Description
Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.
Example 1:
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]
Example 2:
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]
Explanation: [9,4] is also accepted.
Constraints:
1 <= nums1.length, nums2.length <= 10000 <= nums1[i], nums2[i] <= 1000
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?
Solutions
Solution 1: Hash Table
We can use a hash table cnt to count the occurrences of each element in the array nums1. Then, we iterate through the array nums2. If an element x is in cnt and the occurrence of x is greater than 0, we add x to the answer and then decrement the occurrence of x by one.
After the iteration is finished, we return the answer array.
The time complexity is O(m + n), and the space complexity is O(m). Here, m and n are the lengths of the arrays nums1 and nums2, respectively.
PythonJavaC++GoTypeScriptRustJavaScriptC#PHP
class Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
cnt = Counter(nums1)
ans = []
for x in nums2:
if cnt[x]:
ans.append(x)
cnt[x] -= 1
return ans(code-box)
class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
int[] cnt = new int[1001];
for (int x : nums1) {
++cnt[x];
}
List<Integer> ans = new ArrayList<>();
for (int x : nums2) {
if (cnt[x]-- > 0) {
ans.add(x);
}
}
return ans.stream().mapToInt(Integer::intValue).toArray();
}
}(code-box)
class Solution {
public:
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
unordered_map<int, int> cnt;
for (int x : nums1) {
++cnt[x];
}
vector<int> ans;
for (int x : nums2) {
if (cnt[x]-- > 0) {
ans.push_back(x);
}
}
return ans;
}
};(code-box)
func intersect(nums1 []int, nums2 []int) (ans []int) {
cnt := map[int]int{}
for _, x := range nums1 {
cnt[x]++
}
for _, x := range nums2 {
if cnt[x] > 0 {
ans = append(ans, x)
cnt[x]--
}
}
return
}(code-box)
function intersect(nums1: number[], nums2: number[]): number[] {
const cnt: Record<number, number> = {};
for (const x of nums1) {
cnt[x] = (cnt[x] || 0) + 1;
}
const ans: number[] = [];
for (const x of nums2) {
if (cnt[x]-- > 0) {
ans.push(x);
}
}
return ans;
}(code-box)
use std::collections::HashMap;
impl Solution {
pub fn intersect(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
let mut cnt = HashMap::new();
for &x in &nums1 {
*cnt.entry(x).or_insert(0) += 1;
}
let mut ans = Vec::new();
for &x in &nums2 {
if let Some(count) = cnt.get_mut(&x) {
if *count > 0 {
ans.push(x);
*count -= 1;
}
}
}
ans
}
}(code-box)
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number[]}
*/
var intersect = function (nums1, nums2) {
const cnt = {};
for (const x of nums1) {
cnt[x] = (cnt[x] || 0) + 1;
}
const ans = [];
for (const x of nums2) {
if (cnt[x]-- > 0) {
ans.push(x);
}
}
return ans;
};(code-box)
public class Solution {
public int[] Intersect(int[] nums1, int[] nums2) {
Dictionary<int, int> cnt = new Dictionary<int, int>();
foreach (int x in nums1) {
if (cnt.ContainsKey(x)) {
cnt[x]++;
} else {
cnt[x] = 1;
}
}
List<int> ans = new List<int>();
foreach (int x in nums2) {
if (cnt.ContainsKey(x) && cnt[x] > 0) {
ans.Add(x);
cnt[x]--;
}
}
return ans.ToArray();
}
}(code-box)
class Solution {
/**
* @param Integer[] $nums1
* @param Integer[] $nums2
* @return Integer[]
*/
function intersect($nums1, $nums2) {
$cnt = [];
foreach ($nums1 as $x) {
if (isset($cnt[$x])) {
$cnt[$x]++;
} else {
$cnt[$x] = 1;
}
}
$ans = [];
foreach ($nums2 as $x) {
if (isset($cnt[$x]) && $cnt[$x] > 0) {
$ans[] = $x;
$cnt[$x]--;
}
}
return $ans;
}
}(code-box)
