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
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
Solution
Method 1 - Using Hashmap
Here is the approach:
- HashMap for Sums: Use a HashMap to store the sums of every pair of elements from
A
andB
. - Checking Complements: For each pair of elements from
nums3
andnums4
, check if the negative of their sum exists in the HashMap. - Counting Matches: If the sum exists, increment the count by the number of times that sum appears in the HashMap.
Code
Java
public class Solution {
public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
Map<Integer, Integer> map = new HashMap<>(); // HashMap to store sums of elements from nums1 and nums2
int n = nums1.length;
int count = 0; // Counter to keep track of valid quadruples
// Compute all possible sums of pairs from arrays nums1 and nums2
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int sum = nums1[i] + nums2[j];
map.put(sum, map.getOrDefault(sum, 0) + 1);
}
}
// Compute sums of pairs from arrays nums3 and nums4 and check for complements in the map
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int sum = nums3[i] + nums4[j];
if (map.containsKey(-sum)) {
count += map.get(-sum); // Increment count by the number of times the complement appears
}
}
}
return count; // Return the total count of valid quadruples
}
public static void main(String[] args) {
Solution sol = new Solution();
int[] nums1 = {1, 2};
int[] nums2 = {-2, -1};
int[] nums3 = {-1, 2};
int[] nums4 = {0, 2};
int result = sol.fourSumCount(nums1, nums2, nums3, nums4);
System.out.println("Number of quadruplets: " + result); // Expected output: 2
}
}
Python
class Solution:
def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
freq_map = defaultdict(int) # Dictionary to store sums of elements from nums1 and nums2
n = len(nums1)
count = 0 # Counter to keep track of valid quadruples
# Compute all possible sums of pairs from arrays nums1 and nums2
for num1 in nums1:
for num2 in nums2:
freq_map[num1 + num2] += 1
# Compute sums of pairs from arrays nums3 and nums4 and check for complements in the dictionary
for num3 in nums3:
for num4 in nums4:
target = -(num3 + num4)
if target in freq_map:
count += freq_map[target] # Increment count by the number of times the complement appears
return count # Return the total count of valid quadruples
Complexity
- ⏰ Time complexity:
O(n^2)
since we compute sums of elements pairs for two arrays and then check pairs from the other two arrays. - 🧺 Space complexity:
O(n^2)
for storing sums in the HashMap