Count Increasing Quadruplets

Count Increasing Quadruplets

What is this problem about?

The "Count Increasing Quadruplets" coding problem asks you to find the number of indices $(i, j, k, l)$ such that $i < j < k < l$ and the values at these indices follow the pattern: $nums[i] < nums[k] < nums[j] < nums[l]$. Note the specific swap in the middle: $nums[k]$ must be smaller than $nums[j]$, even though $k$ comes after $j$.

Why is this asked in interviews?

This "Hard" problem is a favorite at companies like SAP and Deutsche Bank because it tests the ability to optimize a counting problem using precomputation. A naive $O(n^4)$ approach is trivial but useless for large arrays. To solve this, you must use Prefix Sums or Dynamic Programming to count "potential" pairs on the left and right, which requires strong analytical thinking.

Algorithmic pattern used

The pattern is Precomputation with Prefix Sums or a 2D BIT. The core idea is to iterate through all possible pairs $(j, k)$. For each fixed $j$ and $k$ where $nums[k] < nums[j]$, you need to know:

1. How many $i < j$ satisfy $nums[i] < nums[k]$.
2. How many $l > k$ satisfy $nums[l] > nums[j]$. By precomputing these counts in $O(n^2)$, you can calculate the total quadruplets in $O(n^2)$ by simply multiplying these two counts for every valid $(j, k)$ pair.

Example explanation

Consider an array where we find a pair $j=1, k=2$ such that $nums[k]=2$ and $nums[j]=5$.

• We look at indices $i < 1$. Suppose we find one index where $nums[i] = 1$ (which is $< 2$). Count_left = 1.
• We look at indices $l > 2$. Suppose we find two indices where $nums[l] = 6$ and $7$ (both are $> 5$). Count_right = 2.
• The number of quadruplets using this specific $(j, k)$ pair is $1 imes 2 = 2$.

Common mistakes candidates make

The most common mistake is trying to solve the conditions $nums[i] < nums[j] < nums[k] < nums[l]$ (a standard increasing quadruplet), failing to notice the $nums[k] < nums[j]$ twist. Another error is trying to use a BIT/Segment tree in a way that leads to $O(n^3)$ or $O(n^2 \log n)$, which might still be too slow compared to the $O(n^2)$ prefix sum approach.

Interview preparation tip

When faced with counting multi-index patterns, try fixing the "middle" elements (like $j$ and $k$). Once they are fixed, the problem often reduces to counting elements on the left and right that satisfy independent conditions, which is much easier to optimize with precomputation.