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VistasCompare two arrays of objects in Javascript
I have a list of co-ordinates of squares of a grid:
Each square is defined by a class:
class Square {
constructor(x, y) {
this.x = x;
this.y = y;
}
}
This is how the app works: First there is a list of randomly generated squares shown, and then the user has to click the squares from memory. (Not necessarily in order)
An example of how the question/answer array would be: (just an example, they are actually randomly generated)
const arr = [new Square(1, 1), new Square(2, 1)]
now, whenever the user clicks on a box, it goes into another array:
var selectedBlocks = [new Square(2, 1), new Square(1, 1)]
In this case, since the squares selected are equal, the function should return true.
What I've tried - I can't manage to get it without a double-for loop O(n^2). Is it possible to optimise for atleast O(n)?
3 Respuestas
Responde la pregunta0
You can use Array.some to find is item already selected or not
class Square {
constructor(x, y) {
this.x = x;
this.y = y;
}
}
const arr = [new Square(1, 1), new Square(2, 1)];
const selectedBlocks = [new Square(2, 1), new Square(1, 1)];
const isSelected = (block) => {
return selectedBlocks.some(
(item) => item.x === block.x && item.y === block.y
);
};
console.log(isSelected(new Square(2, 1)));
console.log(isSelected(new Square(2, 2)));
console.log(isSelected(new Square(1, 1)));0
I forgot that there were few built in methods and i wrote this i guess (nlogn) Time complexity code
class Square {
constructor(x, y) {
this.x = x;
this.y = y;
}
pri(){
return this.x + this.y;
}
}
let inp = [new Square(2,1), new Square(2,3), new Square(1,3)]
let opt = [new Square(1,3), new Square(2,1), new Square(2,3)]
inp.sort((a, b) => {
return a.pri() - b.pri();
});
opt.sort((a, b) => {
return a.pri() - b.pri();
});
// assuminng they are of same len
let flag = true
for(i=0; i<inp.length ; i++){
if(inp[i].x != opt[i].x || inp[i].y != opt[i].y){
flag = false
break
}
}
console.log(flag)