Number of permutation with K inversions

12 Jan, 2022
Given an array, an inversion is defined as a pair a[i], a[j] such that a[i] > a[j] and i < j. We are given two numbers N and k, we need to tell how many permutations of the first N number have exactly K inversion.
Examples: 
Input  : N = 3, K = 1
Output : 2
Explanation :  
Total Permutation of first N number,
123, 132, 213, 231, 312, 321
Permutation with 1 inversion : 132 and 213

Input  : N = 4, K = 2
Output : 5
A Naïve way to solve this problem is noting down all permutations then checking the count of inversion in them but iterating through permutation itself will take O(N!) time, which is too large. 
We can solve this problem using a dynamic programming approach. Below is the recursive formula.  
If N is 0, Count(0, K) = 0

If K is 0, Count(N, 0) = 1 (Only sorted array)

In general case, 
If we have N number and require K inversion, 
Count(N, K) = Count(N - 1, K) + 
              Count(N – 1, K - 1) + 
              Count(N – 1, K – 2) + 
              .... + 
              Count(N – 1, 0)
How does the above recursive formula work? If we have N number and want to have K permutation and suppose all permutation of (N – 1) number are written somewhere, the new number (Nth number and largest) need to be placed in all permutation of (N – 1) number and those whose inversion count becomes K after adding this number should be added in our answer. Now take those sets of permutation of (N – 1) number which has let (K – 3) inversion, now we can place this new largest number at position 3 from last, then inversion count will be K, so count(N – 1, K – 3) should be added to our answer, the same argument can be given for another inversion also and we will reach to above recursion as the final answer.
The below code is written following the above recursion in a memorization way.  
C++
Java
Python3
C#
PHP
Javascript
// C++ program to find number of permutation// with K inversion using Memoization#include <bits/stdc++.h>using namespace std; // Limit on N and Kconst int M = 100; // 2D array memo for stopping// solving same problem againint memo[M][M]; // method recursively calculates// permutation with K inversionint numberOfPermWithKInversion(int N, int K){         // base cases    if (N == 0)        return 0;    if (K == 0)        return 1;     // if already solved then    // return result directly    if (memo[N][K] != 0)        return memo[N][K];     // calling recursively all subproblem    // of permutation size N - 1    int sum = 0;    for (int i = 0; i <= K; i++)    {         // Call recursively only        // if total inversion        // to be made are less        // than size        if (i <= N - 1)            sum += numberOfPermWithKInversion(N - 1,                                              K - i);    }     // store result into memo    memo[N][K] = sum;     return sum;} // Driver codeint main(){    int N = 4;    int K = 2;    cout << numberOfPermWithKInversion(N, K);    return 0;}
// Java program to find number of permutation with// K inversion using Memoization import java.io.*; class GFG {         // Limit on N and K    static int M = 100;     // 2D array memo for stopping solving same problem    // again    static int memo[][] = new int[M][M];     // method recursively calculates permutation with    // K inversion    static int numberOfPermWithKInversion(int N, int K)    {                 // base cases        if (N == 0)            return 0;        if (K == 0)            return 1;         // if already solved then return result directly        if (memo[N][K] != 0)            return memo[N][K];         // calling recursively all subproblem of        // permutation size N - 1        int sum = 0;        for (int i = 0; i <= K; i++) {                         // Call recursively only if total inversion            // to be made are less than size            if (i <= N - 1)                sum += numberOfPermWithKInversion(N - 1,                                                  K - i);        }         // store result into memo        memo[N][K] = sum;         return sum;    }     // Driver code to test above methods    public static void main(String[] args)    {        int N = 4;        int K = 2;        System.out.println(numberOfPermWithKInversion(N, K));    }} // This code is contributed by vt_m.
