NYUer | LeetCode454 4Sum II

LeetCode454 4Sum II

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Author: Stefan Su
Create time: 2022-10-31 15:16:25
Location: New York City, NY, USA

Description Medium

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

Constrains

• 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

Analysis

1. Define a unordered_hash_map. The key is sum of each pair in nums1 and nums2. The value is the time of each pair occurs.
2. Go through nums1 and nums2. Calculate sum of each pair and time of occurrence. Store in the unordered_hash_map.
3. Define an integer count, which is used to count the time of n1 + n2 + n3 + n4 = 0 (also n1 + n2 = - (n3 + n4))
4. And then, go through nums3 and nums4, if 0 - (n3 + n4) occurs in unordered_hash_map, count += unordered_hash_map[0 - (n3 + n4)]
5. Return count

Solution

• unordered hash map version

class Solution(object):
    def fourSumCount(self, nums1, nums2, nums3, nums4):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :type nums3: List[int]
        :type nums4: List[int]
        :rtype: int
        """
        unordered_hash_map = dict()

        # use unordered hash map to store the sum of each element in nums1 and nums2
        # key is sum, value is how much time the value of the sum occurs
        for n1 in nums1:
            for n2 in nums2:
                if n1 + n2 in unordered_hash_map:
                    unordered_hash_map[n1 + n2] += 1
                else:
                    unordered_hash_map[n1 + n2] = 1
        
        # if [0 - (n1 + n2)] exists in (-nums3[i] - nums4[j]), then increment count by 1
        count = 0
        for n3 in nums3:
            for n4 in nums4:
                key = -n3 - n4
                if key in unordered_hash_map:
                    count += unordered_hash_map[key]

        return count

Hopefully, this blog can inspire you when solving LeetCode454. For any questions, please comment below.