem = nums[i + j]
+        max_vals[i] = current_elem
+    return max_vals
+
+    # remove all elems from a queue
+    # max_vals = []
+    # q = deque()
+    # # remove all elems from a queue
+    # for i, n in enumerate(nums):
+    #     while len(q) > 0 and n > q[-1]:
+    #         q.pop()
+    #     q.append(n)
+    #     # calc the window range
+    #     window_range = i - k + 1
+    #     # as long as our windows range == k, then we will add elements to the queue
+    # if window_range >= 0:
+    #     # add the max elem (in this case first in the queue) to the max_vals
+    #     max_vals.append(q[0])
+    #     # check num on the left needs to be evicted
+    #     # if so take it out of the start of the queue
+    #     if nums[window_range] == q[0]:
+    #         q.popleft()
+    # return max_vals
+
 
-    pass
 
 
 if __name__ == '__main__':