Count the Number of Houses at a Certain Distance II

Count the Number of Houses at a Certain Distance II

What is this problem about?

Building on its predecessor, the Count the Number of Houses at a Certain Distance II interview question asks the same question but with much larger constraints ($n$ up to $10^5$). You have $n$ houses in a line and one additional "shortcut" between house $x$ and $y$. For every distance $k \in [1, n]$, you must count how many pairs of houses $(i, j)$ have a shortest path of length exactly $k$.

Since $n$ is large, an $O(N^2)$ BFS approach is no longer feasible. You need an $O(N)$ or $O(N \log N)$ solution.

Why is this asked in interviews?

Oracle uses this "Hard" version to see if a candidate can move from a graph algorithm to a purely mathematical and constructive approach. It tests knowledge of the prefix sum interview pattern (specifically difference arrays) and the ability to manually solve cases based on geometry. It evaluations whether you can categorize pairs of houses into different "zones" (e.g., both left of the shortcut, one in the shortcut loop, etc.) and calculate their distance distributions analytically.

Algorithmic pattern used

This problem uses Difference Arrays and Case-Based Math.

1. Model the shortcut: Assume $x < y$. The houses between $x$ and $y$ form a cycle.
2. Difference Array: Use an array diff where diff[k] tracks the change in frequency for distance $k$.
3. Categorize Pairs:
   • Pairs $(i, j)$ where neither uses the shortcut: distance is $|i-j|$.
   • Pairs $(i, j)$ where the shortcut might be used: distance is $\min(|i-j|, |i-x| + 1 + |j-y|)$.
4. For each house $i$, you can mathematically determine the ranges of $j$ where each path is shorter and update the diff array for those distance ranges.
5. Finally, compute prefix sums of diff to get the actual counts.

Example explanation

$n=5, x=2, y=4$.

• The shortcut $(2, 4)$ creates a "shortcut" for pairs like $(1, 5)$.
• Distance $1 o 5$: $|1-5|=4$. With shortcut: $|1-2| + 1 + |5-4| = 1 + 1 + 1 = 3$.
• Shortest is 3.
• Instead of checking each pair, you'd calculate how many pairs fall into the "distance 3" bucket using range updates on your difference array.

Common mistakes candidates make

• Complexity: Staying stuck on $O(N^2)$ logic.
• Cycle Math: Incorrectly calculating distances within the cycle $(x o \dots o y o x)$.
• Symmetry: Forgetting to multiply by 2 if the problem asks for ordered pairs $(i, j)$ where $i e j$.

Interview preparation tip

Practice using Difference Arrays for range update problems. If a problem asks you to calculate something for every possible value of $k$, and the values are determined by intervals, a difference array can often reduce the complexity by a full factor of $N$.