How can I find & groups pairs from an array while changing the original array?

Question

So for ex. I have this array with these values :

let arr = [1, 2, 3, 3, 3, 4, 4];

How can I get a new arr as :

let newArr = [3, 3, 4, 4];

while at the same time the original array should be changed as :

let arr = [1, 2, 3];

one 3 is left behind cause there is only one pair of 3.

I tried this code below ( a for loop & splice ) but it is not working properly.

let result = [];

for (let i = 0; i < sorted.length; i++) {
    if (sorted[i] === sorted[i + 1]) {
        let pair = sorted.splice(i, 1);
        pair.forEach(el => result.push(el));
    }
}

You’re mutating the array while iterating without adjusting the index. `console.log({ i, sorted: sorted.slice(), current: sorted[i] });` would help with debugging. `--i;` decrements a number by one. — Sebastian Simon, Jan 13 '22 at 05:14
I wanted to ask if he's sorting the array or if the input is already sorted. — Ramesh Reddy, Jan 13 '22 at 05:16

Amila Senadheera · Accepted Answer · 2022-01-15T05:11:20.497

1

Few things corrected,

You need to splice two items from the point you see duplicates. Try like this.
Iteration stop condition should be i < sorted.length - 1 as we consider two consecutive items.
Use Array.prototype.concat to join duplicates to the result array.

let sorted = [1, 2, 3, 3, 3, 4, 4];
let result = [];

for (let i = 0; i < sorted.length - 1; i++) {
    if (sorted[i] === sorted[i + 1]) {
        let pair = sorted.splice(i, 2);
        result = result.concat(pair);
        i--;
    }
}

console.log(sorted)
console.log(result)

edited Jan 15 '22 at 05:11

answered Jan 13 '22 at 05:40

Amila Senadheera

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@Arya, check this out !! – Amila Senadheera Jan 13 '22 at 05:46
thanks man. Ya, I was splicing 2 items. Splicing 1 was a typo. But why should the stopping condition be length - 1 ? – Rabten Jan 13 '22 at 07:04
Because last item has no another item to compare with. But the code works fine as it returns undefined instead of throwing a runtime error – Amila Senadheera Jan 13 '22 at 07:11
1

Right it does work fine. Thanks for your time Amila, I'll use the condition just as a good practice – Rabten Jan 13 '22 at 07:12
Are you suggesting I should not decrement ?? – Rabten Jan 14 '22 at 23:22
@Rabten, you need to shift back once(i--) a duplicate is found as we are changing the array using splice. Because there could be a pair we missing. Updated the answer! – Amila Senadheera Jan 15 '22 at 05:13

How can I find & groups pairs from an array while changing the original array?

1 Answers1