Day 1: check if a list has consecutive 3

Author: AimenYaseen

Day 1/__init__.py

@@ -0,0 +1,17 @@
+def has_33(input_list):
+    length = len(input_list)
+    for i in range(0, length):
+        if input_list[i] == 3:
+            if i + 1 < length:
+                if input_list[i + 1] == 3:
+                    return True
+
+    return False
+
+
+print(has_33([1, 2, 3]))  # False
+print(has_33([1, 3, 3]))  # True
+print(has_33([1, 3, 1, 3]))  # False
+print(has_33([3, 1, 3]))  # False
+print(has_33([3, 3, 1]))  # True
+print(has_33([1, 2, 2, 3]))  # False

Day 1/has_33.py

@@ -0,0 +1,17 @@
+# to check two consecutive values use list range
+# [i: i+2] == [3, 3]
+
+def has_33(input_list):
+    length = len(input_list) - 1
+    for i in range(0, length):
+        if input_list[i: i + 2] == [3, 3]:
+            return True
+    return False
+
+
+print(has_33([1, 2, 3]))  # False
+print(has_33([1, 3, 3]))  # True
+print(has_33([1, 3, 1, 3]))  # False
+print(has_33([3, 1, 3]))  # False
+print(has_33([3, 3, 1]))  # True
+print(has_33([1, 2, 2, 3]))  # False

Day 1/has_33_advanced.py

@@ -0,0 +1,14 @@
+def has_33(input_list):
+    length = len(input_list) - 1
+    for i in range(0, length):
+        if input_list[i] == 3 and input_list[i + 1] == 3:
+            return True
+    return False
+
+
+print(has_33([1, 2, 3]))  # False
+print(has_33([1, 3, 3]))  # True
+print(has_33([1, 3, 1, 3]))  # False
+print(has_33([3, 1, 3]))  # False
+print(has_33([3, 3, 1]))  # True
+print(has_33([1, 2, 2, 3]))  # False