The core challenge of this problem is to count the number of permutations of length N that have exactly K inversions. An inversion is defined as a pair of indices (i, j) such that i < j and p(i) > p(j). This problem is significant in combinatorics and has applications in sorting algorithms and sequence analysis.
Potential pitfalls include misunderstanding the definition of inversions and not accounting for all possible permutations.
To solve this problem, we can use dynamic programming. The idea is to build a table where dp[n][k] represents the number of permutations of length n with exactly k inversions.
1. **Naive Solution**: Generate all permutations of length N and count the inversions for each. This approach is not optimal due to its factorial time complexity.
2. **Optimized Solution**: Use dynamic programming to build the solution iteratively. This approach is more efficient and avoids the need to generate all permutations explicitly.
1. Initialize a 2D list dp where dp[n][k] will store the number of permutations of length n with exactly k inversions.
2. Base case: dp[0][0] = 1 (There is one permutation of length 0 with 0 inversions).
3. Fill the dp table using the relation:
dp[n][k] = dp[n-1][k] + dp[n-1][k-1] + ... + dp[n-1][k-(n-1)]
4. The final answer will be dp[N][K].
def count_permutations_with_k_inversions(N, K):
# Initialize the dp table with zeros
dp = [[0] * (K + 1) for _ in range(N + 1)]
# Base case: one permutation of length 0 with 0 inversions
dp[0][0] = 1
# Fill the dp table
for n in range(1, N + 1):
for k in range(K + 1):
dp[n][k] = 0
for i in range(min(k, n - 1) + 1):
dp[n][k] += dp[n - 1][k - i]
return dp[N][K]
# Example usage
N = 4
K = 2
print(count_permutations_with_k_inversions(N, K)) # Output: 5
The time complexity of this approach is O(N^2 * K) because we have two nested loops iterating over N and K, and an inner loop that can run up to N times. The space complexity is O(N * K) due to the storage required for the dp table.
1. N = 0 or K = 0: The function should handle these gracefully, returning 1 if K = 0 and N = 0, and 0 otherwise.
2. K > N * (N - 1) / 2: This is impossible, so the function should return 0.
To test the solution comprehensively, consider the following test cases:
assert count_permutations_with_k_inversions(4, 2) == 5
assert count_permutations_with_k_inversions(3, 0) == 1
assert count_permutations_with_k_inversions(3, 3) == 1
assert count_permutations_with_k_inversions(5, 10) == 1
assert count_permutations_with_k_inversions(5, 11) == 0
1. Break down the problem into smaller subproblems and solve them iteratively.
2. Use dynamic programming to store intermediate results and avoid redundant calculations.
3. Practice similar problems to improve your understanding of combinatorial algorithms and dynamic programming.
Understanding and solving problems involving permutations and inversions is crucial for mastering combinatorial algorithms. By using dynamic programming, we can efficiently solve these problems and gain deeper insights into sequence analysis and sorting algorithms.
1. Inversion (Discrete Mathematics) - Wikipedia