# Optimizing the number of loop iterations in JavaScript

In the following code, a programmer checks if the array contains the number `3`:

`let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]; let exists = false; for (let elem of arr) { if (elem === 3) { exists = true; } } console.log(exists);`

What's wrong with his decision? It seems that no extra operations are done in the loop. The problem, however, is that after it is determined that the number `3` is in the array, the loop still continues to iterate until the end of the array.

It would be most irrational if the number `3` is found somewhere at the beginning of the array, and the array itself is, say, `1000` elements long. You get a thousand useless extra iterations of the loop! Not optimal.

We optimize code by stopping the loop in time:

`let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]; let exists = false; for (let elem of arr) { if (elem === 3) { exists = true; break; } } console.log(exists);`

The following code calculates how many array elements must be added to make the sum greater than `10`. Perform optimization:

`let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]; let sum = 0; let i = 1; for (let elem of arr) { sum += elem; if (sum <= 10) { i++; } } console.log(i);`

The following code outputs even numbers from a given range. Perform optimization:

`for (let i = 0; i <= 100; i++) { if (i % 2 === 0) { console.log(i); } }`

The following code outputs numbers divisible by `2` and `3` at the same time. Perform optimization:

`for (let i = 0; i <= 100; i++) { if (i % 2 === 0 && i % 3 === 0) { console.log(i); } }`

The following code searches for all Fridays the `13`th in the current year. Perform optimization:

`for (let i = 0; i <= 11; i++) { let curr = new Date; let last = new Date(curr.getFullYear(), i + 1, 0).getDate(); for (let j = 1; j <= last; j++) { let date = new Date(curr.getFullYear(), i, j); if (date.getDate() === 13 && date.getDay() === 5) { console.log(date); } } }`
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