The shallow ordered check is length then every with strict equality. Strict equality (===) gets you NaN !== NaN, which is rarely what you want; swap to Object.is(v, b[i]) if NaN equality matters in your data. For unordered comparison, sort copies of both inputs first (do not sort the originals) and compare position-by-position. The big watch-out is object elements: [{id:1}] vs another [{id:1}] is unequal because the inner objects are different references. For structural object equality at the leaf level, swap to JSON.stringify comparison or js-object-deep-equal.

new Set(arr).size !== arr.length is the cleanest "does this array have duplicates?" check. The Set constructor walks the input once and skips repeats, so a smaller size means at least one repeat existed. To get the actual duplicate VALUES, walk the array yourself and track two sets: one for items already seen, one for items already flagged as duplicates. Both forms are linear time. For non-primitive items (objects), Set compares by reference, so two structurally-equal but differently-allocated objects are NOT considered duplicates by this method.

When the items are objects, you cannot drop them into a Set and expect dedup by content. The Map keyed by a function-extracted value is the canonical answer: walk the array, keep the first (or last) occurrence per key, then return the values. Insertion order is preserved by both Map and Set, so the output keeps the original order of first appearance. For dedup by multiple fields, build a composite key string (u => ${u.tenant}:${u.id}``); for primitives, fall back to [...new Set(arr)].