Codility Lesson2 - OddOccurrencesInArray

Codility Lesson2 - OddOccurrencesInArray

https://app.codility.com/demo/results/trainingCGBH6B-Q9C/

Test results - Codility

 

문제

A non-empty 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:

def 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.

Write an efficient algorithm for the following assumptions:

• 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.

 

정답

def solution(A):
    for i in range(len(A)):
        found=A.count(A[i])

        if found >1:
            continue
        else:
            result=A[i]
       
    
    return result

 

결과

-> for 문을 돌리니 timeout

---

정답-Retry

import collections

def solution(A):
    dict={}
    dict=collections.Counter(A)
    for key,value in dict.items():
        if value==1:
            result=key
            return result

 

결과-Retry

-> pair는 꼭 1개가 아니라 홀수면 해당됨

---

정답-Retry2

import collections

def solution(A):
    dict={}
    dict=collections.Counter(A)
    for key,value in dict.items():
        if value%2!=0:
            result=key

    return result

 

결과-Retry2