Added Leetcode solution 0454-4Sum-II

dsa-problems/leetcode-problems/0400-0499.md

@@ -338,7 +338,7 @@ export const problems = [
     "problemName": "454. 4Sum II",
     "difficulty": "Medium",
     "leetCodeLink": "https://leetcode.com/problems/4sum-ii",
-    "solutionLink": "#"
+    "solutionLink": "/dsa-solutions/lc-solutions/0400-0499/4Sum-II"
   },
   {
     "problemName": "455. Assign Cookies",
@@ -619,4 +619,4 @@ export const problems = [
     collectionLink="https://leetcode.com/study-plan/programming-skills"
 />
 
-Now, you can see the list of problems in a table format. You can click on the problem link to view the problem on the LeetCode website. You can also click on the solution link to view the solution of the problem.
+Now, you can see the list of problems in a table format. You can click on the problem link to view the problem on the LeetCode website. You can also click on the solution link to view the solution of the problem.

dsa-solutions/lc-solutions/0400-0499/0454-4Sum-II.md

@@ -0,0 +1,131 @@
+---
+id: 4Sum-II
+title: 4Sum II (LeetCode)
+sidebar_label: 0454-4Sum-II
+tags:
+  - Hash Table
+  - Arrays
+description: "Given a string s, sort it in decreasing order based on the frequency of the characters. The frequency of a character is the number of times it appears in the string."
+sidebar_position: 0451
+---
+## Problem Description
+
+Given four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples (i, j, k, l) such that:
+
+0 <= i, j, k, l < n
+nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0
+
+## Examples
+
+Example 1:
+
+Input: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]
+Output: 2
+Explanation:
+The two tuples are:
+1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
+2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0
+Example 2:
+
+Input: nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
+Output: 1
+
+## Constraints
+
+n == nums1.length
+n == nums2.length
+n == nums3.length
+n == nums4.length
+1 <= n <= 200
+-228 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 228
+
+## Approach
+
+Create an empty map called "mp" to store integer keys and their corresponding counts.
+
+Iterate over each element "k" in the "nuns3" vector.
+
+For each "k", iterate over each element "l" in the "nums4" vector.
+
+Add the sum of "k" and "l" as a key in the map "mp" and increment its count by 1.
+
+Initialize a variable "count" to 0.
+
+Iterate over each element "i" in the "nums1" vector.
+
+For each "i", iterate over each element "j" in the "nums2" vector.
+
+Find the value associated with the key -(i + j) in the map "mp" and add it to the "count".
+
+Return the value of "count" as the result.
+
+### Solution Code
+
+#### C++
+
+```c++
+class Solution {
+public:
+    int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
+
+        unordered_map<int,int> mp;
+
+
+        for(int k : nums3)
+            for(int l : nums4)
+                mp[k + l]++;
+
+
+        int count = 0;
+        for(int i : nums1)
+            for(int j : nums2)
+                count += mp[-(i + j)];
+
+        return count;
+    }
+};
+```
+
+#### java
+```java
+
+class Solution {
+    public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
+        Map<Integer, Integer> map = new HashMap<>();
+        for (int n1 : nums1) {
+            for (int n2 : nums2) {
+                map.put(n1 + n2, map.getOrDefault(n1 + n2, 0) + 1);
+            }
+        }
+
+        int count = 0;
+        for (int n3 : nums3) {
+            for (int n4 : nums4) {
+                count += map.getOrDefault(-(n3 + n4), 0);
+            }
+        }
+        return count;
+    }
+}
+
+```
+#### Python
+```Python
+class Solution:
+    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
+        dc1=defaultdict(lambda:0)
+        for a in nums1:
+            for b in nums2:
+                dc1[a+b]+=1
+        ans=0
+        for c in nums3:
+            for d in nums4:
+                ans+=dc1[-c-d]
+        return ans
+
+```
+
+## Conclusion
+
+- 1. Time complexity:O(n^2)
+- 2. Space complexity:O(n^2)