algoadvance

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

Clarifying Questions

1. Input Constraints:
   • Can the integers in the arrays be negative?
     • Yes.
   • What is the range of the length n of each array?
     • Typically, the problem constraints might be 1 <= n <= 200.
2. Output:
   • Return a single integer representing the number of valid tuples (i, j, k, l).

Strategy

1. Brute Force Approach:
   • Iterate through all possible tuples (i, j, k, l) and check if the sum of their corresponding values in the arrays equals 0.
   • Time Complexity: (O(n^4)), which will be inefficient for larger values of n (e.g., if n = 200, then (200^4 = 1.6 \times 10^{10}) operations).
2. Optimized Approach:
   • We can reduce the complexity to (O(n^2)) using a hash map.
   • Create a hash map to store the sums of all pairs from nums1 and nums2 and their counts.
   • Then for each pair in nums3 and nums4, check if the negative of their sum exists in the hash map.

Code

Here is the implementation of the optimized approach:

from collections import defaultdict

def fourSumCount(nums1, nums2, nums3, nums4):
    count_map = defaultdict(int)
    
    # Store sums of all pairs from nums1 and nums2
    for a in nums1:
        for b in nums2:
            count_map[a + b] += 1
    
    count = 0
    
    # Check sums of all pairs from nums3 and nums4
    for c in nums3:
        for d in nums4:
            target = -(c + d)
            if target in count_map:
                count += count_map[target]
    
    return count

# Example usage
nums1 = [1, 2]
nums2 = [-2, -1]
nums3 = [-1, 2]
nums4 = [0, 2]
print(fourSumCount(nums1, nums2, nums3, nums4))  # Output: 2

Time Complexity

• Preprocessing Step:
  • Storing sums of pairs (a+b) in count_map from nums1 and nums2 takes (O(n^2)).
• Query Step:
  • Checking sums of pairs (c+d) from nums3 and nums4 against count_map also takes (O(n^2)).
• Overall Time Complexity: (O(n^2)).

This approach leverages hash map lookups for efficient counting and ensures we can handle larger inputs more effectively.

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