package stevesun.algorithms;

import java.util.HashMap;
import java.util.Map;

public class _4SumII {

    /**Although it's tagged with BinarySearch and HashTable, I didn't come up with a good BinarySearch solution, then looked at this post:
     * https://discuss.leetcode.com/topic/67658/simple-java-solution-with-explanation*/
    public static int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        Map<Integer, Integer> map = new HashMap();
        int result = 0, len = A.length;
        for (int i = 0; i < len; i++){
            for (int j = 0; j < len; j++){
                int sum = A[i] + B[j];
                if (map.containsKey(sum)){
                    map.put(sum, map.get(sum)+1);
                } else {
                   map.put(sum, 1); 
                }
            }
        }
        
        for (int i = 0; i < len; i++){
            for (int j = 0; j < len; j++){
                int sum = -(C[i] + D[j]);
                if(map.containsKey(sum)) result += map.get(sum);
            }
        }
        
        return result;
    }

    public static void main(String...args){
        int[] A = new int[]{1,2};
        int[] B = new int[]{-2,-1};
        int[] C = new int[]{-1,2};
        int[] D = new int[]{0,2};
        
        System.out.println(fourSumCount(A, B, C, D));
    }
}

@@ -76,258 +76,630 @@
 p;%0A%0A
-public class _4SumII %7B%0A%0A    /**Although it's tagged with BinarySearch and HashTable, I didn't come up with a good BinarySearch solution, then looked at this post:%0A     * https://discuss.leetcode.com/topic/67658/simple-java-solution-with-explanation*/
+/**Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A%5Bi%5D + B%5Bj%5D + C%5Bk%5D + D%5Bl%5D is zero.%0A%0A To make problem a bit easier, all A, B, C, D have same length of N where 0 %E2%89%A4 N %E2%89%A4 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.%0A%0A Example:%0A%0A Input:%0A A = %5B 1, 2%5D%0A B = %5B-2,-1%5D%0A C = %5B-1, 2%5D%0A D = %5B 0, 2%5D%0A%0A Output:%0A 2%0A%0A Explanation:%0A The two tuples are:%0A 1. (0, 0, 0, 1) -%3E A%5B0%5D + B%5B0%5D + C%5B0%5D + D%5B1%5D = 1 + (-2) + (-1) + 2 = 0%0A 2. (1, 1, 0, 0) -%3E A%5B1%5D + B%5B1%5D + C%5B0%5D + D%5B0%5D = 2 + (-1) + (-1) + 0 = 0*/%0A%0Apublic class _4SumII %7B%0A
 %0A   

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