4Sum II

Author: anantkaushik

454-4sum-ii.py

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+"""
+Problem Link: https://leetcode.com/problems/4sum-ii/
+
+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
+
+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
+
+Constraints:
+n == nums1.length
+n == nums2.length
+n == nums3.length
+n == nums4.length
+1 <= n <= 200
+-228 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 228
+"""
+# O(N^2)
+class Solution:
+ def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
+ cur_sum = {}
+ 
+ for num1 in nums1:
+ for num2 in nums2:
+ cur_sum[num1+num2] = cur_sum.get(num1+num2, 0) + 1
+ 
+ count = 0 
+ for num3 in nums3:
+ for num4 in nums4:
+ count += cur_sum.get(-num3-num4, 0)
+ 
+ return count