# Intersection of Two Arrays

Given two arrays, write a function to compute their intersection.

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]``````

Note:

• Each element in the result must be unique.

• The result can be in any order.

## Analysis & Solution

@divingboy89

### Use two hash sets

Time complexity: O(n)

``````public class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
Set<Integer> set = new HashSet<>();
Set<Integer> intersect = new HashSet<>();
for (int i = 0; i < nums1.length; i++) {
set.add(nums1[i]);
}
for (int i = 0; i < nums2.length; i++) {
if (set.contains(nums2[i])) {
intersect.add(nums2[i]);
}
}
int[] result = new int[intersect.size()];
int i = 0;
for (Integer num : intersect) {
result[i++] = num;
}
return result;
}
}``````

### Sort both arrays, use two pointers

Time complexity: O(nlogn)

Seems fastest in OJ (2ms)

``````public class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
Set<Integer> set = new HashSet<>();
Arrays.sort(nums1);
Arrays.sort(nums2);
int i = 0;
int j = 0;
while (i < nums1.length && j < nums2.length) {
if (nums1[i] < nums2[j]) {
i++;
} else if (nums1[i] > nums2[j]) {
j++;
} else {
set.add(nums1[i]);
i++;
j++;
}
}
int[] result = new int[set.size()];
int k = 0;
for (Integer num : set) {
result[k++] = num;
}
return result;
}
}``````

Time complexity: O(nlogn)

``````public class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
Set<Integer> set = new HashSet<>();
Arrays.sort(nums2);
for (Integer num : nums1) {
if (binarySearch(nums2, num)) {
set.add(num);
}
}
int i = 0;
int[] result = new int[set.size()];
for (Integer num : set) {
result[i++] = num;
}
return result;
}

public boolean binarySearch(int[] nums, int target) {
int low = 0;
int high = nums.length - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
if (nums[mid] == target) {
return true;
}
if (nums[mid] > target) {
high = mid - 1;
} else {
low = mid + 1;
}
}
return false;
}
}``````

Java 8 Using `Stream`

``````Set<Integer> set = Arrays.stream(nums2).boxed().collect(Collectors.toSet());
return Arrays.stream(nums1).distinct().filter(e -> set.contains(e)).toArray();``````

## Reference

https://leetcode.com/problems/intersection-of-two-arrays/discuss/81969/Three-Java-Solutions

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