Is there a better way to make this function return true as expected

Question

I have a JS logic that I am trying to implement, and now pondering why it is not working: I have two arrays with at least one similar element in them, but implementing the logic returned false - but I was expecting the opposite result. Please take a look:

 const array_one = [
  "Apparel",
  "Footwear",
];

const array_two = [
  "Soap",
  "Footwear",
];


const checkArray = (arr1, arr2) => {
    if (arr1.length !== arr2.length) {
      return false;
    } else {
      for (let i = 0; i < arr1.length; i++) {
        if (arr1[i] !== arr2[i]) {
          return false;
        }
      }
    }
    return true;
  };

  console.log(checkArray(array_one, array_two)); //logs false to the console, was expecting 'true'.

what do you want your function to do? its not clear what you want — Daniel A. White, Feb 17 '23 at 19:58
Your code checks two arrays for ***equality*** and since they are different you get `false`. — PM 77-1, Feb 17 '23 at 20:02
I want the function to compare elements in each array and see if one exists in another. — Ptrades, Feb 17 '23 at 20:02
You are returning as soon as you find a mismatch. You should return true if/when you find a match, but otherwise return false. — mykaf, Feb 17 '23 at 20:13

XMehdi01 · Answer 1 · 2023-02-17T20:14:46.763

Using some method to check if any value from one array exists in another array:

const array_one = [ "Apparel", "Footwear", ]; const array_two = [ "Soap", "Footwear", ];
const checkArray = (arr1, arr2) => {
  if (arr1.length !== arr2.length) {
    return false;
  } else {
    return arr1.some(value => arr2.includes(value));
  }
};

console.log(checkArray(array_one, array_two)); //true

Using filter method:

const array_one = [ "Apparel", "Footwear", ]; const array_two = [ "Soap", "Footwear", ];
const checkArray = (arr1, arr2) => {
  if (arr1.length !== arr2.length) {
    return false;
  } else {
    return arr1.filter(value => arr2.includes(value)).length > 0;
  }
};

console.log(checkArray(array_one, array_two)); //true

Using the some method with the Set which is more performant since it has an average time complexity of O(n) for the has method, and does not require iterating through the entire array.

const array_one = [ "Apparel", "Footwear", ]; const array_two = [ "Soap", "Footwear", ];

const checkArray = (arr1, arr2) => {
  if (arr1.length !== arr2.length) {
    return false;
  } else {
    const set = new Set(arr2);
    return arr1.some(value => set.has(value));
  }
};

console.log(checkArray(array_one, array_two)); //true

Using the reduce method:

const array_one = [ "Apparel", "Footwear", ]; const array_two = [ "Soap", "Footwear", ];

const checkArray = (arr1, arr2) => {
  if (arr1.length !== arr2.length) {
    return false;
  } else {
    return arr1.reduce((accumulator, value) => accumulator || arr2.includes(value), false);
  }
};

console.log(checkArray(array_one, array_two)); //true

Note: methods 1. filter and 2. reduce are less performant , since they create new arrays, accumulate values, which can be memory-intensive and slow for larger arrays.

Thank you. Very much appreciated. I have soooo much to learn "Me Lord" :) — Ptrades, Feb 17 '23 at 20:05

score 0 · Answer 2 · answered Feb 17 '23 at 20:24

Just flip your logic around.

Make false the default assumption
Create a Set for each input Array for O(1) lookup
Check if the current item in the set exists in the other and immediately return true

const
  array_one   = [ "Apparel" , "Footwear" ],
  array_two   = [ "Soap"    , "Footwear" ],
  array_three = [ "Apple"   , "Orange"   ];

const checkArray = (arr1, arr2) => {
  if (arr1.length !== arr2.length) {
    return false; // lengths must match
  }
  const
    set1 = new Set(arr1),
    set2 = new Set(arr2);
  for (const [value] of set1.entries()) {
    if (set2.has(value)) {
      return true; // short-circuit with true
    }
  }
  return false; // default assumption
};

console.log(checkArray(array_one, array_two));   // true
console.log(checkArray(array_one, array_three)); // false

If you think Set is overkill, you can always use Array.prototype.includes:

const
  array_one   = [ "Apparel" , "Footwear" ],
  array_two   = [ "Soap"    , "Footwear" ],
  array_three = [ "Apple"   , "Orange"   ];

const checkArray = (arr1, arr2) => {
  if (arr1.length !== arr2.length) {
    return false; // lengths must match
  }
  for (let i = 0; i < arr1.length; i++) {
    if (arr2.includes(arr1[i])) {
      return true; // short-circuit with true
    }
  };
  return false; // default assumption
};

console.log(checkArray(array_one, array_two));   // true
console.log(checkArray(array_one, array_three)); // false

Is there a better way to make this function return true as expected

2 Answers2