Codility 2.1 Arrays OddOccurrencesInArray
Problem
A non-empty zero-indexed array A consisting of N integers is given. The array contains an odd number of elements, and each element of the array can be paired with another element that has the same value, except for one element that is left unpaired.
For example, in array A such that:
A[0] = 9 A[1] = 3 A[2] = 9 A[3] = 3 A[4] = 9 A[5] = 7 A[6] = 9
- the elements at indexes 0 and 2 have value 9,
- the elements at indexes 1 and 3 have value 3,
- the elements at indexes 4 and 6 have value 9,
- the element at index 5 has value 7 and is unpaired.
Write a function:
int solution(A);
that, given an array A consisting of N integers fulfilling the above conditions, returns the value of the unpaired element.
For example, given array A such that:
A[0] = 9 A[1] = 3 A[2] = 9 A[3] = 3 A[4] = 9 A[5] = 7 A[6] = 9
the function should return 7, as explained in the example above.
Assume that:
- N is an odd integer within the range [1..1,000,000];
- each element of array A is an integer within the range [1..1,000,000,000];
- all but one of the values in A occur an even number of times.
Complexity:
- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).
Solving Process
The array will be iterated by the unit of two elements. After sorting an array, Checking the paired element and if it is not paired, then the element’s value is the element the method returns.
It is like sorting rocks that have several height levels and spotting the only unpaired rock.
Code Solution(mine)
def solution(A):
if len(A) == 1:
return A[0]
A = sorted(A)
for i in range(0 , len (A) , 2):
if i+1 == len(A):
return A[i]
if A[i] != A[i+1]:
return A[i]Big-O Calculation
def solution(A):
if len(A) == 1:
return A[0]
A = sorted(A) # O(n*log(N) or N)
for i in range(0 , len (A) , 2): # O(N)
if i+1 == len(A):
return A[i]
if A[i] != A[i+1]:
return A[i]
# O(N*log(N) or O(N))Other People’s code solution
def solution(A): result = 0 for number in A: result ^= number return result
#O(N)
Learning Points
- XOR bitwise operator can detect overlapping values.
- https://codesays.com/2015/solution-to-odd-occurrences-in-array-by-codility/ Solution link