# Python3 program to find number of permutation# with K inversion using Memoization # Limit on N and KM = 100 # 2D array memo for stopping # solving same problem againmemo = [[0 for i in range(M)] for j in range(M)]  # method recursively calculates# permutation with K inversiondef numberOfPermWithKInversion(N, K):     # Base cases    if (N == 0): return 0    if (K == 0): return 1     # If already solved then    # return result directly    if (memo[N][K] != 0):        return memo[N][K]     # Calling recursively all subproblem    # of permutation size N - 1    sum = 0    for i in range(K + 1):             # Call recursively only if        # total inversion to be made        # are less than size        if (i <= N - 1):            sum += numberOfPermWithKInversion(N - 1, K - i)         # store result into memo    memo[N][K] = sum     return sum # Driver codeN = 4; K = 2print(numberOfPermWithKInversion(N, K)) # This code is contributed by Anant Agarwal.
// C# program to find number of// permutation with K inversion// using Memoizationusing System; class GFG{ // Limit on N and Kstatic int M = 100; // 2D array memo for stopping// solving same problem againstatic int [,]memo = new int[M, M]; // method recursively calculates// permutation with K inversionstatic int numberOfPermWithKInversion(int N,                                      int K){         // base cases    if (N == 0)        return 0;    if (K == 0)        return 1;     // if already solved then    // return result directly    if (memo[N, K] != 0)        return memo[N, K];     // calling recursively all    // subproblem of permutation    // size N - 1    int sum = 0;    for (int i = 0; i <= K; i++)    {                 // Call recursively only if        // total inversion to be        // made are less than size        if (i <= N - 1)            sum += numberOfPermWithKInversion(N - 1,                                              K - i);    }     // store result into memo    memo[N, K] = sum;     return sum;} // Driver Codestatic public void Main (){    int N = 4;    int K = 2;    Console.WriteLine(numberOfPermWithKInversion(N, K));}} // This code is contributed by ajit
<?php// PHP program to find number of permutation// with K inversion using Memoization // method recursively calculates// permutation with K inversionfunction numberOfPermWithKInversion($N, $K){         $memo = array();         // base cases    if ($N == 0)        return 0;    if ($K == 0)        return 1;     // if already solved then    // return result directly    if ($memo[$N][$K] != 0)        return $memo[$N][$K];     // calling recursively all subproblem    // of permutation size N - 1    $sum = 0;    for ($i = 0; $i <= $K; $i++)    {         // Call recursively only        // if total inversion        // to be made are less        // than size        if ($i <= $N - 1)            $sum += numberOfPermWithKInversion($N - 1,                                               $K - $i);    }     // store result into memo    $memo[$N][$K] = $sum;     return $sum;} // Driver code$N = 4;$K = 2;echo numberOfPermWithKInversion($N, $K); // This code is contributed// by Akanksha Rai(Abby_akku)?>
<script> // Javascript program to find number of// permutation with K inversion using// Memoization // Limit on N and Klet M = 100; // 2D array memo for stopping solving// same problem againlet memo = new Array(M);for(let i = 0; i < M; i++){    memo[i] = new Array(M);    for(let j = 0; j < M; j++)    {        memo[i][j] = 0;    }} // Method recursively calculates permutation// with K inversionfunction numberOfPermWithKInversion(N, K){         // base cases    if (N == 0)        return 0;    if (K == 0)        return 1;     // If already solved then return    // result directly    if (memo[N][K] != 0)        return memo[N][K];     // Calling recursively all subproblem of    // permutation size N - 1    let sum = 0;    for(let i = 0; i <= K; i++)    {                 // Call recursively only if total inversion        // to be made are less than size        if (i <= N - 1)            sum += numberOfPermWithKInversion(                N - 1, K - i);    }         // Store result into memo    memo[N][K] = sum;     return sum;} // Driver codelet N = 4;let K = 2; document.write(numberOfPermWithKInversion(N, K)); // This code is contributed by divyesh072019  </script>
5
Time Complexity: O(N*N*K)
Optimized Approach: Using Tabulation and cumulative sum
C++
// C++ program to find number of permutation// with K inversions #include <bits/stdc++.h>using namespace std; int numberOfPermWithKInversions(int N, int K) {    vector<vector<int>> dp(N+1,vector<int>(K+1));       // As for k=0, number of permutations is 1 for every N      for(int i = 1; i <= N; i++)        dp[i][0] = 1;       // Using Dynamic Programming with cumulative sum    for(int i = 1; i <= N; i++)    {        for(int j = 1; j <= K; j++)        {              // This is same as val = dp[i-1][j] - dp[i-1][j-i]              // i.e. dp[i-1][j........j-i], just taking care of              // boundaries            int val = dp[i-1][j];            if(j >= i)                val -= dp[i-1][j-i];             dp[i][j] = dp[i][j-1] + val;        }    }         // And, in the end calculate the dp[n][k]      // which is dp[n][k]-dp[n][k-1]    int ans = dp[N][K];    if(K >= 1)        ans -= dp[N][K-1];     return ans;} int main() {    int N = 4;    int K = 2;           cout << numberOfPermWithKInversions(N,K) << "\n";    return 0;}
5
Time Complexity: O(N*K)
This article is contributed by Utkarsh Trivedi and Ankit Kumar Sharma. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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