Problem:- Consider an array of integers, , where every element in occurs exactly twice except for one unique element.Find and print the unique element.
Input :- 4 9 95 93 57 4 57 93 9 #Input array with 95 as unique element.
Output:- 95
Background of XOR operator ( ^ ) :- Bit-wise operator which results zero, if XOR operator is applied between same bits otherwise gives 1.
We can use XOR operation to reset all bit for two numbers , if they are same. i.e : 3 ^ 3 = 0
Sample output:-
>>>
Enter size of array (Input)
9
Enter elements separated by space
4 9 95 93 57 4 57 93 9
Unique element is
95
Input :- 4 9 95 93 57 4 57 93 9 #Input array with 95 as unique element.
Output:- 95
Background of XOR operator ( ^ ) :- Bit-wise operator which results zero, if XOR operator is applied between same bits otherwise gives 1.
XOR Operator operation |
Sample program to print unique element using XOR :-
#!/bin/python def lonely_integer(a): storage = a[0] for i in range(1, a.__len__()): storage = storage ^ a[i] #print (storage) return storage print "Enter size of array (Input) " n = int(raw_input().strip()) print "Enter elements separated by space " a = map(int,raw_input().strip().split(' ')) print "Unique element is " print lonely_integer(a)
Sample output:-
>>>
Enter size of array (Input)
9
Enter elements separated by space
4 9 95 93 57 4 57 93 9
Unique element is
95
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