LeetCode in Kotlin

1093. Statistics from a Large Sample

Medium

You are given a large sample of integers in the range [0, 255]. Since the sample is so large, it is represented by an array count where count[k] is the number of times that k appears in the sample.

Calculate the following statistics:

• minimum: The minimum element in the sample.
• maximum: The maximum element in the sample.
• mean: The average of the sample, calculated as the total sum of all elements divided by the total number of elements.
• median:
  • If the sample has an odd number of elements, then the median is the middle element once the sample is sorted.
  • If the sample has an even number of elements, then the median is the average of the two middle elements once the sample is sorted.
• mode: The number that appears the most in the sample. It is guaranteed to be unique.

Return the statistics of the sample as an array of floating-point numbers [minimum, maximum, mean, median, mode]. Answers within 10-5 of the actual answer will be accepted.

Example 1:

Input: count = [0,1,3,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

Output: [1.00000,3.00000,2.37500,2.50000,3.00000]

Explanation: The sample represented by count is [1,2,2,2,3,3,3,3].

The minimum and maximum are 1 and 3 respectively.

The mean is (1+2+2+2+3+3+3+3) / 8 = 19 / 8 = 2.375.

Since the size of the sample is even, the median is the average of the two middle elements 2 and 3, which is 2.5.

The mode is 3 as it appears the most in the sample.

Example 2:

Input: count = [0,4,3,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

Output: [1.00000,4.00000,2.18182,2.00000,1.00000]

Explanation: The sample represented by count is [1,1,1,1,2,2,2,3,3,4,4].

The minimum and maximum are 1 and 4 respectively.

The mean is (1+1+1+1+2+2+2+3+3+4+4) / 11 = 24 / 11 = 2.18181818… (for display purposes, the output shows the rounded number 2.18182).

Since the size of the sample is odd, the median is the middle element 2.

The mode is 1 as it appears the most in the sample.

Constraints:

• count.length == 256
• 0 <= count[i] <= 109
• 1 <= sum(count) <= 109
• The mode of the sample that count represents is unique.

Solution

class Solution {
    fun sampleStats(count: IntArray): DoubleArray {
        var l = 0
        var r = 255
        var nl = 0
        var nr = 0
        var mn = 256
        var mx = -1
        var mid1 = 0
        var mid2 = 0
        var mode = 0
        var avg = 0.0
        while (l <= r) {
            while (count[l] == 0) {
                l++
            }
            while (count[r] == 0) {
                r--
            }
            if (nl < nr) {
                avg += count[l].toDouble() * l
                nl += count[l]
                if (count[l] > count[mode]) {
                    mode = l
                }
                mx = Math.max(mx, l)
                mn = Math.min(mn, l)
                mid1 = l
                l++
            } else {
                avg += count[r].toDouble() * r
                nr += count[r]
                if (count[r] > count[mode]) {
                    mode = r
                }
                mx = Math.max(mx, r)
                mn = Math.min(mn, r)
                mid2 = r
                r--
            }
        }
        avg /= (nl + nr).toDouble()
        // Find median
        val mid: Double = if (nl < nr) {
            mid2.toDouble()
        } else if (nl > nr) {
            mid1.toDouble()
        } else {
            (mid1 + mid2).toDouble() / 2
        }
        return doubleArrayOf(mn.toDouble(), mx.toDouble(), avg, mid, mode.toDouble())
    }
}

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