new soln

Author: shreymehul

2215.FindDifference.cs

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+// 2215. Find the Difference of Two Arrays
+// Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:
+
+// answer[0] is a list of all distinct integers in nums1 which are not present in nums2.
+// answer[1] is a list of all distinct integers in nums2 which are not present in nums1.
+// Note that the integers in the lists may be returned in any order.
+
+// Example 1:
+// Input: nums1 = [1,2,3], nums2 = [2,4,6]
+// Output: [[1,3],[4,6]]
+// Explanation:
+// For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nums1[2] = 3 are not present in nums2. Therefore, answer[0] = [1,3].
+// For nums2, nums2[0] = 2 is present at index 1 of nums1, whereas nums2[1] = 4 and nums2[2] = 6 are not present in nums2. Therefore, answer[1] = [4,6].
+// Example 2:
+// Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]
+// Output: [[3],[]]
+// Explanation:
+// For nums1, nums1[2] and nums1[3] are not present in nums2. Since nums1[2] == nums1[3], their value is only included once and answer[0] = [3].
+// Every integer in nums2 is present in nums1. Therefore, answer[1] = [].
+
+// Constraints:
+// 1 <= nums1.length, nums2.length <= 1000
+// -1000 <= nums1[i], nums2[i] <= 1000
+
+public class Solution {
+    public IList<IList<int>> FindDifference(int[] nums1, int[] nums2) {
+        HashSet<int> set1 = new();
+        HashSet<int> set2 = new();
+
+        foreach(int num in nums1){
+            set1.Add(num);
+        }
+        foreach(int num in nums2){
+            set2.Add(num);
+        }
+        foreach(var item in set1){
+            if(set2.Contains(item)){
+                set2.Remove(item);
+                set1.Remove(item);
+            }
+        }
+        
+        return new List<IList<int>>(){set1.ToList(), set2.ToList()};
+    }
+}
+
+
+public class Solution {
+    public IList<IList<int>> FindDifference(int[] nums1, int[] nums2) {
+        HashSet<int> set1 = new();
+        HashSet<int> set2 = new();
+
+        foreach(int num in nums1){
+            set1.Add(num);
+        }
+        foreach(int num in nums2){
+            set2.Add(num);
+        }
+        List<int> result1 = new();
+        foreach(var item in set1){
+            if(!set2.Contains(item)){
+                result1.Add(item);
+            }
+        }
+        List<int> result2 = new();
+        foreach(var item in set2){
+            if(!set1.Contains(item)){
+                result2.Add(item);
+            }
+        }
+        return new List<IList<int>>(){result1, result2};
+    }
+}

2248.Intersection.cs

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+// 2248. Intersection of Multiple Arrays
+// Given a 2D integer array nums where nums[i] is a non-empty array of distinct positive integers, return the list of integers that are present in each array of nums sorted in ascending order.
+
+// Example 1:
+// Input: nums = [[3,1,2,4,5],[1,2,3,4],[3,4,5,6]]
+// Output: [3,4]
+// Explanation: 
+// The only integers present in each of nums[0] = [3,1,2,4,5], nums[1] = [1,2,3,4], and nums[2] = [3,4,5,6] are 3 and 4, so we return [3,4].
+// Example 2:
+// Input: nums = [[1,2,3],[4,5,6]]
+// Output: []
+// Explanation: 
+// There does not exist any integer present both in nums[0] and nums[1], so we return an empty list [].
+ 
+
+// Constraints:
+// 1 <= nums.length <= 1000
+// 1 <= sum(nums[i].length) <= 1000
+// 1 <= nums[i][j] <= 1000
+// All the values of nums[i] are unique.
+
+public class Solution {
+    public IList<int> Intersection(int[][] nums) {
+        HashSet<int> set = new HashSet<int>(nums[0]);
+
+        for(int  i = 1; i < nums.Length; i++){
+            HashSet<int> tempSet = new();
+            for(int j =0; j < nums[i].Length; j++){
+                if(set.Contains(nums[i][j]))
+                    tempSet.Add(nums[i][j]);
+            }
+            set = tempSet;
+        }
+        var result = set.ToList();
+        result.Sort();
+        return result;
+    }
+}

2540.GetCommon.cs

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+// 2540. Minimum Common Value
+// Given two integer arrays nums1 and nums2, sorted in non-decreasing order, return the minimum integer common to both arrays. If there is no common integer amongst nums1 and nums2, return -1.
+
+// Note that an integer is said to be common to nums1 and nums2 if both arrays have at least one occurrence of that integer.
+
+// Example 1:
+// Input: nums1 = [1,2,3], nums2 = [2,4]
+// Output: 2
+// Explanation: The smallest element common to both arrays is 2, so we return 2.
+// Example 2:
+// Input: nums1 = [1,2,3,6], nums2 = [2,3,4,5]
+// Output: 2
+// Explanation: There are two common elements in the array 2 and 3 out of which 2 is the smallest, so 2 is returned.
+
+// Constraints:
+// 1 <= nums1.length, nums2.length <= 105
+// 1 <= nums1[i], nums2[j] <= 109
+// Both nums1 and nums2 are sorted in non-decreasing order.
+
+public class Solution {
+    public int GetCommon(int[] nums1, int[] nums2) {
+        int i = 0, j = 0;
+        while(i < nums1.Length && j < nums2.Length){
+            if(nums1[i] == nums2[j])
+                return nums1[i];
+            else if(nums1[i] > nums2[j])
+                j++;
+            else
+                i++;
+        }
+        return -1;
+    }
+}