Create 4SumII.java

Author: pyusha1006

4SumII.java

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+/**
+Leetcode Question 454
+Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.
+
+To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.
+
+Example:
+
+Input:
+A = [ 1, 2]
+B = [-2,-1]
+C = [-1, 2]
+D = [ 0, 2]
+
+Output:
+2
+
+Explanation:
+The two tuples are:
+1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
+2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0
+**/
+
+// Time Complexity : O(n^2)
+// Space Complexity : O(n)
+/** The solution considers 2 arrays at a time and finds all possible sum from the two arrays and stores the same in a map where
+* sum is stored in key and value stores the frequency of occurence of that sum.Next, we take other two arrays and for all
+* possible sums in those two arrays we check if the map contains a compliment key of the sum,if yes, we increase the count.
+**/
+class Solution {
+    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
+        if(A.length == 0 && B.length == 0 && C.length == 0 && D.length == 0)
+            return 0;
+        
+        int count = 0;
+        Map<Integer,Integer> sumAndOcurrences = new HashMap<>();
+        for(int aIndex = 0;aIndex<A.length;aIndex++){
+           for(int bIndex = 0;bIndex<B.length;bIndex++){
+               int sum = A[aIndex]+B[bIndex];
+               //creating a map to set the sum and frequency
+                sumAndOcurrences.put(sum,sumAndOcurrences.getOrDefault(sum,0)+1);
+            }
+        }
+        
+        for(int cIndex = 0;cIndex<C.length;cIndex++){
+           for(int dIndex = 0;dIndex<D.length;dIndex++){
+               // suming the elements of rest two arrays and checking if the complement of that exists in the 
+               //hashmap created above.
+                count += sumAndOcurrences.getOrDefault(-1 * (C[cIndex]+D[dIndex]),0);
+            }
+        }
+        return count;
+    }
+}