Coding problem · Easy · Basics

Intersection of Two Arrays

Easy Algorithms & data structures

Return the values that appear in both a and b, without duplicates, sorted in ascending order.

Example 1

Input: a = [1,2,2,1], b = [2,2]
Output: [2]

Example 2

Input: a = [4,9,5], b = [9,4,9,8,4]
Output: [4,9]

Hints

Hint 1

Put one array's values in a set.

Hint 2

Keep values of the other array that are in the set, then remove duplicates and sort.

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Solution and explanation

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Time O(n + m + k log k) · Space O(n)

Build a set from a, collect values of b that are in it into a result set (removing duplicates), then sort.

Reference solution

function intersection(a, b) {
  const inA = new Set(a);
  return [...new Set(b.filter((x) => inA.has(x)))].sort((x, y) => x - y);
}

function intersection(a: number[], b: number[]): number[] {
  const inA = new Set(a);
  return [...new Set(b.filter((x) => inA.has(x)))].sort((x, y) => x - y);
}

def intersection(a: list[int], b: list[int]) -> list[int]:
    return sorted(set(a) & set(b))

<?php

function intersection(array $a, array $b): array
{
    $out = array_values(array_unique(array_intersect($a, $b)));
    sort($out);
    return $out;
}

class Solution {
    public int[] intersection(int[] a, int[] b) {
        java.util.Set<Integer> inA = new java.util.HashSet<>();
        for (int x : a) inA.add(x);
        java.util.TreeSet<Integer> both = new java.util.TreeSet<>();
        for (int x : b) if (inA.contains(x)) both.add(x);
        return both.stream().mapToInt(Integer::intValue).toArray();
    }
}

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