yokolet's notelets

  1. Arrays
  2. 4 Sum II

Arrays

4 Sum II

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

  • 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)
Medium
Array
Hash Table