/**
 * 100/100
 * @param A
 * @return
 */
fun solution(A: IntArray): Int {
    val N = A.size
    var count = 1
    var candidate = A[0]

    // 1. Find a candidate, note: if there is a leader it will for sure be candidate
    for (i in 1 until N) {
        if (count == 0) candidate = A[i]
        if (A[i] == candidate) {
            count++
        } else {
            count--
        }
    }

    // 2. Check if candidate is a leader?
    count = 0
    // Count how many times a candidate appears
    for (i in 0 until N) {
        if (candidate == A[i]) count++
    }
    // If it appears less than half there is no solution, return 0
    if (count < N / 2) return 0

    // 3. Count EquiLeaders
    var countLeft = 0
    var countRight = count
    var equiLeaders = 0
    for (i in 0 until N) {
        if (A[i] == candidate) {
            countLeft++
            countRight--
        }
        if (countLeft > (i + 1) / 2
            && countRight > (N - i - 1) / 2
        ) {
            equiLeaders++
        }
    }
    return equiLeaders
}

/**
 * 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:
 *
 * class Solution { public int solution(int[] 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].
 */
/**
 * 100/100
 * @param A
 * @return
 */
fun solution(A: IntArray): Int {
    val N = A.size
    var count = 1
    var candidate = A[0]

    // 1. Find a candidate, note: if there is a leader it will for sure be candidate
    for (i in 1 until N) {
        if (count == 0) candidate = A[i]
        if (A[i] == candidate) {
            count++
        } else {
            count--
        }
    }

    // 2. Check if candidate is a leader?
    count = 0
    // Count how many times a candidate appears
    for (i in 0 until N) {
        if (candidate == A[i]) count++
    }
    // If it appears less than half there is no solution, return 0
    if (count < N / 2) return 0

    // 3. Count EquiLeaders
    var countLeft = 0
    var countRight = count
    var equiLeaders = 0
    for (i in 0 until N) {
        if (A[i] == candidate) {
            countLeft++
            countRight--
        }
        if (countLeft > (i + 1) / 2
            && countRight > (N - i - 1) / 2
        ) {
            equiLeaders++
        }
    }
    return equiLeaders
}

/**
 * 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:
 *
 * class Solution { public int solution(int[] 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].
 */