#3193

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Count the Number of Inversions

Count the number of valid permutations satisfying inversion constraints using state transition dynamic programming.

Problem Statement

You are given an integer n and a 2D array requirements, where requirements[i] = [endi, cnti] represents the end index and the inversion count for a specific subarray of length endi+1. Your task is to determine the number of permutations of the array [0, 1, 2, ..., n-1] that satisfy the inversion conditions specified by all the entries in the requirements array.

A pair of indices (i, j) from an array nums is considered an inversion if i < j and nums[i] > nums[j]. For each requirement requirements[i], the number of inversions in the first endi+1 elements of the permutation should be exactly cnti. Return the total number of valid permutations that satisfy all the inversion conditions.

Examples

Example 1

Input: n = 3, requirements = [[2,2],[0,0]]

Output: 2

The two permutations are:

Example 2

Input: n = 3, requirements = [[2,2],[1,1],[0,0]]

Output: 1

The only satisfying permutation is [2, 0, 1] :

Example 3

Input: n = 2, requirements = [[0,0],[1,0]]

Output: 1

The only satisfying permutation is [0, 1] :

Constraints

2 <= n <= 300
1 <= requirements.length <= n
requirements[i] = [endi, cnti]
0 <= endi <= n - 1
0 <= cnti <= 400
The input is generated such that there is at least one i such that endi == n - 1.
The input is generated such that all endi are unique.

Solution Approach

Dynamic Programming with State Transitions

Use dynamic programming (DP) to solve the problem. Let dp[i][j] denote the number of ways to form a permutation of length i with exactly j inversions. The state transition can be derived from previous states, accounting for the number of inversions added by each new element.

Efficient Inversion Calculation

The key challenge is efficiently calculating the number of inversions in any prefix of the permutation. A Fenwick Tree or similar data structure can help keep track of the inversion counts and allow quick updates as new elements are added to the permutation.

Handling Multiple Requirements

For each requirement, adjust the DP table by ensuring that the required number of inversions is respected. Each requirement can be treated as a constraint on the number of inversions within a specific prefix, and dynamic programming is used to propagate valid counts across the table.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach used for dynamic programming and inversion counting. With an efficient inversion counting technique, the complexity can be reduced to O(n^2 * m), where n is the array length and m is the maximum inversion count, although optimizations can improve this further.

What Interviewers Usually Probe

Candidate understands the concept of dynamic programming for state transitions.
Candidate can efficiently calculate and update inversion counts using data structures like Fenwick Tree.
Candidate identifies the challenge of handling multiple requirements and can incorporate them into the DP approach.

Common Pitfalls or Variants

Common pitfalls

Not accounting for all inversion constraints, which can lead to incorrect permutation counts.
Inefficient inversion counting leading to excessive time complexity.
Failure to properly update the DP table when new elements are added, causing incorrect counts of valid permutations.

Follow-up variants

Modify the problem to handle permutations with additional constraints on element values.
Add a requirement to count the number of inversions for subsets of the array, not just prefixes.
Allow the inversion count to be variable for each prefix, testing how the algorithm handles different ranges of inversion constraints.

FAQ

How can dynamic programming help solve 'Count the Number of Inversions'?

Dynamic programming helps by storing intermediate results for permutation lengths and inversion counts, avoiding recomputation and improving efficiency.

What is the best data structure to track inversions in this problem?

A Fenwick Tree or Binary Indexed Tree (BIT) is optimal for tracking inversions because it allows for efficient updates and prefix sum queries.

How do multiple inversion constraints affect the dynamic programming solution?

Multiple constraints must be integrated into the DP solution by ensuring that the inversion counts match the specified requirements for each prefix.

What is the time complexity of solving this problem?

The time complexity can be O(n^2 * m), where n is the array length and m is the maximum inversion count. Optimizations may reduce this complexity.

What are the common pitfalls in solving 'Count the Number of Inversions'?

Common pitfalls include inefficient inversion counting, failure to correctly update the DP table, and missing constraints for specific prefixes.

terminal

Solution

Solution 1: Dynamic Programming

We define $f[i][j]$ as the number of permutations of $[0..i]$ with $j$ inversions. Consider the relationship between the number $a_i$ at index $i$ and the previous $i$ numbers. If $a_i$ is smaller than $k$ of the previous numbers, then each of these $k$ numbers forms an inversion pair with $a_i$, contributing to $k$ inversions. Therefore, we can derive the state transition equation:

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class Solution:
    def numberOfPermutations(self, n: int, requirements: List[List[int]]) -> int:
        req = [-1] * n
        for end, cnt in requirements:
            req[end] = cnt
        if req[0] > 0:
            return 0
        req[0] = 0
        mod = 10**9 + 7
        m = max(req)
        f = [[0] * (m + 1) for _ in range(n)]
        f[0][0] = 1
        for i in range(1, n):
            l, r = 0, m
            if req[i] >= 0:
                l = r = req[i]
            for j in range(l, r + 1):
                for k in range(min(i, j) + 1):
                    f[i][j] = (f[i][j] + f[i - 1][j - k]) % mod
        return f[n - 1][req[n - 1]]

Continue Practicing

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