Added solution to hackerrank

hackerrank/easy-lonely-integer.md

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+Lonely Integer (HackerRank)
+===============================
+### Easy
+You will be given an array of integers. All of the integers except one occur twice. That one is unique in the array.
+
+Given an array of integers, find and print the unique element.
+
+For example, [1,2,3,4,3,2,1], the unique element is 4.
+
+Function Description
+
+Complete the lonelyinteger function in the editor below. It should return the integer which occurs only once in the input array.
+
+lonelyinteger has the following parameter(s):
+
+a: an array of integers
+Input Format
+
+The first line contains a single integer, , denoting the number of integers in the array.
+The second line contains  space-separated integers describing the values in .
+
+Constraints
+
+It is guaranteed that  is an odd number and that there is one unique element.
+, where .
+Output Format
+
+Print the unique integer in the array.
+
+```
+Sample Input 0
+1
+1
+Sample Output 0
+1
+Explanation 0
+There is only one element in the array, thus it is unique.
+```
+```
+Sample Input 1
+3
+1 1 2
+Sample Output 1
+2
+Explanation 1
+We have two 's, and  is unique.
+```
+```
+Sample Input 2
+5
+0 0 1 2 1
+Sample Output 2
+2
+Explanation 2
+We have two 's, two 's, and one .  is unique.
+```
+Solution
+========
+
+```python
+#!/bin/python3
+
+import math
+import os
+import random
+import re
+import sys
+
+# Complete the lonelyinteger function below.
+def lonelyinteger(a):
+    result = a[0]
+    for i in range(1, len(a)):
+        result ^= a[i]
+    return result
+
+if __name__ == '__main__':
+    fptr = open(os.environ['OUTPUT_PATH'], 'w')
+
+    n = int(input())
+
+    a = list(map(int, input().rstrip().split()))
+
+    result = lonelyinteger(a)
+
+    fptr.write(str(result) + '\n')
+
+    fptr.close()
+
+
+```