LeetCode #3017 — HARD

Count the Number of Houses at a Certain Distance II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given three positive integers n, x, and y.

In a city, there exist houses numbered 1 to n connected by n streets. There is a street connecting the house numbered i with the house numbered i + 1 for all 1 <= i <= n - 1 . An additional street connects the house numbered x with the house numbered y.

For each k, such that 1 <= k <= n, you need to find the number of pairs of houses (house₁, house₂) such that the minimum number of streets that need to be traveled to reach house₂ from house₁ is k.

Return a 1-indexed array result of length n where result[k] represents the total number of pairs of houses such that the minimum streets required to reach one house from the other is k.

Note that x and y can be equal.

Example 1:

Input: n = 3, x = 1, y = 3
Output: [6,0,0]
Explanation: Let's look at each pair of houses:
- For the pair (1, 2), we can go from house 1 to house 2 directly.
- For the pair (2, 1), we can go from house 2 to house 1 directly.
- For the pair (1, 3), we can go from house 1 to house 3 directly.
- For the pair (3, 1), we can go from house 3 to house 1 directly.
- For the pair (2, 3), we can go from house 2 to house 3 directly.
- For the pair (3, 2), we can go from house 3 to house 2 directly.

Example 2:

Input: n = 5, x = 2, y = 4
Output: [10,8,2,0,0]
Explanation: For each distance k the pairs are:
- For k == 1, the pairs are (1, 2), (2, 1), (2, 3), (3, 2), (2, 4), (4, 2), (3, 4), (4, 3), (4, 5), and (5, 4).
- For k == 2, the pairs are (1, 3), (3, 1), (1, 4), (4, 1), (2, 5), (5, 2), (3, 5), and (5, 3).
- For k == 3, the pairs are (1, 5), and (5, 1).
- For k == 4 and k == 5, there are no pairs.

Example 3:

Input: n = 4, x = 1, y = 1
Output: [6,4,2,0]
Explanation: For each distance k the pairs are:
- For k == 1, the pairs are (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), and (4, 3).
- For k == 2, the pairs are (1, 3), (3, 1), (2, 4), and (4, 2).
- For k == 3, the pairs are (1, 4), and (4, 1).
- For k == 4, there are no pairs.

Constraints:

2 <= n <= 10⁵
1 <= x, y <= n

Step 01

Brute Force Baseline

Problem summary: You are given three positive integers n, x, and y. In a city, there exist houses numbered 1 to n connected by n streets. There is a street connecting the house numbered i with the house numbered i + 1 for all 1 <= i <= n - 1 . An additional street connects the house numbered x with the house numbered y. For each k, such that 1 <= k <= n, you need to find the number of pairs of houses (house1, house2) such that the minimum number of streets that need to be traveled to reach house2 from house1 is k. Return a 1-indexed array result of length n where result[k] represents the total number of pairs of houses such that the minimum streets required to reach one house from the other is k. Note that x and y can be equal.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

3
1
3

Example 2

5
2
4

Example 3

4
1
1

Core Insight

What unlocks the optimal approach

If there were no additional street connecting house <code>x</code> to house <code>y</code>, there would be <code>2 * (n - i)</code> pairs of houses at distance <code>i</code>.
The shortest distance between house <code>i</code> and house <code>j</code> (<code>j < i</code>) is along one of these paths: - <code>i -> j</code> - <code>i -> y---x -> j</code>
Try to change the distances calculated by path <code>i ->j</code> to the other path.
Can we use prefix sums to compute the answer?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3017: Count the Number of Houses at a Certain Distance II
class Solution {
    public long[] countOfPairs(int n, int x, int y) {
        --x;
        --y;
        if (x > y) {
            int temp = x;
            x = y;
            y = temp;
        }
        long[] diff = new long[n];
        for (int i = 0; i < n; ++i) {
            diff[0] += 1 + 1;
            ++diff[Math.min(Math.abs(i - x), Math.abs(i - y) + 1)];
            ++diff[Math.min(Math.abs(i - y), Math.abs(i - x) + 1)];
            --diff[Math.min(Math.abs(i - 0), Math.abs(i - y) + 1 + Math.abs(x - 0))];
            --diff[Math.min(Math.abs(i - (n - 1)), Math.abs(i - x) + 1 + Math.abs(y - (n - 1)))];
            --diff[Math.max(x - i, 0) + Math.max(i - y, 0) + ((y - x) + 0) / 2];
            --diff[Math.max(x - i, 0) + Math.max(i - y, 0) + ((y - x) + 1) / 2];
        }
        for (int i = 0; i + 1 < n; ++i) {
            diff[i + 1] += diff[i];
        }
        return diff;
    }
}

