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Daily Challenge #16 - Number of People on the Bus

dev.to staff on July 16, 2019

Today's challenge comes from aryan-firouzian on Codewars. You are to figure out how many people are on the bus given the following scenario: The...

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Tobias SN • Jul 16 '19 • Edited

JS:

result = arr.reduce((a, [wentOn, wentOff]) => {
    return a + wentOn - wentOff;
});

Stephen Ramthun • Jul 17 '19

Nice one, although you're missing the second argument to reduce which specifies the initial value for a.

Tobias SN • Jul 17 '19

According to the MDN Web Docs, that argument is optional.

Florin Pop • Jul 17 '19

Right :)

Corey Alexander • Jul 17 '19

Rust Solution

This also checks to make sure that more people don't get off the stop than were on + got on that stop. This assumes no passengers were on the bus initially.

pub fn riders_left(bus_stops: &[(u32, u32)]) -> Result<u32, &str> {
    bus_stops.iter().fold(Ok(0), |sum, (on, off)| {
        let new_sum = sum? + on;

        if off > &new_sum {
            Err("Somehow more people left the bus than got on")
        } else {
            Ok(new_sum - off)
        }
    })
}

#[cfg(test)]
mod tests {
    use crate::*;

    #[test]
    fn it_returns_zero_for_an_empty_array() {
        assert_eq!(riders_left(&[]), Ok(0))
    }

    #[test]
    fn it_for_a_semi_complicated_array_example() {
        assert_eq!(riders_left(&[(10, 0), (5, 9), (6, 6), (2, 6)]), Ok(2));
    }

    #[test]
    fn it_for_a_semi_complicated_vec_example() {
        assert_eq!(riders_left(&vec![(10, 0), (5, 9), (6, 6), (2, 6)]), Ok(2));
    }

    #[test]
    fn it_for_a_broken_example() {
        assert_eq!(
            riders_left(&vec![(10, 0), (5, 9), (6, 6), (2, 12)]),
            Err("Somehow more people left the bus than got on")
        );
    }
}

Payton Bice • Jul 16 '19 • Edited

PHP 7.4:

$bus = fn(array $stops, int $riders = 0): int => array_reduce($stops, fn(int $count, array $data): int => ($count + $data[0]) - $data[1], $riders);

var_dump($bus([[1, 0], [2, 1], [5, 3], [4, 2]])); //=> int(6)

E. Choroba • Jul 16 '19 • Edited

#! /usr/bin/perl
use warnings;
use strict;
use feature qw{ say };

use List::Util qw{ sum };

sub bus { sum(map $_->[0] - $_->[1], @_) }

use Test::More tests => 1;
my @people = ([10, 0], [5, 3], [12, 1], [1, 3], [0, 3], [1, 7]);
is bus(@people), 12;

BTW, it seems today's post didn't make it into the series.
Update: It's been fixed.

Kasper Meyer • Jul 16 '19 • Edited

Ruby 2.6

require "minitest/autorun"

class BusStopSimulator
  def self.call bus_stops
    bus_stops.map { |stop| stop.reduce(:-) }.sum
  end
end

class BusStopSimulatorTest < MiniTest::Test
  def test_four_bus_stops
    assert_equal 7, BusStopSimulator.call([[8,4],[1,0],[2,5],[6,1]])
  end
end

Oleksii Filonenko • Jul 17 '19

Elixir:

defmodule Bus do
  @spec passengers([{non_neg_integer, non_neg_integer}]) :: non_neg_integer
  def passengers(list) do
    list
    |> Enum.map(fn {went_on, went_off} -> went_on - went_off end)
    |> Enum.sum()
  end
end

Alvaro Montoro • Jul 21 '19 • Edited

CSS (kind of):

For the list of of integer arrays, we can use an actual HTML list with <li> that contain CSS variables with the number of people going up (--up) and down (--down) the bus. Then using CSS counters we can keep a count of the people who are on the bus at each moment.

Live demo on CodePen.

Yozen Hernandez • Jul 17 '19

Didn't see anyone else try it in C, so here's my solution in C:

#include "stdio.h"
#include "stdlib.h"

unsigned int bus( unsigned int ( *data )[2], size_t n ) {
    unsigned int sum = 0;

    for ( size_t i = 0; i < n; ++i ) {
        sum += ( data[i][0] - data[i][1] );
    }

    return sum;
}

int main( int argc, char const *argv[] ) {
    // Read in data, somehow
    printf( "Passengers remaining after last stop: %u\n",
      bus( ( unsigned int[][2] ){{1, 0}, {2, 1}, {5, 3}, {4, 2}}, 4 ) );
    // 6
    return 0;
}

Sascha • Jul 16 '19

Python:

bus_array = [(10,0), (2,1), (1,11)]
passengers_left = sum([entered - left for entered, left in bus_array])

Ahmed Abumostafa • Jul 17 '19

const stops = [
 [10, 0],
 [5, 3],
 [4, 4],
 [3, 8],
 [0, 2],
]

const inBus = stops.reduce(
(n, [gotIn, gotOut]) => n + gotIn - gotOut,
0
)

Yozen Hernandez • Jul 17 '19

# Read in data, somehow. Likely into a data.frame
bus_data <- data.frame(on = c(1, 2, 5, 4), off = c(0, 1, 3, 2))
sum(bus_data$on - bus_data$off)
# [1] 6

peter279k • Jun 17 '20

Here is the simple solution with Python:

def number(bus_stops):
    people = 0
    for items in bus_stops:
        people += items[0]
        people -= items[1]

    return people

Anmol Gautam • Jul 17 '19

Haskell

passengers :: [(Int, Int)] -> Int
passengers xs = sum $ map (\x -> fst x - snd x) xs

main = print $ passengers [(1, 0), (2, 1), (5, 3), (4, 2)]

maxmattone • Jul 18 '19

Smalltalk --- because it seems nobody did it yet:

result := arr sum: [ :aStop | aStop first - aStop second ]

Craig McIlwrath • Aug 20 '19

Haskell:

type BusStop = (Int, Int)

getChange :: BusStop -> Int
getChange stop = fst stop - snd stop

peopleOnBus :: [BusStop] -> Int
peopleOnBus = sum . map getChange

K.V.Harish • Oct 6 '19

My solution in js

const totalPeopleInBus = (arr) => arr.reduce((total, [enter, exit]) => total + enter - exit, 0);