I added Intersection of two arrays

ArraysSolutions/Intersection of two arrays.md

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+
+## Solution of Intersection Of Two Arrays
+
+#### Time Complexity - O (n*m)
+
+#### Space Complexity - O(n)
+
+```java
+
+class Solution {
+ public int[] intersection(int[] nums1, int[] nums2) {
+ HashSet<Integer> hs = new HashSet<>();
+ for (int i = 0; i <nums1.length ; i++) {
+ for (int j = 0; j <nums2.length ; j++) {
+
+ if(nums1[i]==nums2[j]){
+
+ hs.add(nums2[j]);
+}
+}
+}
+
+ int []nums3 = new int[hs.size()];
+ int h=0;
+ for(Integer n: hs){
+ nums3[h++]=n;
+ }
+ return nums3;
+
+
+
+}
+}
+
+
+```
+
+### Solution 2: Two Pointer Approach 
+
+#### Time Complexity - O (nlog(n))
+
+#### Space Complexity - O(n)
+
+```java
+
+class Solution {
+ public int[] intersection(int[] nums1, int[] nums2) {
+ HashSet<Integer> hs = new HashSet<>();
+ Arrays.sort(nums1);
+ Arrays.sort(nums2);
+ int i=0;
+ int j=0;
+ while(i<nums1.length&&j<nums2.length){
+ if(nums1[i]==nums2[j]){
+ hs.add(nums1[i]);
+ i++;j++;
+}
+ else{
+ if(nums1[i]<nums2[j])i++;
+ else j++;
+}
+
+}
+
+ int []nums3 = new int[hs.size()];
+ int h=0;
+ for(Integer n: hs){
+ nums3[h++]=n;
+}
+ return nums3;
+
+
+
+}
+}
+
+```
+
+<div align="right"> 
+ Contributed By <a href="https://github.com/Love-Garg-19"> Love Garg</a>
+</div>

ArraysSolutions/Remove all occurrence of Given number.md

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+## Solution of Remove all occurrences of Given Number
+
+#### Time Complexity - O (n)
+
+#### Space Complexity - O(1)
+
+```java
+
+class Solution {
+ public int removeElement(int[] nums, int val) {
+
+ int i = 0;
+ int n = nums.length;
+ while (i < n) {
+ if (nums[i] == val) {
+ nums[i] = nums[n - 1];
+ n--;
+ } else {
+ i++;
+ }
+ }
+ return n;
+ }
+}
+
+```
+<div align="right"> 
+ Contributed By <a href="https://github.com/Love-Garg-19"> Love Garg</a>
+</div>