LeetCode in Kotlin

2090. K Radius Subarray Averages

Medium

You are given a 0-indexed array nums of n integers, and an integer k.

The k-radius average for a subarray of nums centered at some index i with the radius k is the average of all elements in nums between the indices i - k and i + k (inclusive). If there are less than k elements before or after the index i, then the k-radius average is -1.

Build and return an array avgs of length n where avgs[i] is the k-radius average for the subarray centered at index i.

The average of x elements is the sum of the x elements divided by x, using integer division. The integer division truncates toward zero, which means losing its fractional part.

• For example, the average of four elements 2, 3, 1, and 5 is (2 + 3 + 1 + 5) / 4 = 11 / 4 = 2.75, which truncates to 2.

Example 1:

Input: nums = [7,4,3,9,1,8,5,2,6], k = 3

Output: [-1,-1,-1,5,4,4,-1,-1,-1]

Explanation:

• avg[0], avg[1], and avg[2] are -1 because there are less than k elements before each index.

• The sum of the subarray centered at index 3 with radius 3 is: 7 + 4 + 3 + 9 + 1 + 8 + 5 = 37.

  Using integer division, avg[3] = 37 / 7 = 5.

• For the subarray centered at index 4, avg[4] = (4 + 3 + 9 + 1 + 8 + 5 + 2) / 7 = 4.

• For the subarray centered at index 5, avg[5] = (3 + 9 + 1 + 8 + 5 + 2 + 6) / 7 = 4.

• avg[6], avg[7], and avg[8] are -1 because there are less than k elements after each index.

Example 2:

Input: nums = [100000], k = 0

Output: [100000]

Explanation:

• The sum of the subarray centered at index 0 with radius 0 is: 100000.

  avg[0] = 100000 / 1 = 100000.

Example 3:

Input: nums = [8], k = 100000

Output: [-1]

Explanation:

• avg[0] is -1 because there are less than k elements before and after index 0.

Constraints:

• n == nums.length
• 1 <= n <= 105
• 0 <= nums[i], k <= 105

Solution

class Solution {
    fun getAverages(nums: IntArray, k: Int): IntArray {
        // initialize result array with -1
        val res = IntArray(nums.size)
        res.fill(-1)
        if (nums.size <= k * 2) {
            // return if not enough elements
            return res
        }
        var sum: Long = 0
        val range = 2 * k + 1L
        // take sum of all elements from 0 to k*2 index
        for (i in 0..2 * k) {
            sum += nums[i].toLong()
        }
        // update first valid avg
        res[k] = (sum / range).toInt()
        // update other valid averages using sliding window
        for (i in k + 1 until nums.size - k) {
            sum = sum - nums[i - k - 1] + nums[i + k]
            res[i] = (sum / range).toInt()
        }
        return res
    }
}

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