Added python solution for day 7 (#77)

* Add @vishalshirke7 as a contributor

* Update @vishalshirke7 as a contributor

* Added python solutions for Day 4

* modified readme

* Add @Vishal

* reverted adding as a contributor

* Update README.md

* Update CONTRIBUTORS.md

* Update CONTRIBUTORS.md

* Added python solutions for day 7

CONTRIBUTORS.md

@@ -1,5 +1,7 @@
+
 [![All Contributors](https://img.shields.io/badge/all_contributors-15-orange.svg?style=flat-square)](#contributors)
 
+
 ## Contributors
 
 Thanks goes to these wonderful people ([emoji key](https://github.com/kentcdodds/all-contributors#emoji-key)):

day4/README.md

@@ -273,6 +273,8 @@ for i in b:
 print(count)
 ```
 
+<hr/>
+
 ## C++ Implementation
 
 ### [NumVowelsPartA.cpp](./C++/NumVowelsPartA.cpp)

day7/Python/onewordaway.py

@@ -0,0 +1,29 @@
+"""
+  @author : vishalshirke7
+  @date : 28/12/2018
+"""
+
+
+def check_one_edit_away(str1, str2):
+    l1, l2 = len(str1), len(str2)
+    if abs(l1 - l2) > 1:
+        return False
+    if l1 == l2:
+        count = 0
+        for i in range(l1):
+            if str1[i] != str2[i]:
+                count += 1
+        if count > 1:
+            return False
+        else:
+            return True
+    else:
+        if len(set(str1) & set(str2)) == min(l1, l2):
+            return True
+        else:
+            return False
+
+
+check_one_edit_away(*(input().split()))
+
+

day7/README.md

@@ -121,4 +121,48 @@ oneEditAway ("abcdef", "abcde");  // true
 oneEditAway ("aaa", "abc");  // false
 oneEditAway ('bc', 'abc'); // true
 oneEditAway ('abc', 'abced'); // false
+```
+
+## Python Implementatiom
+
+### [Solution 1](./Python/onewordaway.py)
+
+```python
+"""
+  @author : vishalshirke7
+  @date : 28/12/2018
+  
+  Method - There are 3 test cases
+      1) Difference of length between 2 characters is more than one, then answer is False
+      2) If length of both strings is equal then we'll have to check for 2 cases
+          2.1) if count of different characters at each position is > 1 then answer is False
+          2.2) if count of different characters at each position is < 2 then answer is True
+      3) if difference in length of both strings in 1 then we check for intersection of 
+         both strings and if it is equal to min(length1, length2) then answer is true 
+         as we can add a character to minimum length string or remove one from bigger length string
+"""
+
+
+def check_one_edit_away(str1, str2):
+    l1, l2 = len(str1), len(str2)
+    if abs(l1 - l2) > 1:
+        return False
+    if l1 == l2:
+        count = 0
+        for i in range(l1):
+            if str1[i] != str2[i]:
+                count += 1
+        if count > 1:
+            return False
+        else:
+            return True
+    else:
+        if len(set(str1) & set(str2)) == min(l1, l2): 
+            # check for intersection of two strings
+            return True
+        else:
+            return False
+
+
+check_one_edit_away(*(input().split())) # here list unpacking is used
 ```