코딜리티 EquiLeader

python 의 collections 에 most_common 과 부분합을 연관지어 풀면 쉽게 풀리는 문제이다.

Task description

A non-empty array A consisting of N integers is given.

The leader of this array is the value that occurs in more than half of the elements of A.

An equi leader is an index S such that 0 ≤ S < N − 1 and two sequences A[0], A[1], ..., A[S] and A[S + 1], A[S + 2], ..., A[N − 1] have leaders of the same value.

For example, given array A such that:

A[0] = 4 A[1] = 3 A[2] = 4 A[3] = 4 A[4] = 4 A[5] = 2

we can find two equi leaders:

• 0, because sequences: (4) and (3, 4, 4, 4, 2) have the same leader, whose value is 4.
• 2, because sequences: (4, 3, 4) and (4, 4, 2) have the same leader, whose value is 4.

The goal is to count the number of equi leaders.

Write a function:

def solution(A)

that, given a non-empty array A consisting of N integers, returns the number of equi leaders.

For example, given:

A[0] = 4 A[1] = 3 A[2] = 4 A[3] = 4 A[4] = 4 A[5] = 2

the function should return 2, as explained above.

Write an efficient algorithm for the following assumptions:

• N is an integer within the range [1..100,000];
• each element of array A is an integer within the range [−1,000,000,000..1,000,000,000]

import collections
def solution(A):
    answer=0
    # write your code in Python 3.6
    ans=collections.Counter(A).most_common(1)[0]
    if ans[1]<=len(A)//2:
        return 0
    comons=[0]*len(A)
    for i in range(len(A)):
        if i==0 and  A[i] == ans[0]:
            comons[0]=1
        elif A[i] == ans[0]:
            comons[i]=comons[i-1]+1
        else:
            comons[i]=comons[i-1]
    for i in range(len(A)-1):
        before=comons[i]
        after=comons[len(A)-1]-before
        afterLen= len(A)-i-1
        beforeLen=i+1
        if before<= beforeLen//2 or after<=afterLen//2:
            continue
        else:
            answer+=1
    
    return answer