How do you solve Count the Number of Inversions on LeetCode?

Count the Number of Inversions (LeetCode #3193) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k) time complexity and O(n * k) space complexity. Key insight: Dynamic programming can significantly reduce the time complexity for counting permutations with specific properties.

What is the brute force solution for Count the Number of Inversions?

The brute force approach for Count the Number of Inversions is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute-force approach generates all possible permutations of the array and checks each permutation against the inversion requirements. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n * k).

What algorithm or data structure is used to solve Count the Number of Inversions?

Count the Number of Inversions uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Combinatorial Counting.

What are the interview tips for Count the Number of Inversions?

Always clarify the problem requirements and constraints before diving into coding. Discuss your thought process and approach with the interviewer to demonstrate your understanding. Practice explaining your code and logic clearly, as communication is key in interviews.

What are common mistakes when solving Count the Number of Inversions?

Failing to account for the modulo operation when counting large numbers. Not properly initializing the DP table or misunderstanding the base cases.

#3193

Count the Number of Inversions

Hard

ArrayDynamic ProgrammingDynamic ProgrammingCombinatorial Counting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n * k)
Space	O(n)	O(n * k)

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Intuition

Time O(n * k)Space O(n * k)

Instead of generating permutations, we use dynamic programming to count valid permutations directly based on the inversion requirements. This is efficient and avoids the combinatorial explosion of the brute-force approach.

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Algorithm

3 steps

1Step 1: Initialize a DP table where dp[i][j] represents the number of ways to arrange i elements with j inversions.
2Step 2: Fill the DP table using the relation: dp[i][j] = dp[i-1][j] + dp[i-1][j-1] + ... + dp[i-1][0].
3Step 3: For each requirement, sum the valid configurations from the DP table that match the inversion count.

solution.py15 lines

1def count_valid_permutations(n, requirements):
2    MOD = 10**9 + 7
3    dp = [[0] * (400 + 1) for _ in range(n + 1)]
4    dp[0][0] = 1
5    for i in range(1, n + 1):
6        for j in range(400 + 1):
7            dp[i][j] = dp[i - 1][j]
8            if j > 0:
9                dp[i][j] = (dp[i][j] + dp[i][j - 1]) % MOD
10            if j >= i:
11                dp[i][j] = (dp[i][j] - dp[i - 1][j - i]) % MOD
12    result = 1
13    for end, cnt in requirements:
14        result = (result * dp[end + 1][cnt]) % MOD
15    return result

ℹ

Complexity note: This complexity arises from filling the DP table where n is the number of elements and k is the maximum number of inversions (400 in this case).

1Dynamic programming can significantly reduce the time complexity for counting permutations with specific properties.
2Understanding how to build and use a DP table is crucial for solving combinatorial problems efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.