Count Special Quadruplets

Count Special Quadruplets

What is this problem about?

The "Count Special Quadruplets interview question" is an array-based counting problem. You are given an array of integers and need to find the number of distinct quadruplets of indices $(a, b, c, d)$ such that $a < b < c < d$ and nums[a] + nums[b] + nums[c] == nums[d]. This is a variation of the classic "3Sum" problem, but with the target value being an element of the array itself located at a later index.

Why is this asked in interviews?

Companies like Microsoft and Meta ask the "Count Special Quadruplets coding problem" to test a candidate's ability to optimize nested loops. While a brute-force $O(N^4)$ solution is possible for small arrays, interviewers often look for $O(N^3)$ or even $O(N^2)$ solutions using hash maps. It evaluates "Hash Table interview pattern" and "Enumeration" skills.

Algorithmic pattern used

This problem can be solved using Optimized Enumeration with a Hash Map.

1. Iterative Search: The condition is $nums[a] + nums[b] = nums[d] - nums[c]$.
2. Reverse Pass Logic:
   • Iterate through index $b$ from right to left.
   • For a fixed $b$, we look at all $a < b$ to find sums $nums[a] + nums[b]$.
   • Simultaneously, we maintain a map of all differences $nums[d] - nums[c]$ where $d > c > b$.
3. Simplification ($O(N^3)$): Most candidates solve this in $O(N^3)$ by fixing $a, b, c$ and counting how many $nums[d]$ equal the sum in the suffix of the array.

Example explanation

Array: [1, 2, 3, 6]

• $a=0, b=1, c=2$: $1 + 2 + 3 = 6$.
• Check if index 3 is '6'. Yes. Result: 1. Array: [1, 1, 1, 3, 5]
• $a=0, b=1, c=2$: $1+1+1=3$. $3$ is at index 3. (Count = 1)
• No other quadruplet sums to a value at a later index. Result: 1.

Common mistakes candidates make

• Index Order: Forgetting the $a < b < c < d$ constraint and including pairs that appear out of order.
• Brute Force TLE: Using four nested loops on an array of size 5000 (though usually the constraints for this specific problem are small enough for $O(N^3)$).
• Duplicate Values: Failing to handle multiple occurrences of the same sum.

Interview preparation tip

Whenever you need to find a sum of elements matching another element, think about splitting the equation ($A+B+C=D ightarrow A+B = D-C$). This "Two Sum" style transformation is a powerful way to reduce the complexity of counting problems.