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

Constraints:

• n == nums1.length
• n == nums2.length
• n == nums3.length
• n == nums4.length
• 1 <= n <= 200
• -2**28 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 2**28

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

How to Solve

This is a variation of 2 Sum problem. Use a hash table and process two and two arrays. It will be two paths solution. The first loop takes the first and second arrays, and count nums1[i] + nums2[j]. The second loop takes the third and fourth arrays, and find 0 - (nums3[k] + nums4[l]), and count up the value.

Solution

• C++
• Java
• JavaScript
• Python
• Ruby

• class FourSumTwo {
  public:
      int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
          unordered_map<int, int> m;
          for (int n1 : nums1) {
            for (int n2 : nums2) {
              m[n1 + n2]++;
            }
          }
          int count = 0;
          for (int n3 : nums3) {
            for (int n4 : nums4) {
              int target = 0 - (n3 + n4);
              if (m.count(target)) {
                count += m[target];
              }
            }
          }
          return count;
      }
  };
  

• class FourSumTwo {
      public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
          Map<Integer, Integer> m = new HashMap();
          for (int n1 : nums1) {
            for (int n2 : nums2) {
              m.put(n1 + n2, m.getOrDefault(n1 + n2, 0) + 1);
            }
          }
          int count = 0;
          for (int n3 : nums3) {
            for (int n4 : nums4) {
              count += m.getOrDefault(0 - (n3 + n4), 0);
            }
          }
          return count;
      }
  }
  

• /**
   * @param {number[]} nums1
   * @param {number[]} nums2
   * @param {number[]} nums3
   * @param {number[]} nums4
   * @return {number}
   */
  var fourSumCount = function(nums1, nums2, nums3, nums4) {
      m  = new Map();
      nums1.forEach((n1) => {
        nums2.forEach((n2) => {
          m.set(n1 + n2, (m.get(n1 + n2) || 0) + 1);
        });
      });
      let count = 0;
      nums3.forEach((n3) => {
        nums4.forEach((n4) => {
          count += (m.get(0 - (n3 + n4)) || 0);
        });
      });
      return count;
  };
  

• class FourSumTwo:
      def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
          m = defaultdict(int)
          for n1 in nums1:
            for n2 in nums2:
              m[n1 + n2] += 1
          count = 0
          for n3 in nums3:
            for n4 in nums4:
              count += m[0 - (n3 + n4)]
          return count
  

• # @param {Integer[]} nums1
  # @param {Integer[]} nums2
  # @param {Integer[]} nums3
  # @param {Integer[]} nums4
  # @return {Integer}
  def four_sum_count(nums1, nums2, nums3, nums4)
      m = {}
      nums1.each do |n1|
        nums2.each do |n2|
          m[n1 + n2] = (m[n1 + n2] || 0) + 1;
        end
      end
      count = 0
      nums3.each do |n3|
        nums4.each do |n4|
          count += (m[0 - (n3 + n4)] || 0);
        end
      end
      count
  end
  

Complexities

• Time: O(n^2)
• Space: O(n^2)