Set intersection

3 snippets across 3 stacks — Interview Prep, JavaScript, Python

Also written as Set intersection

DSAInterview Prep

Intersection of Collections

syntax

Convert one collection to a set, iterate the other and check membership.
For sorted arrays: use two pointers instead.

example

// JavaScript
function intersection(arr1, arr2) {
  const set1 = new Set(arr1);
  const result = [];
  const seen = new Set();
  for (const val of arr2) {
    if (set1.has(val) && !seen.has(val)) {
      result.push(val);
      seen.add(val);
    }
  }
  return result;
}

# Python
def intersection(arr1, arr2):
    return list(set(arr1) & set(arr2))

# With duplicates preserved:
def intersect_with_dupes(arr1, arr2):
    from collections import Counter
    c1, c2 = Counter(arr1), Counter(arr2)
    return list((c1 & c2).elements())

output

intersection([1,2,2,1], [2,2]) → [2]
intersect_with_dupes([1,2,2,1], [2,2]) → [2, 2]

Note Set approach: O(n + m) time, O(min(n,m)) space. Sorted two-pointer approach: O(n log n + m log m) time, O(1) extra space. Ask the interviewer: unique results or preserve duplicates? Are inputs sorted?

JSJavaScript

Set Methods (ES2025)

JS · Modern Features

syntax

setA.union(setB)
setA.intersection(setB)
setA.difference(setB)
setA.symmetricDifference(setB)
setA.isSubsetOf(setB)
setA.isSupersetOf(setB)
setA.isDisjointFrom(setB)

example

const frontend = new Set(["js", "css", "html", "ts"]);
const backend = new Set(["js", "python", "go", "ts"]);

console.log(frontend.union(backend));
// Set {"js", "css", "html", "ts", "python", "go"}

console.log(frontend.intersection(backend));
// Set {"js", "ts"}

console.log(frontend.difference(backend));
// Set {"css", "html"}

console.log(frontend.isSubsetOf(backend)); // false

Note ES2025. All methods return new Sets (non-mutating). Works with any iterable argument, not just Sets. Replaces manual loop-based set operations.

PYPython

Set Operations

PY · Tuples & Sets

syntax

a | b  (union)
a & b  (intersection)
a - b  (difference)
a ^ b  (symmetric difference)

example

frontend = {"alice", "bob", "carol"}
backend = {"bob", "carol", "dave"}

print(frontend | backend)
print(frontend & backend)
print(frontend - backend)
print(frontend ^ backend)

output

{'alice', 'bob', 'carol', 'dave'}
{'bob', 'carol'}
{'alice'}
{'alice', 'dave'}

Note Set operations are extremely fast (O(min(len(a), len(b))) for intersection). Use them instead of nested loops for membership checks.