Running Totals (Scan)

Problem

Print the running totals of an array: the first number, then the sum of the first two numbers, then the sum of the first three, and so on, separated by spaces. Write it in a functional way with a fold (reduce) that carries the total so far.

Constraints

  • 1 <= n <= 5000
  • |value| <= 10^6

Examples

Input

4
1 2 3 4

Output

1 3 6 10

Solutions

Best approachTime O(n)Space O(n)

A fold (reduce) gives one final value. A scan gives every step. Python has itertools.accumulate for this. The spread-based JavaScript version copies the array each step, so it is O(n^2); the Java and Python versions are O(n).

Same code works in Java810172125Basic features only, so nothing changes between versions.
/*
 * Problem : Running totals.
 * Input   : n, then n integers.
 * Output  : n totals.
 * Approach: Fold with state.
 * Time    : O(n)
 * Space   : O(n)
 * Java    : 8 or newer.
 */
import java.util.Scanner;
import java.util.Arrays;
import java.util.stream.Collectors;
import java.util.*;
public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static int[] solve(int[] a) {
        // Step 1: The accumulator is a one-element array that holds the total so far. A lambda cannot change a normal variable.
        long[] total = {0};
        // Step 2: The stream turns each value into the new running total. mapToInt gives an IntStream again.
        return java.util.Arrays.stream(a).map(value -> {
            total[0] += value;
            return (int) total[0];
        }).toArray();
    }

    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(Arrays.stream(solve(a)).mapToObj(String::valueOf).collect(Collectors.joining(" ")));
    }
}
Hints (1)
  1. What is the difference between reduce and scan?