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Count Special Quadruplets

Number: 2122

Difficulty: Easy

Paid? No

Companies: N/A

Problem Description

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

  • a, b, c, d are indices satisfying a < b < c < d
  • nums[a] + nums[b] + nums[c] == nums[d]

Key Insights

  • The quadruplets must respect increasing index order (a < b < c < d).
  • Brute force using four nested loops is acceptable given nums.length is at most 50.
  • Alternatively, a hash table approach can optimize the inner search by precomputing sums.
  • The constraints allow a straightforward solution without the need for heavy optimization.

Space and Time Complexity

Time Complexity: O(n^4) using a brute-force approach; however, n is at most 50 so this is acceptable. Space Complexity: O(1) with constant extra space (ignoring input storage).

Solution

The solution involves enumerating all possible quadruplets (a, b, c, d) by using four nested loops such that a < b < c < d. For each quadruplet, we check if the sum of elements at indices a, b, and c equals the element at index d. If it does, we increment our count. All code samples provided below use this approach with comprehensive inline comments.

Code Solutions

# Python solution for Count Special Quadruplets

def countQuadruplets(nums):
    count = 0  # Initialize count of valid quadruplets to 0
    n = len(nums)  # Length of the input array
    
    # Iterate through all possible quadruplets with indices a, b, c, d
    for a in range(n - 3):
        for b in range(a + 1, n - 2):
            for c in range(b + 1, n - 1):
                for d in range(c + 1, n):
                    # Check if the current quadruplet satisfies the condition
                    if nums[a] + nums[b] + nums[c] == nums[d]:
                        count += 1  # Increment count if condition is satisfied
    return count

# Example usage:
nums = [1, 1, 1, 3, 5]
print(countQuadruplets(nums))  # Expected output: 4
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