Intersection of Two Arrays II - Hash Table - Easy - LeetCode
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Intersection of Two Arrays II - Hash Table - Easy - LeetCode

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  • 1The function computes the intersection of two arrays, returning elements that appear in both with their respective counts.
  • 2The algorithm utilizes hash tables to efficiently track occurrences of elements in each array.
  • 3Time complexity is O(n) and space complexity is also O(n), making it suitable for large input sizes.

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"The function computes the intersection of two arrays, returning elements that appear in both with their respective counts."

Intersection of Two Arrays II - Hash Table - Easy - LeetCode

Given two arrays, write a function to compute their intersection.

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] 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?

public class Solution {
 public int[] Intersect(int[] nums1, int[] nums2) {
 var map1 = new Dictionary<int,int>();
 var map2 = new Dictionary<int,int>();
 for(int i=0;i<nums1.Length;i++){
 if(map1.ContainsKey(nums1[i])){
 map1[nums1[i]]++;
 }
 else{
 map1.Add(nums1[i],1);
 }
 }
 
 for(int i=0;i<nums2.Length;i++){
 if(map2.ContainsKey(nums2[i])){
 map2[nums2[i]]++;
 }
 else{
 map2.Add(nums2[i],1);
 }
 }
 
 var list = new List<int>();
 foreach(var item in map1){
 if(map2.ContainsKey(item.Key)){
 int minCount = Math.Min(item.Value,map2[item.Key]);
 for(int i=0;i<minCount;i++){
 list.Add(item.Key);
 }
 }
 }
 
 return list.ToArray();
 }
}

Time Complexity: O(n)

Space Complexity: O(n)

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sumitc91

Published on 28 September 2020 · 1 min read · 181 words

Part of AskGif Blog · coding

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