Day 0: Mean, Median, and Mode HackerRank Solution

Day 0: Mean, Median, and Mode HackerRank Solution
Source : https://www.hackerrank.com/challenges/s10-basic-statistics

Objective
In this challenge, we practice calculating the mean, median, and mode. Check out the Tutorial tab for learning materials and an instructional video!

Task
Given an array, , of integers, calculate and print the respective mean, median, and mode on separate lines. If your array contains more than one modal value, choose the numerically smallest one.

Note: Other than the modal value (which will always be an integer), your answers should be in decimal form, rounded to a scale of decimal place (i.e., , format).

Input Format

The first line contains an integer, , denoting the number of elements in the array.
The second line contains space-separated integers describing the array's elements.

Constraints

• , where is the element of the array.

Output Format

Print lines of output in the following order:

1. Print the mean on a new line, to a scale of decimal place (i.e., , ).
2. Print the median on a new line, to a scale of decimal place (i.e., , ).
3. Print the mode on a new line; if more than one such value exists, print the numerically smallest one.

Sample Input

1064630 11735 14216 99233 14470 4978 73429 38120 51135 67060

Sample Output

43900.644627.54978

Explanation

Mean:
We sum all elements in the array, divide the sum by , and print our result on a new line.

Median:
To calculate the median, we need the elements of the array to be sorted in either non-increasing or non-decreasing order. The sorted array . We then average the two middle elements:

and print our result on a new line.

Mode:
We can find the number of occurrences of all the elements in the array:

 4978 : 111735 : 114216 : 114470 : 138120 : 151135 : 164630 : 167060 : 173429 : 199233 : 1

Every number occurs once, making the maximum number of occurrences for any number in . Because we have multiple values to choose from, we want to select the smallest one, , and print it on a new line.

Source : https://www.hackerrank.com/challenges/s10-basic-statistics

Solution

	// github.com/RodneyShag
	import java.util.Scanner;
	import java.util.Arrays;
	import java.util.HashMap;
	// Time Complexity: O(n log n) due to sorting
	public class Solution {
	public static void main(String[] args) {
	/* Save input */
	Scanner scan = new Scanner(System.in);
	int size = scan.nextInt();
	int [] array = new int[size];
	for (int i = 0; i < size; i++) {
	array[i] = scan.nextInt();
	}
	scan.close();
	/* Sort array: O(n log n) runtime */
	Arrays.sort(array);
	/* Calculate Mean */
	int total = 0;
	for (int num : array) {
	total += num;
	}
	double mean = (double) total / size;
	/* Calculate Median */
	double median;
	if (size % 2 == 0) {
	median = (array[size / 2 - 1] + array[size / 2]) / 2.0;
	} else {
	median = array[size / 2];
	}
	/* Calculate Mode - if there's a tie, choose the smaller number */
	HashMap<Integer, Integer> map = new HashMap();
	int maxOccurrences = 0;
	int mode = Integer.MAX_VALUE;
	for (int num : array) {
	map.merge(num, 1, Integer::sum);
	int occurrences = map.get(num);
	if (occurrences > maxOccurrences || (occurrences == maxOccurrences && num < mode)) {
	maxOccurrences = occurrences;
	mode = num;
	}
	}
	/* Print results */
	System.out.println(mean);
	System.out.println(median);
	System.out.println(mode);
	}
	}