EquiLeader.md

Task description A non-empty zero-indexed 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:

function solution(A);

that, given a non-empty zero-indexed 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.

Assume that:

• 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].

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments). Elements of input arrays can be modified.

function solution(A) {
    var pos = 0;
    var count = 0;

    for (var i = 0; i < A.length; i++) {
        if (A[pos] == A[i]) {
            count++;
        } else {
            count--;
            if (count == 0) {
                pos = i;
                count++;
            }
        }
    }

    var ret = 0;
    var cand = A[pos];

    var E = [];
    var N = [];

    var ec = 0;
    var nc = 0;
    for (var i = 0; i < A.length; i++) {
        if (A[i] == cand) {
            ec++;
        } else {
            nc++;
        }
        E[i] = ec;
        N[i] = nc;
    }

    for (var i = 0; i < A.length; i++) {
        if (E[i] > N[i] && ((nc - N[i]) < (ec - E[i]))) {
            ret++;
        }
    }

    return ret;
}