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Java-based LeetCode algorithm problem solutions, regularly updated
package g1001_1100.s1093_statistics_from_a_large_sample;
// #Medium #Math #Two_Pointers #Probability_and_Statistics
// #2022_02_21_Time_1_ms_(86.00%)_Space_43.8_MB_(22.00%)
/**
* 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**.
**/
public class Solution {
public double[] sampleStats(int[] count) {
int l = 0;
int r = 255;
int nl = 0;
int nr = 0;
int mn = 256;
int mx = -1;
int mid1 = 0;
int mid2 = 0;
int mode = 0;
double avg = 0;
double mid;
while (l <= r) {
while (count[l] == 0) {
l++;
}
while (count[r] == 0) {
r--;
}
if (nl < nr) {
avg += (double) count[l] * 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 += (double) count[r] * 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;
// Find median
if (nl < nr) {
mid = mid2;
} else if (nl > nr) {
mid = mid1;
} else {
mid = (double) (mid1 + mid2) / 2;
}
return new double[] {mn, mx, avg, mid, mode};
}
}
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