Given an array arr[] consisting of permutation of the first N natural numbers, the task is to find the expected number of swaps to sort the given array where the probability of swapping any inversion pair is equal and the probability of swapping any non-inversion pair is 0.
Examples:
Input: arr[] = {3, 2, 1}
Output: 2.3333
Explanation: The given array can be sorted in non-decreasing order using sequences of swaps as shown in the image below.
The expected numbers of swaps of each node can be calculated by 1 + (average of expected number of swaps of all the child nodes).
Since the nodes 3, 9, 10, 11, and 12 are already sorted, the number of expected swaps to sort them are 0.
The expected number of swaps of nodes 5, 6, 7, and 8 will be 1.
Similarly, the expected number of swaps of nodes 2 and 4 will be 2.
Hence, the expected number of swaps of node 1 is 1 + (2 + 2 + 0)/3 = 1 + 1.3333 = 2.3333Input: arr[] = {1, 3, 2}
Output: 1.0
Explanation: Since there is only 1 inversion pair (arr[1], arr[2]), the expected number of swaps is 1.
Approach: The given problem can be solved with the help of Recursion and Backtracking using Memoization which is based on the following observations:
- An inversion in the array is defined as two indices (i, j) such that i < j and arr[i] > arr[j]. The expected number of swaps can be calculated recursively after swapping the integers of each valid inversion one at a time and recursively calling for the permutation after the swap until the whole permutation is sorted.
- Suppose there are K inversions in the current permutation and after swapping the ith inversion, the expected number of swaps to sort the permutation is denoted by Pi. Therefore, the expected number of swaps after swapping all possible inversions will be (P1 + P2 + … + PK)/K.
Using the above observations, the given problem can be solved by the following steps:
- Create a map, say memo that stores the expected number of swaps for a given permutation.
- Create a Recursive Function expectedSwaps(), which takes a permutation as an argument and returns the expected number of swaps.
- In the expectedSwaps() function, if the expected swaps of the current permutation are already calculated, return the answer. Otherwise, iterate over each valid inversion and swap the index of the current inversion and recursively call for the permutation after swapping.
- Find the sum of the expected swaps after each valid swap in a variable, say res, and the count of inversions in a variable, say K.
- After completing the above steps, print the value of res / K as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Stores result of each state of the// array in a memoization tablemap<vector<int>, double> memo;// Recursive function to find expected// number of swaps to sort given arraydouble expectedSwaps(vector<int> a){ // The value for the given permutation // is already calculated if (memo.count(a)) { return memo[a]; } // Stores the expected number of ways double res = 0; // Stores the total operations done int K = 0; // Iterate for all possible pairs of // Inversions for (int i = 0; i < a.size(); i++) { for (int j = i + 1; j < a.size(); j++) { // For the current inversion // find the expected value if (a[i] > a[j]) { swap(a[i], a[j]); // Recursively Call res += 1 + expectedSwaps(a); // Backtrack swap(a[i], a[j]); K++; } } } // If permutation is already sorted if (K == 0) res = 0; // Calculate expected swaps for // current permutation else res /= K; // Return answer return memo[a] = res;}// Driver Codeint main(){ int N = 3; vector<int> arr = { 3, 2, 1 }; cout << expectedSwaps(arr); return 0;} |
Java
// Java code for the above approach:import java.util.*;public class Main { // Stores result of each state of the // array in a memoization table static HashMap <ArrayList <Integer>, Double> memo; // Recursive function to find expected // number of swaps to sort given array static double expectedSwaps(ArrayList <Integer> a) { // The value for the given permutation // is already calculated if (memo.containsKey(a)) { return memo.get(a); } // Stores the expected number of ways double res = 0; // Stores the total operations done int K = 0; // Iterate for all possible pairs of // Inversions for (int i = 0; i < a.size(); i++) { for (int j = i + 1; j < a.size(); j++) { // For the current inversion // find the expected value if (a.get(i) > a.get(j)) { Collections.swap(a, i, j); // Recursively Call res += 1 + expectedSwaps(a); // Backtrack Collections.swap(a, i, j); K++; } } } // If permutation is already sorted if (K == 0) res = 0; // Calculate expected swaps for // current permutation else res /= K; // Return answer memo.put(a,res); return res; } // Driver Code public static void main(String args[]) { int N = 3; ArrayList <Integer> arr = new ArrayList <Integer> (Arrays.asList( 3, 2, 1 )); memo = new HashMap <ArrayList <Integer>, Double> (); // Function calling System.out.println( expectedSwaps(arr) ); }}// This code has been contributed by Sachin Sahara (sachin801) |
