Simple "opposite" check for string values in Javascript

Question

EDIT: So, it seems like there's not really a "better" way of doing this, but a lot of folks are pointing out that a hashmap approach might be more readable/composable.

Given the following function:

const isOpposites = (a, b) => {
  return (
    (a === "SOUTH" && b === "NORTH") ||
    (b === "SOUTH" && a === "NORTH") ||
    (a === "EAST" && b === "WEST") ||
    (a === "WEST" && b === "EAST")
  )
};

Is there a simpler way (one-liner, maybe) of checking whether a and b are opposites?

*There are only four possible values for a and b, they are: "NORTH", "SOUTH","EAST" and "WEST"

User81646 · Accepted Answer · 2021-04-07 02:30:24Z

1

Not a one liner, but probably more readable:

const opposites = {
    "NORTH": "SOUTH",
    "SOUTH": "NORTH",
    "EAST": "WEST",
    "WEST": "EAST"
}

const isOpposites = (a, b) => (opposites[a] === b)

answered Apr 7, 2021 at 2:30

User81646

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CertainPerformance · Accepted Answer · 2021-04-07 02:28:42Z

1

One option would be to have an object mapping the directions to a number, and have the opposite directions have the same number:

const directions = {
  SOUTH: 1,
  NORTH: 1,
  EAST: 2,
  WEST: 2,
};
const isOpposites = (a, b) => {
  return a !== b && directions[a] === directions[b];
};

answered Apr 7, 2021 at 2:28

CertainPerformance

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1 Comment

Kaiido Over a year ago

Just a note, it doesn't need to be a number, if one needs it might be a good idea to use this dict for something else too, for instance "y" and "x" or or "V" and "H" could be used to determine which axis would be used, other values might also have other meanings.

Pushpendu Ghosh · Accepted Answer · 2021-04-07 02:44:42Z

0

For this specific problem , there are many other ways to solve it but it's the best solution which you have provided. Clearly the space and time complexity is O(n). It's perfect. You don't need any change;

answered Apr 7, 2021 at 2:44

Pushpendu Ghosh

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Comments

tadman · Accepted Answer · 2021-04-07 02:30:23Z

-1

Could always use modulo:

const CARDINALS = [
  "NORTH",
  "EAST",
  "SOUTH",
  "WEST"
];

function isOpposites(a, b) {
  return CARDINALS.indexOf(a) == (CARDINALS.indexOf(b) + 2) % 4;
}

Of course a simple look-up table like this could also work:

const OPPOSITES = {
  NORTH: 'SOUTH',
  EAST: 'WEST',
  SOUTH: 'NORTH',
  WEST: 'EAST'
};

function isOpposites(a, b) {
  return OPPOSITES[a] === b;
}

answered Apr 7, 2021 at 2:30

tadman

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