What is the best way to verify an iterative if statement within a for loop?

I am currently working on checking an iterative if condition within a for loop. This involves generating some expressions for the if conditions iteratively, as demonstrated in the example below:

let union = [1,2,3,4,5,6,7,8,9]
let control = [2,4,6]
let result = []
for(let i=0; i<union.length; i++){
  if(union[i] % 2 == 0 && union[i] !== control[i]){
    result.push(union[i])
  }
}

In this example, union[i] !== control[i] is the conditional expression that needs to be validated/generated iteratively. To put it simply,

The result array should only include even numbers from the union array and not any element from the control array.

Therefore, the resulting array should be [8]

Answer №1

To populate the result array with only even numbers from the combined array and without including any elements from the comparison array.

let combination = [3,6,9,12,15,18]
let comparison = [6,12]
let output = []

combination.forEach(num => {
    if(num % 2 == 0 && !comparison.includes(num)) {
        output.push(num);
    }
});

The .includes function is used to verify if an item is present in the array.

An alternative approach could be using a simple filter method:

const output = combination.filter(num => num % 2 == 0 && !comparison.includes(num));

Answer №2

Consider this alternative approach:

const originalArray = [1,2,3,4,5,6,7,8,9]
const controlArray = [2,4,6]
let finalResult = []

const uniqueArray = originalArray.filter(function(value) {
  return !controlArray.includes(value);
});

for(let j=0; j<uniqueArray.length; j++){
  if(uniqueArray[j] % 2 === 0){
    finalResult.push(uniqueArray[j])
  }
}

Answer №3

To determine if the current union variable is included in control, you can modify the check to see if control.includes(union[i]). If this condition is met, it means that control contains the current union value. Similarly, if control includes the union value, then the union also contains the control variable. Use ! to negate the check and verify if this value is not included.

let union = [1,2,3,4,5,6,7,8,9]
let control = [2,4,6]
let result = []
for(let i=0; i<union.length; i++){
  if(union[i] % 2 == 0 && !control.includes(union[i])){
    result.push(union[i])
  }
}
console.log(result);

Answer №4

Implement Array#filter using a Set.

Your current code has a flaw where you are not checking each item in control against those in union, leading to inefficiencies.

To optimize this, I suggest utilizing a Set instead of repetitive array searches.

const 
  union = [1, 2, 3, 4, 5, 6, 7, 8, 9],
  control = [2, 4, 6],
  result = ((s) => union.filter((u) => !(u&1) && !s.has(u)))(new Set(control));
console.log(result);

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