Hone your understanding and apply modern techniques backed up by hands-on experience. These five engaging projects cover web applications and range from social media website applications to geo-social networking maps.

Top 5 MERN STACK projects to improve your practical understandingīoost your MERN Stack development skills by undertaking interesting beginner projects. User11536834 second example was a surprisingly poor performer.The JavaScript Brief 1. There are some improvements that can be made to the already given answers. It may be worth noting that this solution appears to perform quite well, and can be further optimized by doing the inequality check on the first two elements before invoking find, and by using the third parameter to find to access the array, making the callback a pure function and eliminating the need to reference the array via the closed-over variable, for example: a.find(a != a ? This is a bit code-golfey and not very readable, so I wouldn't recommend it in production, but it may be of some interest as a curiosity. In this case, we can check against the third element instead of the first otherwise we check against the first element as before, thus: a.find(v => a != a ? v != a : v != a) If they do, then the stray is either in the first or second position, so the third element is not a stray.

But what if the stray element comes first in the array? We can check if the first two elements differ. That is, find an element that's different from the first element. We could simply write: a.find(v => v != a) Let's start by handling the case where the stray value is not in the first element. This is a similar idea to the other answers here, but the implementation is a bit different.įirst of all, we can assume that the array's length is at least 3, since it needs to have at least two of the same values and one different value.

Javascript find element in array