CODEWITHSUNDEEP
javascript
leo
leo

Reputation: 121

Check if two following values exist in an array

I have a nested array like this:

[ [5, 10, 5, 15, 5, 10],   [10, 15, 50, 200] ]

So basically I have index 0 that also has an array with the values [5,10,5,15,5,10] and index 1 with an array with values [10,15,50,350].

let array = [[],[]];
let x = 0;
let y = 0;

for (let i = 0; i < 10; i++) {
 x = ... generate random number ...
 y = ... generate random number ...

 while (array[0].includes(x,y) {
  x = ... generate new random number ...
  y = ... generate new random number ...
 }

 array[0].push(x,y);
}

Is there a way for me to find if for example index 0 in the array already contains the two generated values in the same order they are being pushed?

For example, I'm adding value 5,15 to the array. And the array already contains index 0 with 5 and index 1 with, 15 in that order. So I want to generate a new number if the numbers already exist in that order.

I've tried looking for solutions, but haven't found anything that helps me with what I wanna do.

Upvotes: 2

Views: 122

Answers (3)

Jostein S
Jostein S

Reputation: 466

I recommend making a function that checks for a specific sequence of numbers, this way if you need it to find a X number sequence you're ready.

const max = 100;
let array = [[], []];
let x = 0, 
    y = 0;

for (let i = 0; i < 1000; i++) {
    x = Math.floor(Math.random() * max);
    y = Math.floor(Math.random() * max);

    while (containsSequence(array[0], [x, y])) {
        console.log(`Sequence ${x},${y} already exists.`);
        x = Math.floor(Math.random() * max);
        y = Math.floor(Math.random() * max);
    }

    array[0].push(x, y);
}

/** 
 * @param {Array<Number>} arr 
 * @param {Array<Number>} sequence */
function containsSequence(arr, sequence) {
    if(arr.length === 0) { return false; }
    const first = sequence[0];
    let index = arr.indexOf(first, 0);

    while(index > -1) {
        if(sequence.every((v, i) => arr[index + i] === v)) {
            return true;
        }

        index = arr.indexOf(first, index);
        if(index == -1) { return false; }
        index++;
    }

    return false;
}

Upvotes: 1

codingStarter
codingStarter

Reputation: 364

You are putting x and y one after each other in the list, if you connect them to an object or an array, you can iterate over the outer array more easily:

for (let i = 0; i < 10; i++) {
 x = ... generate random number ...
 y = ... generate random number ...

 while (array[0].filter(i=>i.x == x && i.y == y).length) {
  x = ... generate new random number ...
  y = ... generate new random number ...
 }

 array[0].push({x,y});
}

it will also make it more simle. Of course,only if you can make this change to the array structure

Upvotes: 1

Stitt
Stitt

Reputation: 494

Assuming that this is just a general structural question as mentioned in the comments, you can adapt your code as below.

for (let i = 0; i < 10; i++) {
  int index = 1;
  while (index < array[0].length){
    if ((array[0][i][index-1].includes(x) && array[0][i][index].includes(y)) {
      x = ... generate new random number ...
      y = ... generate new random number ...
      index = 0;
    }
    else{
      index++;
    }
  }
  array[0].push(x,y);
 }
}

Upvotes: 0

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