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 various fields such as sorting algorithms, where inversions are used to measure how far a permutation is from being sorted.
Common applications include genetic algorithms, where permutations with specific properties are needed, and in the study of sorting algorithms, where the number of inversions can indicate the complexity of sorting a list.
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 factorial complexity of generating all permutations.
1. Initialize a 2D array dp where dp[n][k] represents the number of permutations of length n with exactly k inversions.
2. Base case: dp[0][0] = 1 (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].
// Function to count permutations with exactly K inversions
function countPermutationsWithKInversions(N, K) {
// Initialize a 2D array dp with zeros
const dp = Array.from({ length: N + 1 }, () => Array(K + 1).fill(0));
// Base case: one permutation of length 0 with 0 inversions
dp[0][0] = 1;
// Fill the dp table
for (let n = 1; n <= N; n++) {
for (let k = 0; k <= K; k++) {
for (let i = 0; i <= Math.min(k, n - 1); i++) {
dp[n][k] += dp[n - 1][k - i];
}
}
}
// Return the number of permutations of length N with exactly K inversions
return dp[N][K];
}
// Example usage
const N = 4;
const K = 2;
console.log(countPermutationsWithKInversions(N, K)); // Output: 5
The time complexity of this approach is O(N^2 * K) because we have nested loops iterating over N, K, and the minimum of K and N. The space complexity is O(N * K) due to the 2D dp array.
1. N = 0 or K = 0: The function should handle these gracefully, returning 1 for dp[0][0] and 0 for other cases.
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:
console.log(countPermutationsWithKInversions(4, 2)); // Output: 5
console.log(countPermutationsWithKInversions(3, 0)); // Output: 1
console.log(countPermutationsWithKInversions(3, 3)); // Output: 1
console.log(countPermutationsWithKInversions(5, 10)); // Output: 1
console.log(countPermutationsWithKInversions(5, 0)); // Output: 1
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 permutations and inversions.
Understanding and solving problems involving permutations and inversions is crucial for various applications in computer science. By using dynamic programming, we can efficiently count permutations with a given number of inversions. Practice and exploration of similar problems will further enhance your problem-solving skills.
1. Inversion (Discrete Mathematics) - Wikipedia