CodingGyan

Summary Statistics in One Pass

medium Generics, Lambdas and Streams

Problem

Print five lines for an array: count=, sum=, min=, max= and average=. The average is the sum divided by the count, rounded down toward negative infinity (floor). Compute all values in ONE pass over the numbers with a single fold (reduce) that carries a small state, or with the library helper for statistics.

Constraints

  • 1 <= n <= 10^5
  • |value| <= 10^6

Examples

Input

3
4 8 15

Output

count=3
sum=27
min=4
max=15
average=9

Solutions

Best approachTime O(n)Space O(1)

Computing sum, min and max in three separate loops reads the data three times. One fold with a small state reads it once, which matters for big files or streams. Java's IntSummaryStatistics does exactly this. Use floor division for negatives: -5 / 2 is -3 with floor and -2 when it is cut toward zero.

Same code works in Java810172125Basic features only, so nothing changes between versions.
/*
 * Problem : Five statistics.
 * Input   : n, then n integers.
 * Output  : Five lines.
 * Approach: One fold.
 * Time    : O(n)
 * Space   : O(1)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static String[] solve(int[] a) {
        // Step 1: IntSummaryStatistics collects count, sum, min and max in one pass over the stream.
        java.util.IntSummaryStatistics stats = java.util.Arrays.stream(a).summaryStatistics();
        // Step 2: The library average is a double. For a whole-number floor we divide the sum ourselves.
        long average = Math.floorDiv(stats.getSum(), stats.getCount());
        return new String[] {
            "count=" + stats.getCount(),
            "sum=" + stats.getSum(),
            "min=" + stats.getMin(),
            "max=" + stats.getMax(),
            "average=" + average
        };
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int size = scanner.nextInt();
        int[] a = new int[size];
        for (int i = 0; i < size; i++) {
            a[i] = scanner.nextInt();
        }
        System.out.println(String.join("\n", solve(a)));
    }
}
Hints (1)
  1. Why is one pass better than three loops for a data stream?

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