Dare2Solve
The problem requires counting permutations of an array such that certain segments of the array have specified inversion counts. An inversion in an array is a pair of indices where the earlier index has a larger value than the later index. Given this, the goal is to find permutations of [0, 1, 2, ..., n - 1] that meet these inversion requirements for specified segments.
The core idea is to use dynamic programming to count the number of valid permutations while ensuring that the specified inversion counts are satisfied. We use a DP table where each entry dp[i][j] represents the number of ways to arrange the first i elements with exactly j inversions.
dp where dp[i][j] stores the number of ways to arrange the first i elements of the permutation such that there are exactly j inversions. Also, create an array req to store the required inversion counts for each segment.req array with the given requirements. For each requirement [endi, cnti], set req[endi + 1] = cnti.dp[1][0] to 1 because there's only one way to arrange a single element with zero inversions.2 to n and for each element, calculate the possible number of inversions. For each possible inversion count j up to 400 (as a heuristic upper bound), update the DP table by considering all positions where the new element can be inserted to create additional inversions.dp[n][req[n]], which gives the number of permutations of the entire array satisfying the inversion count for the final segment.O(n^3). This is because there are n segments to process, and for each segment, we may need to iterate over possible positions and inversion counts up to O(n^2) times in the worst case.O(n). We use a DP table of size n x 401 to store the number of ways to form valid permutations with specific inversion counts.class Solution {
public:
int numberOfPermutations(int n, vector<vector<int>>& requirements) {
vector<vector<int>> dp(n + 1, vector<int>(401, 0));
vector<int> req(n + 1, -1);
for (auto& r : requirements) {
req[r[0] + 1] = r[1];
}
const int mod = 1e9 + 7;
if (req[1] <= 0) {
dp[1][0] = 1;
}
for (int i = 2; i <= n; i++) {
for (int c = 0; c < i; c++) {
for (int j = 0; j + c <= 400; j++) {
if (req[i] != -1 && j + c != req[i]) {
continue;
}
dp[i][j + c] += dp[i - 1][j];
dp[i][j + c] %= mod;
}
}
}
return dp[n][req[n]];
}
};
class Solution:
def numberOfPermutations(self, n: int, requirements: List[List[int]]) -> int:
dp = [[0] * 401 for _ in range(n + 1)]
req = [-1] * (n + 1)
for a, b in requirements:
req[a + 1] = b
if req[1] <= 0:
dp[1][0] = 1
for i in range(2, n + 1):
for c in range(i):
for j in range(401 - c):
if req[i] != -1 and j + c != req[i]:
continue
dp[i][j + c] += dp[i - 1][j]
dp[i][j + c] %= 1000000007
return dp[n][req[n]]
class Solution {
public int numberOfPermutations(int n, int[][] requirements) {
int[][] dp = new int[n + 1][401];
int[] req = new int[n + 1];
Arrays.fill(req, -1);
for (int[] requirement : requirements) {
int a = requirement[0];
int b = requirement[1];
req[a + 1] = b;
}
if (req[1] <= 0) {
dp[1][0] = 1;
}
for (int i = 2; i <= n; i++) {
for (int c = 0; c < i; c++) {
for (int j = 0; j + c <= 400; j++) {
if (req[i] != -1 && j + c != req[i]) continue;
dp[i][j + c] += dp[i - 1][j];
dp[i][j + c] %= 1000000007;
}
}
}
return dp[n][req[n]];
}
}
/**
* @param {number} n
* @param {number[][]} requirements
* @return {number}
*/
var numberOfPermutations = function (n, requirements) {
let dp = Array.from({ length: n + 1 }, () => new Array(401).fill(0));
let req = new Array(n + 1).fill(-1);
for (let [a, b] of requirements) {
req[a + 1] = b;
}
if (req[1] <= 0) dp[1][0] = 1;
for (let i = 2; i <= n; i++) {
for (let c = 0; c < i; c++) {
for (let j = 0; j + c <= 400; j++) {
if (req[i] !== -1 && j + c !== req[i]) continue;
dp[i][j + c] += dp[i - 1][j];
dp[i][j + c] %= 1e9 + 7;
}
}
}
return dp[n][req[n]];
};
func numberOfPermutations(n int, requirements [][]int) int {
dp := make([][]int, n+1)
for i := range dp {
dp[i] = make([]int, 401)
}
req := make([]int, n+1)
for i := range req {
req[i] = -1
}
for _, r := range requirements {
a := r[0]
b := r[1]
req[a+1] = b
}
const mod = 1e9 + 7
if req[1] <= 0 {
dp[1][0] = 1
}
for i := 2; i <= n; i++ {
for c := 0; c < i; c++ {
for j := 0; j+c <= 400; j++ {
if req[i] != -1 && j+c != req[i] {
continue
}
dp[i][j+c] += dp[i-1][j]
dp[i][j+c] %= mod
}
}
}
return dp[n][req[n]]
}
public class Solution {
public int NumberOfPermutations(int n, int[][] requirements) {
int[,] dp = new int[n + 1, 401];
int[] req = new int[n + 1];
Array.Fill(req, -1);
foreach (var requirement in requirements) {
int a = requirement[0];
int b = requirement[1];
req[a + 1] = b;
}
if (req[1] <= 0) {
dp[1, 0] = 1;
}
for (int i = 2; i <= n; i++) {
for (int c = 0; c < i; c++) {
for (int j = 0; j + c <= 400; j++) {
if (req[i] != -1 && j + c != req[i]) continue;
dp[i, j + c] += dp[i - 1, j];
dp[i, j + c] %= 1000000007;
}
}
}
return dp[n, req[n]];
}
}
In conclusion, the problem of counting permutations of an array that meet specific inversion requirements for certain segments can be effectively solved using dynamic programming. By leveraging a DP table to track the number of ways to arrange elements with a given number of inversions, and by carefully managing the inversion requirements for each segment, we can efficiently compute the desired count of valid permutations. The approach ensures that only valid configurations are considered, adhering to the constraints provided. The final result is obtained from the DP table, providing the number of permutations that satisfy the given conditions, modulo (10^9 + 7).
This solution offers a structured and scalable way to tackle permutation problems with constraints, making it a powerful technique for similar combinatorial challenges.