Getting minimum index while ignoring element value in array

Question

Say we have this array:

arr = [[3, 3], [3, 4], [3, 5], [3, 6]]

And we have this conditional:

arr = [[3, 3], [3, 4], [3, 5], [3, 6]]

for(let i = 0; i < arr.length; i++){
    if (arr[i][0] === 3 && arr[i][1] === 4 || arr[i][0] === 3 && arr[i][1] === 5){
        console.log(i)
    }
}

There respective indexes would be 1 and 2, which would display in the console.log() in sequential order.

How would I grab the minimum index (1 in this case), specifically, from that? I cannot store the numbers in an array due to my use case, so I need to establish the minimum within the loop. And you can't really set a min number because you won't know which element index will be the smallest beforehand.

Is this possible or am I overthinking it hard?

Sounds like a good match for findIndex. It returns the index once the first result is found. — Anil
– Anil, Commented Nov 12, 2021 at 18:47
i = arr.findIndex(a => a[0] === 3 && a[1] === 4 || a[0] === 3 && a[1] === 5) — Anil
– Anil, Commented Nov 12, 2021 at 18:48

Spectric · Accepted Answer · 2021-11-12 18:57:05Z

1

You can declare a variable outside the loop, then assign i to it and break the loop if the condition is true:

arr = [[3, 3], [3, 4], [3, 5], [3, 6]]

var ind = 0

for(let i = 0; i < arr.length; i++){
    if (arr[i][0] === 3 && arr[i][1] === 4 || arr[i][0] === 3 && arr[i][1] === 5){
        ind = i;
        break;
    }
}

console.log(ind)

Alternatively, you can use Array.findIndex:

arr = [[3, 3], [3, 4], [3, 5], [3, 6]]

var ind = arr.findIndex(e => e[0] === 3 && e[1] === 4 || e[0] === 3 && e[1] === 5)

console.log(ind)

edited Nov 12, 2021 at 18:57

answered Nov 12, 2021 at 18:48

Spectric

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1 Comment

AttemptedMastery Over a year ago

I knew it was something simple... Thanks a ton - this works perfectly.

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