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| 1 | +/** |
| 2 | +Leetcode Question 454 |
| 3 | +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. |
| 4 | +
|
| 5 | +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. |
| 6 | +
|
| 7 | +Example: |
| 8 | +
|
| 9 | +Input: |
| 10 | +A = [ 1, 2] |
| 11 | +B = [-2,-1] |
| 12 | +C = [-1, 2] |
| 13 | +D = [ 0, 2] |
| 14 | +
|
| 15 | +Output: |
| 16 | +2 |
| 17 | +
|
| 18 | +Explanation: |
| 19 | +The two tuples are: |
| 20 | +1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0 |
| 21 | +2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0 |
| 22 | +**/ |
| 23 | + |
| 24 | +// Time Complexity : O(n^2) |
| 25 | +// Space Complexity : O(n) |
| 26 | +/** 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 |
| 27 | +* sum is stored in key and value stores the frequency of occurence of that sum.Next, we take other two arrays and for all |
| 28 | +* possible sums in those two arrays we check if the map contains a compliment key of the sum,if yes, we increase the count. |
| 29 | +**/ |
| 30 | +classSolution { |
| 31 | +publicintfourSumCount(int[]A,int[]B,int[]C,int[]D) { |
| 32 | +if(A.length ==0 &&B.length ==0 &&C.length ==0 &&D.length ==0) |
| 33 | +return0; |
| 34 | + |
| 35 | +intcount =0; |
| 36 | +Map<Integer,Integer>sumAndOcurrences =newHashMap<>(); |
| 37 | +for(intaIndex =0;aIndex<A.length;aIndex++){ |
| 38 | +for(intbIndex =0;bIndex<B.length;bIndex++){ |
| 39 | +intsum =A[aIndex]+B[bIndex]; |
| 40 | +//creating a map to set the sum and frequency |
| 41 | +sumAndOcurrences.put(sum,sumAndOcurrences.getOrDefault(sum,0)+1); |
| 42 | + } |
| 43 | + } |
| 44 | + |
| 45 | +for(intcIndex =0;cIndex<C.length;cIndex++){ |
| 46 | +for(intdIndex =0;dIndex<D.length;dIndex++){ |
| 47 | +// suming the elements of rest two arrays and checking if the complement of that exists in the |
| 48 | +//hashmap created above. |
| 49 | +count +=sumAndOcurrences.getOrDefault(-1 * (C[cIndex]+D[dIndex]),0); |
| 50 | + } |
| 51 | + } |
| 52 | +returncount; |
| 53 | + } |
| 54 | +} |