algoadvance

Given a 0-indexed integer array nums, return the number of distinct quadruplets (a, b, c, d) such that:

• nums[a] + nums[b] + nums[c] == nums[d], and
• 0 <= a, b, c, d < nums.length, and
• a, b, c, and d are distinct.

Clarifying Questions

1. Are there any constraints on the size of the array nums?
   • Typically, constraints will be provided in the problem description. But for this exercise, we’ll assume practical limits.
2. Can the solution have more time complexity than O(n^4)?
   • We strive for a more efficient solution, but initially focusing on a basic O(n^4) approach might be helpful to establish correctness.
3. Are the integers in the array nums bounded in any specific range?
   • We can assume typical constraints found in coding challenges, e.g., integers within (-10^6, 10^6).

Strategy

1. Brute Force Approach:
   • Iterate over all possible quadruplets (a, b, c, d) ensuring they are distinct indices.
   • Check if nums[a] + nums[b] + nums[c] == nums[d].
2. Optimized Approach:
   • To reduce the complexity, we can use a hashmap to store sums of pairs and then check if there exists another pair that forms the required quadruplet.

Implementation

Brute Force Approach (O(n^4))

def countQuadruplets(nums):
    n = len(nums)
    count = 0
    for a in range(n):
        for b in range(a+1, n):
            for c in range(b+1, n):
                for d in range(c+1, n):
                    if nums[a] + nums[b] + nums[c] == nums[d]:
                        count += 1
    return count

This approach is straightforward but not efficient for larger arrays.

Optimized Approach using Hashmap (O(n^3))

Here is a more optimized approach using a hashmap to store sums of pairs:

def countQuadruplets(nums):
    count = 0
    n = len(nums)
    for i in range(n-3):
        sum_map = {}
        for j in range(i+1, n-1):
            for k in range(j+1, n):
                sum_map[nums[i] + nums[j]] = sum_map.get(nums[i] + nums[j], 0) + 1
            
            for l in range(j+1, n):
                if nums[l] in sum_map:
                    count += sum_map[nums[l]]

    return count

Time Complexity

The brute force approach has a time complexity of (O(n^4)).

The optimized approach reduces this to (O(n^3)):

• The outer loop iterates (O(n)).
• For each possible value of i, the inner two loops iterate (O(n^2)) to fill the hashmap and check values.

In terms of space complexity:

• The hashmap can store up to (O(n^2)) sums, making it (O(n^2)) in the worst-case scenario.

This optimized approach is more practical for larger inputs while ensuring correctness.

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