import java.util.*;
import java.util.stream.Collectors;
/**
* Given two arrays, write a function to compute their intersection.
* <p>
* Example:
* Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].
* <p>
* Note:
* Each element in the result should appear as many times as it shows in both arrays.
* The result can be in any order.
* Follow up:
* What if the given array is already sorted? How would you optimize your algorithm?
* What if nums1's size is small compared to nums2's size? Which algorithm is better?
* What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into
* the memory at once?
* <p>
* Tags: Binary Search, Hash Table, Two Pointers, Sort
* Similar Problems: (E) Intersection of Two Arrays
*/
public class IntersectionofTwoArrays2 {
/**
* Use a hash map to record count of each integer in one array.
* Then go through the other array to find intersection.
* Count should be updated when an intersection is found.
*/
public int[] intersect(int[] nums1, int[] nums2) {
Map<Integer, Integer> count = new HashMap<>();
for (int n : nums1) {
count.put(n, count.containsKey(n) ? count.get(n) + 1 : 1);
}
List<Integer> intersections = new ArrayList<>(Math.min(nums1.length, nums2.length));
for (int n : nums2) {
if (count.containsKey(n)) {
intersections.add(n);
count.put(n, count.get(n) - 1);
if (count.get(n) == 0)
count.remove(n);
}
}
int[] res = new int[intersections.size()];
for (int i = 0; i < intersections.size(); i++) {
res[i] = intersections.get(i);
}
return res;
}
public int[] intersectJava8(int[] nums1, int[] nums2) {
// Create an integer array stream to boxed value and collect to a map grouping by element and count as value
Map<Integer, Long> map = Arrays.stream(nums2).boxed()
.collect(Collectors.groupingBy(i -> i, Collectors.counting()));
// Return by filtering the other array stream
return Arrays.stream(nums1).filter(i -> {
if (!map.containsKey(i) || map.get(i) < 1) return false;
map.put(i, map.get(i) - 1);
return true;
}).toArray();
}
}