<p>Given two integer arrays <code>nums1</code> and <code>nums2</code>, return <em>an array of their intersection</em>. Each element in the result must appear as many times as it shows in both arrays and you may return the result in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,2,2,1], nums2 = [2,2]
<strong>Output:</strong> [2,2]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [4,9,5], nums2 = [9,4,9,8,4]
<strong>Output:</strong> [4,9]
<strong>Explanation:</strong> [9,4] is also accepted.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 1000</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong></p>

<ul>
	<li>What if the given array is already sorted? How would you optimize your algorithm?</li>
	<li>What if <code>nums1</code>&#39;s size is small compared to <code>nums2</code>&#39;s size? Which algorithm is better?</li>
	<li>What if elements of <code>nums2</code> are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?</li>
</ul>

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int>temp;
        map<int,int>mpp;
        int n=nums1.size();
        int m=nums2.size();

           if(n<m){
             for(int i=0;i<m;i++){
                mpp[nums2[i]]++;
             }
             for(int i=0;i<n;i++){
                if(mpp.find(nums1[i])!=mpp.end()){
                    if(mpp[nums1[i]]>=1){
                    temp.push_back(nums1[i]);
                    mpp[nums1[i]]-=1;}
                }
             }
           }
           else{
            for(int i=0;i<n;i++){
                mpp[nums1[i]]++;
             }
             for(int i=0;i<m;i++){
                if(mpp.find(nums2[i])!=mpp.end()){
                    if(mpp[nums2[i]]>=1){
                    temp.push_back(nums2[i]);
                    mpp[nums2[i]]-=1;
                    }
                }
             }

           }
            return temp;
      
    }
};
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