Patience sorting is a sorting algorithm based on card game Patience. In this sorting algorithm, the rules of patience game is used to sort an list of elements based on their values.
Rules of Patience Game:
- Cards with lower value can be placed over the card.
- If there is no possible position for a card, then a new pile can be created.
- Goal is to form as much as few piles possible.
Below is the visualization of the game as follows:
As in the above visualization, It’s clear that cards are only placed when the value of them is less than the highest card of the pile. Otherwise, If there is no such pile then create a new one.
Patience Sorting: There are generally two steps in the patience sorting that is creation of the piles and merging the piles. Below is the illustration of the steps:
- Initialize a 2D array to store the piles.
- Traverse the given array and perform the following operations:
- Iterate over all the piles and check the top of the stack of each pile is less than the current element or not. IF found to be true, then push the current element to the top of the stack.
- Otherwise, create a new pile with the current element as the top of that stack.
- Merge the Piles:The idea is to perform a k-way merge of the p piles, each of which is internally sorted. Iterate over all the piles while count of elements in the pile is greater than or equal to 0 and find the minimum element from the top of each stack and push it into the sorted array.
Below is the visualization of the sorting steps:
Below is the implementation of the above approach:
C++
// C++ program of the above approach #include <bits/stdc++.h> using namespace std; // Function to merge piles in a sorted order vector< int > merge_piles(vector<vector< int > >& v) { // Store minimum element from // the top of stack vector< int > ans; // In every iteration find the smallest element // of top of pile and remove it from the piles // and store into the final array while (1) { // Stores the smallest element // of the top of the piles int minu = INT_MAX; // Stores index of the smallest element // of the top of the piles int index = -1; // Calculate the smallest element // of the top of the every stack for ( int i = 0; i < v.size(); i++) { // If minu is greater than // the top of the current stack if (minu > v[i][v[i].size() - 1]) { // Update minu minu = v[i][v[i].size() - 1]; // Update index index = i; } } // Insert the smallest element // of the top of the stack ans.push_back(minu); // Remove the top element from // the current pile v[index].pop_back(); // If current pile is empty if (v[index].empty()) { // Remove current pile // from all piles v.erase(v.begin() + index); } // If all the piles are empty if (v.size() == 0) break ; } return ans; } // Function to sort the given array // using the patience sorting vector< int > patienceSorting(vector< int > arr) { // Store all the created piles vector<vector< int > > piles; // Traverse the array for ( int i = 0; i < arr.size(); i++) { // If no piles are created if (piles.empty()) { // Initialize a new pile vector< int > temp; // Insert current element // into the pile temp.push_back(arr[i]); // Insert current pile into // all the piles piles.push_back(temp); } else { // Check if top element of each pile // is less than or equal to // current element or not int flag = 1; // Traverse all the piles for ( int j = 0; j < piles.size(); j++) { // Check if the element to be // inserted is less than // current pile's top if (arr[i] < piles[j][piles[j].size() - 1]) { piles[j].push_back(arr[i]); // Update flag flag = 0; break ; } } // If flag is true if (flag) { // Create a new pile vector< int > temp; // Insert current element // into temp temp.push_back(arr[i]); // Insert current pile // into all the piles piles.push_back(temp); } } } // Store the sorted sequence // of the given array vector< int > ans; // Sort the given array ans = merge_piles(piles); // Traverse the array, ans[] for ( int i = 0; i < ans.size(); i++) cout << ans[i] << " " ; return ans; } // Driver Code int main() { vector< int > arr = { 6, 12, 2, 8, 3, 7 }; // Function Call patienceSorting(arr); } |
Java
