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This is a medium difficulty Coding & Algorithms question, commonly asked during Technical Screen rounds at Turing.

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This question is commonly asked for Software Engineer candidates at Turing during technical interviews.

Count permutations with exactly n inversions

Quick Overview

This Coding & Algorithms question evaluates combinatorial counting and algorithmic skills related to permutations and inversion counts, emphasizing dynamic programming and modular arithmetic for large-result handling.

Company: Turing

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: medium

Interview Round: Technical Screen

You are given two integers **m** and **n**. Consider all permutations of the array `[1, 2, ..., m]` (each number from 1 to m appears exactly once). A permutation `perm` (0-indexed) has an **inversion** pair `(x, y)` if: - `0 <= x < y < m`, and - `perm[x] > perm[y]` A permutation is called **special** if it contains **exactly `n` inversions**. Return the **number of special permutations** modulo **10^9 + 7**. #### Examples 1) **Input:** `m = 3, n = 0` **Output:** `1` **Explanation:** Only `[1, 2, 3]` has 0 inversions. 2) **Input:** `m = 3, n = 1` **Output:** `2` **Explanation:** `[1, 3, 2]` and `[2, 1, 3]` each have exactly 1 inversion. #### Constraints - `1 <= m <= 1000` - `0 <= n <= 1000`

Quick Answer: This Coding & Algorithms question evaluates combinatorial counting and algorithmic skills related to permutations and inversion counts, emphasizing dynamic programming and modular arithmetic for large-result handling.

You are given two integers `m` and `n`. Consider all permutations of the array `[1, 2, ..., m]` (each number from 1 to m appears exactly once). A permutation `perm` (0-indexed) has an **inversion** pair `(x, y)` if: - `0 <= x < y < m`, and - `perm[x] > perm[y]` A permutation is called **special** if it contains **exactly `n` inversions**. Return the **number of special permutations** modulo **10^9 + 7**. #### Examples 1) **Input:** `m = 3, n = 0` -> **Output:** `1` (only `[1, 2, 3]` has 0 inversions). 2) **Input:** `m = 3, n = 1` -> **Output:** `2` (`[1, 3, 2]` and `[2, 1, 3]`). #### Constraints - `1 <= m <= 1000` - `0 <= n <= 1000` #### Approach Let `dp[i][j]` be the number of permutations of `i` elements with exactly `j` inversions. Inserting the `i`-th element into a permutation of `i-1` elements can create between `0` and `i-1` new inversions, so `dp[i][j] = sum_{k=0}^{min(j, i-1)} dp[i-1][j-k]`. The inner sum is a sliding window, so maintain a prefix-sum array to evaluate each state in O(1), giving an overall O(m*n) time / O(n) space algorithm.

Constraints

1 <= m <= 1000
0 <= n <= 1000
Answer is returned modulo 10^9 + 7

Examples

Input: (3, 0)

Expected Output: 1

Explanation: Only [1, 2, 3] has 0 inversions.

Input: (3, 1)

Expected Output: 2

Explanation: [1, 3, 2] and [2, 1, 3] each have exactly 1 inversion.

Hints

Define dp[i][j] = number of permutations of i elements with exactly j inversions. The answer is dp[m][n].
Inserting the largest element into a permutation of i-1 elements at the position that leaves k elements to its right adds exactly k new inversions, with 0 <= k <= i-1. So dp[i][j] = sum_{k=0}^{min(j, i-1)} dp[i-1][j-k].
The inner summation is a sliding window of width i over the previous row. Use a prefix-sum array to compute each new[j] in O(1), reducing the total work to O(m*n). Take care with the modulo on subtraction (add MOD or rely on Python's non-negative modulo).

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