Intersection of Two Arrays II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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
C++ Intersection of Two Arrays II LeetCode Solution
class Solution {// Using Map & without sort
public:
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
map<int,int>freq;
vector<int>ans;
for(int i = 0;i<nums1.size();i++){
freq[nums1[i]]++;
}
for(int i = 0;i<nums2.size();i++){
if (freq[nums2[i]] > 0){
freq[nums2[i]]--;
ans.push_back(nums2[i]);
}
}
return ans;
}
};
Java Intersection of Two Arrays II LeetCode Solution
public int[] intersect(int[] nums1, int[] nums2) {
HashMap<Integer, Integer> map = new HashMap<>();
for(int i : nums1){
int freq = map.getOrDefault(i, 0);
map.put(i, freq + 1);
}
ArrayList<Integer> list = new ArrayList<>();
for(int i : nums2){
if(map.get(i) != null && map.get(i) > 0){
list.add(i);
map.put(i, map.get(i) - 1);
}
}
int[] ret = new int[list.size()];
for(int i = 0; i < list.size();i++){
ret[i] = list.get(i);
}
return ret;
}
Python 3 Intersection of Two Arrays II LeetCode Solution
class Solution(object):
def intersect(self, nums1, nums2):
nums1, nums2 = sorted(nums1), sorted(nums2)
pt1 = pt2 = 0
res = []
while True:
try:
if nums1[pt1] > nums2[pt2]:
pt2 += 1
elif nums1[pt1] < nums2[pt2]:
pt1 += 1
else:
res.append(nums1[pt1])
pt1 += 1
pt2 += 1
except IndexError:
break
return res
Array-1180
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Interactive-18
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