// Java program of the above approach import java.util.*; public class Main { // Function to merge piles in a sorted order public static List<Integer> mergePiles(List<List<Integer> > v) { // Store minimum element from // the top of stack List<Integer> ans = new ArrayList<Integer>(); // In every iteration find the smallest element // of top of pile and remove it from the piles // and store into the final array while ( true ) { // Stores the smallest element // of the top of the piles int minu = Integer.MAX_VALUE; // Stores index of the smallest element // of the top of the piles int index = - 1 ; // Calculate the smallest element // of the top of the every stack for ( int i = 0 ; i < v.size(); i++) { // If minu is greater than // the top of the current stack if (!v.get(i).isEmpty() && minu > v.get(i).get(v.get(i).size() - 1 )) { minu = v.get(i).get(v.get(i).size() - 1 ); index = i; } } if (index == - 1 ) { break ; } ans.add(minu); v.get(index).remove(v.get(index).size() - 1 ); // If current pile is empty if (v.get(index).isEmpty()) { // Remove current pile // from all piles v.remove(index); } } return ans; } // Function to sort the given array // using the patience sorting public static List<Integer> patienceSorting(List<Integer> arr) { List<List<Integer> > piles = new ArrayList<List<Integer> >(); // Store all the created piles for ( int i = 0 ; i < arr.size(); i++) { // If no piles are created if (piles.isEmpty()) { // Initialize a new pile List<Integer> temp = new ArrayList<Integer>(); temp.add(arr.get(i)); piles.add(temp); } else { // Check if top element of each pile // is less than or equal to // current element or not int flag = 1 ; for ( int j = 0 ; j < piles.size(); j++) { // Check if the element to be // inserted is less than // current pile's top if (arr.get(i) < piles.get(j).get( piles.get(j).size() - 1 )) { piles.get(j).add(arr.get(i)); flag = 0 ; break ; } } if (flag == 1 ) { List<Integer> temp = new ArrayList<Integer>(); // Insert current element // into temp temp.add(arr.get(i)); // Insert current pile // into all the piles piles.add(temp); } } } // Store the sorted sequence // of the given array List<Integer> ans = mergePiles(piles); for ( int i = 0 ; i < ans.size(); i++) { System.out.print(ans.get(i) + " " ); } return ans; } // Driver Code public static void main(String[] args) { List<Integer> arr = new ArrayList<Integer>(); arr.add( 6 ); arr.add( 12 ); arr.add( 2 ); arr.add( 8 ); arr.add( 3 ); arr.add( 7 ); patienceSorting(arr); } } |
Javascript
<script> // Javascript program of the above approach // Function to merge piles in a sorted order function merge_piles(v) { // Store minimum element from // the top of stack let ans = []; // In every iteration find the smallest element // of top of pile and remove it from the piles // and store into the final array while (1) { // Stores the smallest element // of the top of the piles let minu = Number.MAX_SAFE_INTEGER; // Stores index of the smallest element // of the top of the piles let index = -1; // Calculate the smallest element // of the top of the every stack for (let i = 0; i < v.length; i++) { // If minu is greater than // the top of the current stack if (minu > v[i][v[i].length - 1]) { // Update minu minu = v[i][v[i].length - 1]; // Update index index = i; } } // Insert the smallest element // of the top of the stack ans.push(minu); // Remove the top element from // the current pile v[index].pop(); // If current pile is empty if (v[index].length == 0) { // Remove current pile // from all piles v.splice(index, 1); } // If all the piles are empty if (v.length == 0) break ; } return ans; } // Function to sort the given array // using the patience sorting function patienceSorting(arr) { // Store all the created piles let piles = []; // Traverse the array for (let i = 0; i < arr.length; i++) { // If no piles are created if (piles.length == 0) { // Initialize a new pile let temp = []; // Insert current element // into the pile temp.push(arr[i]); // Insert current pile into // all the piles piles.push(temp); } else { // Check if top element of each pile // is less than or equal to // current element or not let flag = 1; // Traverse all the piles for (let j = 0; j < piles.length; j++) { // Check if the element to be // inserted is less than // current pile's top if (arr[i] < piles[j][piles[j].length - 1]) { piles[j].push(arr[i]); // Update flag flag = 0; break ; } } // If flag is true if (flag) { // Create a new pile let temp = []; // Insert current element // into temp temp.push(arr[i]); // Insert current pile // into all the piles piles.push(temp); } } } // Store the sorted sequence // of the given array let ans = []; // Sort the given array ans = merge_piles(piles); // Traverse the array, ans[] for (let i = 0; i < ans.length; i++) document.write(ans[i] + " " ); return ans; } // Driver Code let arr = [6, 12, 2, 8, 3, 7]; // Function Call patienceSorting(arr); // This code is contributed by _saurabh_jaiswal. </script> |
Python3
