How can we check for an iterative if condition inside a for loop?

Question

I'm trying to check for an if condition iteratively inside a for loop. That means some expressions of if conditions need to be generated iteratively as shown in the below example

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 the above example union[i] !== control[i] is the conditional expression that need to be validated/generated iteratively. In words we can state the problem as

result array should contain only even numbers from the union array and it should not contain any element from the control array

So the result array should be [8]

To start with, control[i] will cause problems, since its length is different to the length of the union array, i.e what value were you expecting from control[6] ? Do a test and see the result — Spangle, Apr 22 '21 at 06:12
Yea that is an erroneous code and it will not give us the expected answer — Lalas M, Apr 22 '21 at 06:14

score 1 · Accepted Answer · answered Apr 22 '21 at 06:13

1

result array should contain only even numbers from the union array and it should not contain any element from the control array

let union = [1,2,3,4,5,6,7,8,9]
let control = [2,4,6]
let result = []

union.forEach(item => {
    if(item % 2 == 0 && !control.includes(item)) {
        result.push(item);
    }
});

The .includes method checks if the item is in the array.

or just a simple filter

const result = union.filter(item => item % 2 == 0 && !control.includes(item));

answered Apr 22 '21 at 06:13

Michael Mano

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Searching in an array again and again is not an efficient solution, try using a `Set`. – Som Shekhar Mukherjee Apr 22 '21 at 06:23
@SomShekharMukherjee I agree for a massive array, But for this solution its fine. i think hes just doing some coda katas. – Michael Mano Apr 22 '21 at 06:26

Som Shekhar Mukherjee · Answer 2 · 2021-04-22T06:18:46.567

Use `Array#filter` with a `Set`.

The problem with your code is for every item in union you need to check every item in control, whether it exists or not, which you are not doing.

And I would recommend using a Set instead of searching in an array repeatedly.

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);

score 0 · Answer 3 · answered Apr 22 '21 at 06:14

1 more option. remove duplicate item between 2 arrays then just check for even numbers.

let union = [1,2,3,4,5,6,7,8,9]
let control = [2,4,6]
let result = []

union = union.filter(function(val) {
  return control.indexOf(val) == -1;
});

for(let i=0; i<union.length; i++){
  if(union[i] % 2 == 0){
    result.push(union[i])
  }
}

score 0 · Answer 4 · answered Apr 22 '21 at 06:17

You can modify the check and see if control.includes(union[i]). This will be true when control includes the current union variable. This is also true for the other way round, meaning if control includes the union variable, the union variable also contains the control variable. Negate the check with ! to see if it doesn't include this value.

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);

How can we check for an iterative if condition inside a for loop?

4 Answers4

Use Array#filter with a Set.

Use `Array#filter` with a `Set`.