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Check if concatenation of any permutation of given list of arrays generates the given array

  • Difficulty Level : Hard
  • Last Updated : 19 May, 2021

Given an array arr[] of N distinct integers and a list of arrays pieces[] of distinct integers, the task is to check if the given list of arrays can be concatenated in any order to obtain the given array. If it is possible, then print “Yes”. Otherwise, print “No”.

Examples:

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Input: arr[] = {1, 2, 4, 3}, pieces[][] = {{1}, {4, 3}, {2}}
Output: Yes
Explanation:
Rearrange the list to {{1}, {2}, {4, 3}}.
Now, concatenating the all the arrays from the list generates the sequence {1, 2, 4, 3} which is same as the given array.



Input: arr[] = {1, 2, 4, 3}, pieces = {{1}, {2}, {3, 4}}
Output: No
Explanation:
There is no possible permutation of given list such that after concatenating the generated sequence becomes equal to the given array.

Naive Approach: The simplest approach is to traverse the given array arr[] and for each element, arr[i], check if there exists any array present in the list such that it starts from arr[i] or not. If found to be true then increment i while elements present in the array found to be equal to arr[i]. If they are not equal, print No. Repeat the above steps until i < N. After traversing the elements of the given array, print Yes.

Time Complexity: O(N2)
Auxiliary Space: O(N)

Efficient Approach: The idea is to use the concept of Hashing by storing the indices of the elements present in the given array arr[] using Map data structure. Follow the steps below to solve the problem:

  1. Create a Map to store the indices of the elements of the given array arr[].
  2. Iterate over each of the arrays present in the list and for each array, follow the steps below:
    • Find the index of its first element in the array arr[] from the Map.
    • Check if the obtained array is a subarray of the array arr[] or not starting from the index found before.
  3. If the subarray is not equal to the array found, print No.
  4. Otherwise, after traversing the given array, print Yes.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if it is possible
// to obtain array by concatenating
// the arrays in list pieces[]
bool check(vector<int>& arr,
    vector<vector<int>>& pieces)
{
     
    // Stores the index of element in
    // the given array arr[]
    unordered_map<int, int> m;
 
    for(int i = 0; i < arr.size(); i++)
        m[arr[i]] = i + 1;
 
    // Traverse over the list pieces
    for(int i = 0; i < pieces.size(); i++)
    {
         
        // If item size is 1 and
        // exists in map
        if (pieces[i].size() == 1 &&
          m[pieces[i][0]] != 0)
        {
            continue;
        }
 
        // If item contains > 1 element
        // then check order of element
        else if (pieces[i].size() > 1 &&
               m[pieces[i][0]] != 0)
        {
            int idx = m[pieces[i][0]] - 1;
 
            idx++;
 
            // If end of the array
            if (idx >= arr.size())
                return false;
 
            // Check the order of elements
            for(int j = 1; j < pieces[i].size(); j++)
            {
                 
                // If order is same as
                // the array elements
                if (arr[idx] == pieces[i][j])
                {
                     
                    // Increment idx
                    idx++;
 
                    // If order breaks
                    if (idx >= arr.size() &&
                     j < pieces[i].size() - 1)
                        return false;
                }
 
                // Otherwise
                else
                {
                    return false;
                }
            }
        }
 
        // Return false if the first
        // element doesn't exist in m
        else
        {
            return false;
        }
    }
 
    // Return true
    return true;
}
 
// Driver Code
int main()
{
     
    // Given target list
    vector<int> arr = { 1, 2, 4, 3 };
 
    // Given array of list
    vector<vector<int> > pieces{ { 1 }, { 4, 3 }, { 2 } };
 
    // Function call
    if (check(arr, pieces))
    {
        cout << "Yes";
    }
    else
    {
        cout << "No";
    }
    return 0;
}
 
