Codility Lesson 8 EquiLeader Python

Question

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].

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Answer

def solution(A):
    count = 0

    right_dict = {}
    right_len = len(A)
    for a in A:
        if a in right_dict:
            right_dict[a] += 1
        else:
            right_dict[a] = 1
    
    left_leader = 0
    left_leader_count = 0
    left_dict = {}
    left_len = 0

    for a in A:
        right_dict[a] -= 1
        right_len -= 1

        if a in left_dict:
            left_dict[a] += 1
        else:
            left_dict[a] = 1
        left_len += 1
        
        if left_dict[a] > left_leader_count:
            left_leader = a
            left_leader_count = left_dict[a]
        
        if left_leader_count > left_len/2 and right_dict[left_leader] > right_len/2:
            count += 1

    return count