Python3
# Python program for the above approach# Stores result of each state of the# array in a memoization tablememo = {}# Recursive function to find expected# number of swaps to sort given arraydef expectedSwaps(a): # The value for the given permutation # is already calculated if (tuple(a) in memo): return memo[tuple(a)] # Stores the expected number of ways res = 0 # Stores the total operations done K = 0 # Iterate for all possible pairs of # Inversions for i in range(len(a)): for j in range(i + 1,len(a)): # For the current inversion # find the expected value if (a[i] > a[j]): temp = a[i] a[i] = a[j] a[j] = temp # Recursively Call res += 1 + expectedSwaps(a) #Backtrack temp = a[i] a[i] = a[j] a[j] = temp K += 1 # If permutation is already sorted if (K == 0): res = 0 # Calculate expected swaps for # current permutation else: res /= K; # Return answer memo[tuple(a)] = res return res# Driver CodeN = 3arr = [ 3, 2, 1 ]print(expectedSwaps(arr))# This code is contributed by rohitsingh07052. |
Javascript
<script>// JavaScript program for the above approach// Stores result of each state of the// array in a memoization tablevar dict = {};// Recursive function to find expected// number of swaps to sort given arrayfunction expectedSwaps(a) { // The value for the given permutation // is already calculated if (dict.hasOwnProperty(a)) { return dict[a]; } // Stores the expected number of ways var res = 0; // Stores the total operations done var K = 0; // Iterate for all possible pairs of // Inversions for (let i = 0; i < a.length; i++) { for (let j = i + 1; j < a.length; j++) { // For the current inversion // find the expected value if (a[i] > a[j]) { var temp = a[i]; a[i] = a[j]; a[j] = temp; // Recursively Call res += 1 + expectedSwaps(a); // Backtrack var temp2 = a[i]; a[i] = a[j]; a[j] = temp2; K++; } } } // If permutation is already sorted if (K == 0) { res = 0; } // Calculate expected swaps for // current permutation else { res /= K; } // Return answer dict[a] = res; return res;}// Driver Codevar N = 3;var arr = [3, 2, 1];document.write(expectedSwaps(arr).toFixed(5));// This code is contributed by rdtank.</script> |
C#
// C# program to implement above approachusing System;using System.Collections.Generic;class GFG{ // Stores result of each state of the // array in a memoization table public static Dictionary<List<int>, double> memo = new Dictionary<List<int>, double>(); public static bool checkEquality(List<int> a,List<int> b){ if(a.Count != b.Count) return false; for(int i = 0 ; i < a.Count ; i++){ if(a[i]!=b[i]) return false; } return true; } public static double containsKey(List<int> a){ foreach(KeyValuePair<List<int>,double> x in memo){ if(checkEquality(a, x.Key)){ return x.Value; } } return -1; } // Recursive function to find expected // number of swaps to sort given array public static double expectedSwaps(List<int> a) { // The value for the given permutation // is already calculated double res = containsKey(a); if (res != -1) { return res; } // Stores the expected number of ways res = 0; // Stores the total operations done int K = 0; // Iterate for all possible pairs of // Inversions for (int i = 0 ; i < a.Count ; i++) { for (int j = i + 1 ; j < a.Count ; j++) { // For the current inversion // find the expected value if (a[i] > a[j]) { int temp = a[i]; a[i] = a[j]; a[j] = temp; // Recursively Call res += 1 + expectedSwaps(a); // Backtrack temp = a[i]; a[i] = a[j]; a[j] = temp; K++; } } } // If permutation is already sorted if (K == 0) res = 0; // Calculate expected swaps for // current permutation else res = res/(double)K; // for(int i=0 ; i<a.Count ; i++){ // Console.Write(a[i] + " "); // } // Console.Write("\n" + res + "\n"); memo[new List<int>(a)] = res; // Return answer return res; } // Driver Code public static void Main(string[] args){ // int N = 3; List<int> arr = new List<int>{ 3, 2, 1 }; Console.Write(expectedSwaps(arr)); }} |
Output
2.33333
Time Complexity: O(N2N!)
Auxiliary Space: O(N!)
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