LeetCode 350: Intersection of Two Arrays II Solution

Master LeetCode problem 350 (Intersection of Two Arrays II), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

350. Intersection of Two Arrays II

Easy

Array Hash Table Two Pointers Binary Search Sorting

Problem Explanation

Explanation

To find the intersection of two arrays, we can use a hashmap to store the frequency of elements in one of the arrays. Then, we iterate through the second array and decrement the frequency in the hashmap whenever we find a common element. We add this element to the result array. By doing this, we ensure that each element appears as many times as it shows in both arrays.

Algorithm:

Create a hashmap to store the frequency of elements in nums1.
Iterate through nums1 and populate the hashmap.
Iterate through nums2:
- If the element is found in the hashmap and its frequency is greater than 0, add it to the result array and decrement its frequency in the hashmap.
Return the result array.

Time Complexity:

The time complexity of this approach is O(m + n) where m and n are the lengths of nums1 and nums2 respectively.

Space Complexity:

The space complexity is O(min(m, n)) where m and n are the lengths of nums1 and nums2 respectively.

Solution Code

import java.util.*;

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        Map<Integer, Integer> map = new HashMap<>();
        List<Integer> result = new ArrayList<>();

        for (int num : nums1) {
            map.put(num, map.getOrDefault(num, 0) + 1);
        }

        for (int num : nums2) {
            if (map.containsKey(num) && map.get(num) > 0) {
                result.add(num);
                map.put(num, map.get(num) - 1);
            }
        }

        int[] resArray = new int[result.size()];
        for (int i = 0; i < result.size(); i++) {
            resArray[i] = result.get(i);
        }

        return resArray;
    }
}

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