LeetCode in Kotlin

3833. Count Dominant Indices

Easy

You are given an integer array nums of length n.

An element at index i is called dominant if: nums[i] > average(nums[i + 1], nums[i + 2], ..., nums[n - 1])

Your task is to count the number of indices i that are dominant.

The average of a set of numbers is the value obtained by adding all the numbers together and dividing the sum by the total number of numbers.

Note: The rightmost element of any array is not dominant.

Example 1:

Input: nums = [5,4,3]

Output: 2

Explanation:

• At index i = 0, the value 5 is dominant as 5 > average(4, 3) = 3.5.
• At index i = 1, the value 4 is dominant over the subarray [3].
• Index i = 2 is not dominant as there are no elements to its right. Thus, the answer is 2.

Example 2:

Input: nums = [4,1,2]

Output: 1

Explanation:

• At index i = 0, the value 4 is dominant over the subarray [1, 2].
• At index i = 1, the value 1 is not dominant.
• Index i = 2 is not dominant as there are no elements to its right. Thus, the answer is 1.

Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100

Solution

class Solution {
    fun dominantIndices(a: IntArray): Int {
        val n = a.size
        var sum: Long = 0
        var cnt = 0
        for (i in n - 1 downTo 1) {
            sum += a[i].toLong()
            if (a[i - 1] > sum / (n - i)) {
                cnt++
            }
        }
        return cnt
    }
}

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