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454, Four Sum II

MikeAbout 1 minhashtablemediumarrayhash table

I Problem

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[j], nums3[k], nums4[l] <= 2²⁸

Related Topics

  • Array
  • Hash Table

II Solution

Approach 1: Use Hash

/// Time Complexity: O(n^2)
///
/// Space Complexity: O(n^2)
pub fn four_sum_count(nums1: Vec<i32>, nums2: Vec<i32>, nums3: Vec<i32>, nums4: Vec<i32>) -> i32 {
    let len = nums1.len();
    let mut map = HashMap::with_capacity(len * len);
    for i in 0..len {
        for j in 0..len {
            map.entry(nums1[i] + nums2[j])
                .and_modify(|v| *v += 1)
                .or_insert(1);
        }
    }

    let mut res = 0;
    for k in 0..len {
        for l in 0..len {
            if let Some(v) = map.get(&(0 - nums3[k] - nums4[l])) {
                res += *v;
            }
        }
    }

    res
}