2956. Find Common Elements Between Two Arrays
Description
You are given two integer arrays nums1
and nums2
of sizes n
and m
, respectively. Calculate the following values:
answer1
: the number of indicesi
such thatnums1[i]
exists innums2
.answer2
: the number of indicesi
such thatnums2[i]
exists innums1
.
Return [answer1,answer2]
.
Example 1:
Example 2:
Input: nums1 = [4,3,2,3,1], nums2 = [2,2,5,2,3,6]
Output: [3,4]
Explanation:
The elements at indices 1, 2, and 3 in nums1
exist in nums2
as well. So answer1
is 3.
The elements at indices 0, 1, 3, and 4 in nums2
exist in nums1
. So answer2
is 4.
Example 3:
Input: nums1 = [3,4,2,3], nums2 = [1,5]
Output: [0,0]
Explanation:
No numbers are common between nums1
and nums2
, so answer is [0,0].
Constraints:
n == nums1.length
m == nums2.length
1 <= n, m <= 100
1 <= nums1[i], nums2[i] <= 100
Solutions
Solution 1: Hash Table or Array
We can use two hash tables or arrays $s1$ and $s2$ to record the elements that appear in the two arrays respectively.
Next, we create an array $ans$ of length $2$, where $ans[0]$ represents the number of elements in $nums1$ that appear in $s2$, and $ans[1]$ represents the number of elements in $nums2$ that appear in $s1$.
Then, we traverse each element $x$ in the array $nums1$. If $x$ has appeared in $s2$, we increment $ans[0]$. After that, we traverse each element $x$ in the array $nums2$. If $x$ has appeared in $s1$, we increment $ans[1]$.
Finally, we return the array $ans$.
The time complexity is $O(n + m)$, and the space complexity is $O(n + m)$. Here, $n$ and $m$ are the lengths of the arrays $nums1$ and $nums2$ respectively.
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