Single Number III + Readme (#17)

* 1513

* Upload: "Flower Planting with No adjacent"-Python

* Upload: #56 Merge Intervals (Python)

* Updated Readme

* Fixed Readme

* Updated Readme (Shift Twosum -> HashMap)

* Upload Single Number III

* Updated SingleNumber III + Readme

Co-authored-by: Gourav Rusiya <[email protected]>

Author: lokendra1704

Python/single-number-iii.py

@@ -0,0 +1,65 @@
+#Time Complexity:   O(n)
+#Space Complexity:  O(1)
+#Speed: 62.81%
+#Memory: 81.52%
+'''
+Xor Intuition: If different bit, result will be 1
+               If same bit, results will be 0
+               Example: 1^0 = 1
+                        1^1 = 0
+                        0^0 = 0
+Let's analyse few properties of Xor bitwise operation:
+1) x^x = 0
+    Example: 5^5 = 0
+    Thing to note is, It works in collection also, For ex:
+    In nums = [1,1,2,2,3,3,4,4], If we do complete xor of the list, result will be 0
+    Code:
+    
+    xor = 0
+    for i in nums:
+        xor^=i
+    print(xor)      # Prints 0
+
+2) x^0 = x
+    Example: 5^0 = 5
+    Note:  If we have a list, which consists of every number repeating twice except a single number which is unique (does
+    not repeat) then the complete xor of the list will give the unique number as output. (Since xor of all repeating number
+    will be 0, and then 0^unique = unique )
+    Example: nums = [1,1,2,2,999,3,3,4,4]
+             ans = functools.reduce(lambda x,y:x^y, nums, 0)
+             print(ans)         #Prints 999
+
+    Now In the Given problem,
+    [1,2,1,3,2,5]
+    => There are two unique elements, so complete xor of list will result in xor of these two unique elements.
+        Let them be x and y
+    => So the complete xor of list will be:
+        Example:   011
+                 ^ 101
+        result=    110
+    => We will take any index in ans which is 1 (means that bit is diff in x and y) and partition the list into two groups
+       based on whether that index bit is on(1) or off(0).
+       In example we will take the rightmost bit which is 1, so the index is: 1
+       Easiest way to do this:
+
+                xor = xor^(-xor)    #gives a number in which only that particular bit is on.
+                for n in nums:
+                    if n & xor:         #If that bit in n is on
+                        n belongs to group 1
+                    else:
+                        n belongs to group 2
+                        
+    => Xor of first group will give first unique element and xor of second group will give second unique element.
+'''
+
+class Solution:
+    def singleNumber(self, nums: List[int]) -> List[int]:
+        xor = reduce(lambda x,y: x^y, nums, 0)          #complete xor of list will result in xor of these two unique elements
+        xor = xor&(-xor)
+        first = second = 0
+        for num in nums:
+            if xor & num:
+                first ^= num
+            else:
+                second ^= num
+        return [first, second]

README.md

@@ -68,6 +68,7 @@ Check out ---> [Sample PR](https://github.com/codedecks-in/LeetCode-Solutions/pu
 | ---- | ------------------------------------------------------------------- | -------------------------------------------------------------------------------- | ------ | ------ | ---------- | --- | --------- |
 | 0136 | [Single Number](https://leetcode.com/problems/single-number/)       | [Java](./Java/single-number.java) <br> [Python](./Python/single-number.py)       | _O(n)_ | _O(1)_ | Easy       |     | Using XOR |
 | 0137 | [Single Number II](https://leetcode.com/problems/single-number-ii/) | | _O(n)_ | _O(1)_ | Medium     |     |           |
+| 0260 | [Single Number III](https://leetcode.com/problems/single-number-iii/) |[Python](./Python/single-number-iii.py) | _O(n)_ | _O(1)_ | Medium     |     |           |
 
 <br/>
 <div align="right">
@@ -153,6 +154,18 @@ Check out ---> [Sample PR](https://github.com/codedecks-in/LeetCode-Solutions/pu
 </div>
 <br/>
 
+## Two Pointer
+
+| #    | Title                                                                                               | Solution                                                | Time     | Space     | Difficulty | Tag   | Note           |
+| ---- | --------------------------------------------------------------------------------------------------- | ------------------------------------------------------- | -------- | --------- | ---------- | ----- | -------------- |
+|5|[Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)|[Python](./Python/5_LongestPalindromicSubstring.py)|_O(N^2)_|_O(N)_|Medium||Expand the Wings|
+
+<br/>
+<div align="right">
+    <b><a href="#algorithms">⬆️ Back to Top</a></b>
+</div>
+<br/>
+
 ## Math
 
 | #   | Title                                                            | Solution                                | Time       | Space  | Difficulty | Tag       | Note |