Added odd_sum and smooth_max solutions

Added odd_sum and smooth_max solutions that I created

Author: Citrus716

2_intermediate/chapter10/solutions/odd_sum.py

@@ -0,0 +1,25 @@
+# Odd Sum
+# Given a 2D list, find the sum of all elements at odd indexes for all
+# the lists at odd indexes. Print this sum times the sum of all
+# first element of all the 1D lists in the 2D list.
+#
+# Ex:[[1,2,3,6],[2,41,2,1]]should have print 42 after the program runs.
+#
+# Write the code below.
+
+two_d_list = [[1,2,3,5,2],[2,3,1,4],[2,3,1,2,21],[21,3,1,41]]
+#two_d_list should print 51 after the program runs.
+
+odd_sum = 0
+for outer_idx in range(1,len(two_d_list),2):
+    for inner_idx in range(1, len(two_d_list[outer_idx]),2):
+        odd_sum += two_d_list[outer_idx][inner_idx]
+
+first_sum = 0
+for inner_list in range(len(two_d_list)):
+    first_sum += two_d_list[inner_list][0]
+
+print(odd_sum* first_sum )
+
+
+        

2_intermediate/chapter10/solutions/smooth_max.py

@@ -0,0 +1,24 @@
+# Given a 2D list, let's call a element "smooth" if index of the
+# element in its 1D list plus the element is even. For example,
+# given the 2D list [[0,4][2,6]], the 1st element of each of the
+# 1D list is considered "smooth" because 0 + 0 is 0 and 0 + 2 is 2
+# (both are even numbers). Find the maximum "smooth" element and
+# print it. Using the example [[0,4][2,6]] again, the maximum 
+# "smooth" element is 2 because 2 is bigger than 0.
+
+two_d_list = [[425,214,412,123],[312,214,123,343]]
+curr_max = None
+
+for outer_idx in range(len(two_d_list)):
+    for inner_idx in range(len(two_d_list[outer_idx])):
+        curr_elem = two_d_list[outer_idx][inner_idx]
+        to_check = curr_elem + inner_idx
+        if to_check % 2 ==  0:
+            if curr_max == None or curr_elem > curr_max:
+                curr_max = curr_elem
+
+print(curr_max)
+
+
+
+