Completed product question

problems/cons_car_cdr.py

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+"""
+cons(a, b) constructs a pair, and car(pair) and cdr(pair) return the first
+and last element of that pair.
+
+    car(cons(3, 4)) == 3
+    cdr(cons(3, 4)) == 4
+
+Given this implementation of cons, implement car and cdr.
+"""
+
+
+def cons(a, b):
+    return lambda f: f(a, b)
+
+
+def car(pair):
+    return pair(lambda a, b: a)
+
+
+def cdr(pair):
+    return pair(lambda a, b: b)
+
+
+# --- tests ---
+
+assert car(cons(3, 4)) == 3
+assert cdr(cons(3, 4)) == 4
+assert car(cons("a", "b")) == "a"
+assert cdr(cons("a", "b")) == "b"
+
+print("All tests passed!")

problems/product.py

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+"""
+Given an array of integers, return a new array such that each element at index i of the new array is the product of all the numbers in the original array except the one at i.
+
+For example, if the input is [1, 2, 3, 4, 5], the expected output is [120, 60, 40, 30, 24].
+
+Follow-up: What if you can't use division?
+"""
+
+def product(arr: list[int]) -> list[int]:
+    # first we calculate prefix/suffix. We add a spot with multiplicative identity (1) for convenience.
+    prefix = [1]*len(arr)
+    suffix = [1]*len(arr)
+
+    p = 1
+    for i in range(0, len(arr)-1):
+        p *= arr[i]
+        prefix[i+1] = p
+
+    p = 1
+    for i in range(len(arr)-1, 0, -1):
+        p *= arr[i]
+        suffix[i-1] = p
+
+    ans = [1]*len(arr)
+    for i in range(len(ans)):
+        ans[i] = prefix[i] * suffix[i]
+    return ans
+
+
+# --- tests ---
+
+assert product([2, 3, 4, 5]) == [60, 40, 30, 24]
+assert product([1, 2, 3, 4, 5]) == [120, 60, 40, 30, 24]
+assert product([1, 2]) == [2, 1]
+
+# zero cases
+assert product([0, 1, 2, 3]) == [6, 0, 0, 0]
+assert product([0, 0, 2, 3]) == [0, 0, 0, 0]
+
+# single element
+assert product([5]) == [1]
+
+print("All tests passed!")