// Accepted solution for LeetCode #3017: Count the Number of Houses at a Certain Distance II
func countOfPairs(n int, x int, y int) []int64 {
	if x > y {
		x, y = y, x
	}
	A := make([]int64, n)
	for i := 1; i <= n; i++ {
		A[0] += 2
		A[min(int64(i-1), int64(math.Abs(float64(i-y)))+int64(x))] -= 1
		A[min(int64(n-i), int64(math.Abs(float64(i-x)))+1+int64(n-y))] -= 1
		A[min(int64(math.Abs(float64(i-x))), int64(math.Abs(float64(y-i)))+1)] += 1
		A[min(int64(math.Abs(float64(i-x)))+1, int64(math.Abs(float64(y-i))))] += 1
		r := max(int64(x-i), 0) + max(int64(i-y), 0)
		A[r+int64((y-x+0)/2)] -= 1
		A[r+int64((y-x+1)/2)] -= 1
	}
	for i := 1; i < n; i++ {
		A[i] += A[i-1]
	}

	return A
}

# Accepted solution for LeetCode #3017: Count the Number of Houses at a Certain Distance II
class Solution:
    def countOfPairs(self, n: int, x: int, y: int) -> List[int]:
        if abs(x - y) <= 1:
            return [2 * x for x in reversed(range(n))]
        cycle_len = abs(x - y) + 1
        n2 = n - cycle_len + 2
        res = [2 * x for x in reversed(range(n2))]
        while len(res) < n:
            res.append(0)
        res2 = [cycle_len * 2] * (cycle_len >> 1)
        if not cycle_len & 1:
            res2[-1] = cycle_len
        res2[0] -= 2
        for i in range(len(res2)):
            res[i] += res2[i]
        if x > y:
            x, y = y, x
        tail1 = x - 1
        tail2 = n - y
        for tail in (tail1, tail2):
            if not tail:
                continue
            i_mx = tail + (cycle_len >> 1)
            val_mx = 4 * min((cycle_len - 3) >> 1, tail)
            i_mx2 = i_mx - (1 - (cycle_len & 1))
            res3 = [val_mx] * i_mx
            res3[0] = 0
            res3[1] = 0
            if not cycle_len & 1:
                res3[-1] = 0
            for i, j in enumerate(range(4, val_mx, 4)):
                res3[i + 2] = j
                res3[i_mx2 - i - 1] = j
            for i in range(1, tail + 1):
                res3[i] += 2
            if not cycle_len & 1:
                mn = cycle_len >> 1
                for i in range(mn, mn + tail):
                    res3[i] += 2
            for i in range(len(res3)):
                res[i] += res3[i]
        return res

// Accepted solution for LeetCode #3017: Count the Number of Houses at a Certain Distance II
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3017: Count the Number of Houses at a Certain Distance II
// class Solution {
//     public long[] countOfPairs(int n, int x, int y) {
//         --x;
//         --y;
//         if (x > y) {
//             int temp = x;
//             x = y;
//             y = temp;
//         }
//         long[] diff = new long[n];
//         for (int i = 0; i < n; ++i) {
//             diff[0] += 1 + 1;
//             ++diff[Math.min(Math.abs(i - x), Math.abs(i - y) + 1)];
//             ++diff[Math.min(Math.abs(i - y), Math.abs(i - x) + 1)];
//             --diff[Math.min(Math.abs(i - 0), Math.abs(i - y) + 1 + Math.abs(x - 0))];
//             --diff[Math.min(Math.abs(i - (n - 1)), Math.abs(i - x) + 1 + Math.abs(y - (n - 1)))];
//             --diff[Math.max(x - i, 0) + Math.max(i - y, 0) + ((y - x) + 0) / 2];
//             --diff[Math.max(x - i, 0) + Math.max(i - y, 0) + ((y - x) + 1) / 2];
//         }
//         for (int i = 0; i + 1 < n; ++i) {
//             diff[i + 1] += diff[i];
//         }
//         return diff;
//     }
// }

// Accepted solution for LeetCode #3017: Count the Number of Houses at a Certain Distance II
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3017: Count the Number of Houses at a Certain Distance II
// class Solution {
//     public long[] countOfPairs(int n, int x, int y) {
//         --x;
//         --y;
//         if (x > y) {
//             int temp = x;
//             x = y;
//             y = temp;
//         }
//         long[] diff = new long[n];
//         for (int i = 0; i < n; ++i) {
//             diff[0] += 1 + 1;
//             ++diff[Math.min(Math.abs(i - x), Math.abs(i - y) + 1)];
//             ++diff[Math.min(Math.abs(i - y), Math.abs(i - x) + 1)];
//             --diff[Math.min(Math.abs(i - 0), Math.abs(i - y) + 1 + Math.abs(x - 0))];
//             --diff[Math.min(Math.abs(i - (n - 1)), Math.abs(i - x) + 1 + Math.abs(y - (n - 1)))];
//             --diff[Math.max(x - i, 0) + Math.max(i - y, 0) + ((y - x) + 0) / 2];
//             --diff[Math.max(x - i, 0) + Math.max(i - y, 0) + ((y - x) + 1) / 2];
//         }
//         for (int i = 0; i + 1 < n; ++i) {
//             diff[i + 1] += diff[i];
//         }
//         return diff;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.