349. Intersection of Two Arrays – Leetcode Solutions

Example 1:

Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2]

Example 2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [9,4]
Explanation: [4,9] is also accepted.

from typing import List

class Solution:
    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
        # Convert the first list to a set to remove duplicates and allow O(1) average-time complexity checks
        set1 = set(nums1)
        
        # Initialize an empty set to store the intersection
        intersection_set = set()
        
        # Iterate through the second list
        for num in nums2:
            # If the number is in the first set, add it to the intersection set
            if num in set1:
                intersection_set.add(num)
        
        # Convert the set back to a list for the result
        return list(intersection_set)

# Example usage:
solution = Solution()
print(solution.intersection([1, 2, 2, 1], [2, 2])) # Output: [2]
print(solution.intersection([4, 9, 5], [9, 4, 9, 8, 4])) # Output: [9, 4]

var intersection = function(nums1, nums2) {
    // Convert the first array to a Set to remove duplicates and allow O(1) average-time complexity checks
    let set1 = new Set(nums1);
    
    // Initialize an empty Set to store the intersection
    let intersectionSet = new Set();
    
    // Iterate through the second array
    for (let num of nums2) {
        // If the number is in the first set, add it to the intersection set
        if (set1.has(num)) {
            intersectionSet.add(num);
        }
    }
    
    // Convert the set back to an array for the result
    return Array.from(intersectionSet);
};

// Example usage:
console.log(intersection([1, 2, 2, 1], [2, 2])); // Output: [2]
console.log(intersection([4, 9, 5], [9, 4, 9, 8, 4])); // Output: [9, 4]

import java.util.HashSet;
import java.util.Set;
import java.util.Arrays;

class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        // Convert the first array to a HashSet to remove duplicates and allow O(1) average-time complexity checks
        Set<Integer> set1 = new HashSet<>();
        for (int num : nums1) {
            set1.add(num);
        }

        // Initialize another HashSet to store the intersection
        Set<Integer> intersectionSet = new HashSet<>();

        // Iterate through the second array
        for (int num : nums2) {
            // If the number is in the first set, add it to the intersection set
            if (set1.contains(num)) {
                intersectionSet.add(num);
            }
        }

        // Convert the intersection set to an array
        int[] result = new int[intersectionSet.size()];
        int i = 0;
        for (int num : intersectionSet) {
            result[i++] = num;
        }

        return result;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        System.out.println(Arrays.toString(solution.intersection(new int[]{1, 2, 2, 1}, new int[]{2, 2}))); // Output: [2]
        System.out.println(Arrays.toString(solution.intersection(new int[]{4, 9, 5}, new int[]{9, 4, 9, 8, 4}))); // Output: [9, 4]
    }
}

using System;
using System.Collections.Generic;
using System.Linq;

public class Solution {
    public int[] Intersection(int[] nums1, int[] nums2) {
        // Convert the first array to a HashSet to remove duplicates and allow O(1) average-time complexity checks
        HashSet<int> set1 = new HashSet<int>(nums1);
        
        // Initialize another HashSet to store the intersection
        HashSet<int> intersectionSet = new HashSet<int>();
        
        // Iterate through the second array
        foreach (int num in nums2) {
            // If the number is in the first set, add it to the intersection set
            if (set1.Contains(num)) {
                intersectionSet.Add(num);
            }
        }
        
        // Convert the intersection set to an array
        return intersectionSet.ToArray();
    }
}