# Python code to implement the above approach # Function to merge piles in a sorted order def merge_piles(v): # Store minimum element from the top of stack ans = [] # In every iteration find the smallest element # of top of pile and remove it from the piles # and store into the final array while True : # Stores the smallest element of the top of the piles minu = float ( "inf" ) # Stores index of the smallest element of the top of the piles index = - 1 # Calculate the smallest element of the top of the every stack for i in range ( len (v)): # If minu is greater than the top of the current stack if minu > v[i][ - 1 ]: # Update minu minu = v[i][ - 1 ] # Update index index = i # Insert the smallest element of the top of the stack ans.append(minu) # Remove the top element from the current pile v[index].pop() # If current pile is empty if not v[index]: # Remove current pile from all piles v.pop(index) # If all the piles are empty if not v: break return ans # Function to sort the given array using the patience sorting def patienceSorting(arr): # Store all the created piles piles = [] # Traverse the array for i in range ( len (arr)): # If no piles are created if not piles: # Initialize a new pile temp = [] # Insert current element into the pile temp.append(arr[i]) # Insert current pile into all the piles piles.append(temp) else : # Check if top element of each pile is less than or equal to # current element or not flag = True # Traverse all the piles for j in range ( len (piles)): # Check if the element to be inserted is less than # current pile's top if arr[i] < piles[j][ - 1 ]: piles[j].append(arr[i]) # Update flag flag = False break # If flag is True if flag: # Create a new pile temp = [] # Insert current element into temp temp.append(arr[i]) # Insert current pile into all the piles piles.append(temp) # Store the sorted sequence of the given array ans = [] # Sort the given array ans = merge_piles(piles) return ans # Driver code arr = [ 6 , 12 , 2 , 8 , 3 , 7 ] # Function call print (patienceSorting(arr)) |
C#
using System; using System.Collections.Generic; using System.Linq; public class Program { // Function to merge piles in a sorted order public static List< int > MergePiles(List<List< int >> v) { // Store minimum element from // the top of stack List< int > ans = new List< int >(); // In every iteration find the smallest element // of top of pile and remove it from the piles // and store into the final array while ( true ) { // Stores the smallest element // of the top of the piles int minu = int .MaxValue; // Stores index of the smallest element // of the top of the piles int index = -1; // Calculate the smallest element // of the top of the every stack for ( int i = 0; i < v.Count; i++) { // If minu is greater than // the top of the current stack if (v[i].Any() && minu > v[i].Last()) { minu = v[i].Last(); index = i; } } if (index == -1) { break ; } ans.Add(minu); v[index].RemoveAt(v[index].Count - 1); // If current pile is empty if (!v[index].Any()) { // Remove current pile // from all piles v.RemoveAt(index); } } return ans; } // Function to sort the given array // using the patience sorting public static List< int > PatienceSorting(List< int > arr) { List<List< int >> piles = new List<List< int >>(); // Store all the created piles for ( int i = 0; i < arr.Count; i++) { // If no piles are created if (!piles.Any()) { // Initialize a new pile List< int > temp = new List< int >(); temp.Add(arr[i]); piles.Add(temp); } else { // Check if top element of each pile // is less than or equal to // current element or not int flag = 1; for ( int j = 0; j < piles.Count; j++) { // Check if the element to be // inserted is less than // current pile's top if (arr[i] < piles[j].Last()) { piles[j].Add(arr[i]); flag = 0; break ; } } if (flag == 1) { List< int > temp = new List< int >(); // Insert current element // into temp temp.Add(arr[i]); // Insert current pile // into all the piles piles.Add(temp); } } } // Store the sorted sequence // of the given array List< int > ans = MergePiles(piles); foreach ( int num in ans) { Console.Write(num + " " ); } return ans; } // Driver Code public static void Main( string [] args) { List< int > arr = new List< int >() { 6, 12, 2, 8, 3, 7 }; PatienceSorting(arr); } } |
2 3 6 7 8 12
Time Complexity: O(N2)
Auxiliary Space: O(N)
Note: The above approach can be optimized by merging of piles using priority_queue.
Time Complexity: O(N * log(N))
Auxiliary Space: O(N)
</br>function playGif(){ </br> var gif = document.getElementById(‘gif’); </br> if (gif.src == “https://media.neveropen.co.za/wp-content/uploads/20200501171647/Patience.gif”){ </br> gif.src = “https://media.neveropen.co.za/wp-content/uploads/20200501175532/base2.jpg”; </br> }else{ </br> gif.src = “https://media.neveropen.co.za/wp-content/uploads/20200501171647/Patience.gif”; </br> } </br>} </br>function playSortingGif(){ </br> var gif1 = document.getElementById(‘sorting’); </br> var gif2 = document.getElementById(‘sorting_gif1’); </br> gif1.style.display = “none”; </br> gif2.style.display = “block”; </br>}</br>function pauseSortingGif(){ </br> var gif1 = document.getElementById(‘sorting’); </br> var gif2 = document.getElementById(‘sorting_gif1’); </br> gif1.style.display = “block”; </br> gif2.style.display = “none”; </br>} </br>
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