adding 1st exercise solution here

Author: RohanSaxena2020

1.py

@@ -0,0 +1,105 @@
+# ### Binary Search Exercise
+# 1. When I try to find number 5 in below list using binary search, it doesn't work and returns me -1 index. Why is that?
+
+#     ```numbers = [1,4,6,9,10,5,7]```
+
+# This is because the array is not sorted in order from lowest to highest. 
+# Once it splits the first time, it starts looking in the [1,4,6] range and doesn't find 5
+    
+# 1. Find index of all the occurances of a number from sorted list
+
+#     ```
+#     numbers = [1,4,6,9,11,15,15,15,17,21,34,34,56]
+#     number_to_find = 15  
+#     ```
+#    This should return 5,6,7 as indices containing number 15 in the array
+
+from util import time_it
+
+@time_it
+def linear_search(numbers_list, number_to_find):
+    for index, element in enumerate(numbers_list):
+        if element == number_to_find:
+            return index
+    return -1
+
+@time_it
+def binary_search(numbers_list, number_to_find):
+    left_index = 0
+    right_index = len(numbers_list) - 1
+    mid_index = 0
+
+    while left_index <= right_index:
+        mid_index = (left_index + right_index) // 2
+        mid_number = numbers_list[mid_index]
+
+        if mid_number == number_to_find:
+            return mid_index
+
+        if mid_number < number_to_find:
+            left_index = mid_index + 1
+        else:
+            right_index = mid_index - 1
+
+    return -1
+
+def binary_search_recursive(numbers_list, number_to_find, left_index, right_index):
+    if right_index < left_index:
+        return -1
+
+    mid_index = (left_index + right_index) // 2
+    if mid_index >= len(numbers_list) or mid_index < 0:
+        return -1
+
+    mid_number = numbers_list[mid_index]
+
+    if mid_number == number_to_find:
+        return mid_index
+
+    if mid_number < number_to_find:
+        left_index = mid_index + 1
+    else:
+        right_index = mid_index - 1
+
+    return binary_search_recursive(numbers_list, number_to_find, left_index, right_index)
+
+#this should run the binary search, find the index, and then recursively run the search on both the right and left side
+def binary_search_multiple(numbers_list, number_to_find):
+
+    index = binary_search(numbers_list,number_to_find)
+    result_indices = [index]
+
+    # find all indices on the left
+    i = index - 1
+    while i>=0:
+        if numbers_list[i] == numbers_list[index]:
+            result_indices.append(i)
+        else:
+            break
+        i = i-1
+
+    # find all indices on the right
+    i = index + 1
+    while i<len(numbers_list):
+        if numbers_list[i] == numbers_list[index]:
+            result_indices.append(i)
+        else:
+            break
+        i = i+1
+
+    return sorted(result_indices)
+
+numbers_list = [12, 15, 17, 19, 21, 21, 21, 21, 24, 45, 67]
+number_to_find = 21
+
+index = binary_search_multiple(numbers_list, number_to_find)
+print(f"Number found at index {index} using binary search")
+
+numbers = [1,4,6,9,11,15,15,15,15,17,21,34,34,56]
+number_to_find = 15
+
+index = binary_search_multiple(numbers, number_to_find)
+print(f"Number found at index {index} using binary search")
+
+#Lesson: I was approaching it wrong. If something isn't working, scratch the approach. 
+#Lesson #2: Try the simplest solution first. Although in this case it's a bit ugly since you're just doing a linear search after your binary search

util.py

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+import time
+def time_it(func):
+    def wrapper(*args, **kwargs):
+        start = time.time()
+        result = func(*args,**kwargs)
+        end = time.time()
+        print(func.__name__ +" took " + str((end-start)*1000) + " mil sec")
+        return result
+    return wrapper