// This code is contributed by akhilsaini

Java




// Java program for the above approach
 
import java.util.*;
 
class GFG {
 
    // Function to check if it is possible
    // to obtain array by concatenating
    // the arrays in list pieces[]
    static boolean check(
        List<Integer> arr,
        ArrayList<List<Integer> > pieces)
    {
        // Stores the index of element in
        // the given array arr[]
        Map<Integer, Integer> m
            = new HashMap<>();
 
        for (int i = 0; i < arr.size(); i++)
            m.put(arr.get(i), i);
 
        // Traverse over the list pieces
        for (int i = 0;
             i < pieces.size(); i++) {
 
            // If item size is 1 and
            // exists in map
            if (pieces.get(i).size() == 1
                && m.containsKey(
                       pieces.get(i).get(0))) {
                continue;
            }
 
            // If item contains > 1 element
            // then check order of element
            else if (pieces.get(i).size() > 1
                     && m.containsKey(
                            pieces.get(i).get(0))) {
 
                int idx = m.get(
                    pieces.get(i).get(0));
 
                idx++;
 
                // If end of the array
                if (idx >= arr.size())
                    return false;
 
                // Check the order of elements
                for (int j = 1;
                     j < pieces.get(i).size();
                     j++) {
 
                    // If order is same as
                    // the array elements
                    if (arr.get(idx).equals(
                            pieces.get(i).get(j))) {
 
                        // Increment idx
                        idx++;
 
                        // If order breaks
                        if (idx >= arr.size()
                            && j < pieces.get(i).size() - 1)
                            return false;
                    }
 
                    // Otherwise
                    else {
                        return false;
                    }
                }
            }
 
            // Return false if the first
            // element doesn't exist in m
            else {
                return false;
            }
        }
 
        // Return true
        return true;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given target list
        List<Integer> arr
            = Arrays.asList(1, 2, 4, 3);
 
        ArrayList<List<Integer> > pieces
            = new ArrayList<>();
 
        // Given array of list
        pieces.add(Arrays.asList(1));
        pieces.add(Arrays.asList(4, 3));
        pieces.add(Arrays.asList(2));
 
        // Function Call
        if (check(arr, pieces)) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
}

Python3




# Python3 program for the above approach
from array import *
 
# Function to check if it is possible
# to obtain array by concatenating
# the arrays in list pieces[]
def check(arr, pieces):
     
    # Stores the index of element in
    # the given array arr[]
    m = {}
     
    for i in range(0, len(arr)):
        m[arr[i]] = i + 1
     
    # Traverse over the list pieces
    for i in range(0, len(pieces)):
         
        # If item size is 1 and
        # exists in map
        if (len(pieces[i]) == 1 and
              m[pieces[i][0]] != 0):
            continue
     
        # If item contains > 1 element
        # then check order of element
        elif (len(pieces[i]) > 1 and
                m[pieces[i][0]] != 0):
            idx = m[pieces[i][0]] - 1
            idx = idx+1
             
            # If end of the array
            if idx >= len(arr):
                return False
             
            # Check the order of elements
            for j in range(1, len(pieces[i])):
                 
                # If order is same as
                # the array elements
                if arr[idx] == pieces[i][j]:
                    # Increment idx
                    idx = idx+1
                     
                    # If order breaks
                    if (idx >= len(arr) and
                           j < len(pieces[i]) - 1):
                        return False
                 
                # Otherwise
                else:
                    return False
         
        # Return false if the first
        # element doesn't exist in m
        else:
            return False
     
    # Return true
    return True
 
# Driver Code
if __name__ == "__main__":
     
    arr = [ 1, 2, 4, 3 ]
     
    # Given array of list
    pieces = [ [ 1 ], [ 4, 3 ], [ 2 ] ]
     
    # Function call
    if check(arr, pieces) == True:
        print("Yes")
    else:
        print("No")
 
# This code is contributed by akhilsaini

C#




// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
 
class GFG{
 
// Function to check if it is possible
// to obtain array by concatenating
// the arrays in list pieces[]
static bool check(List<int> arr,
                  List<List<int>> pieces)
{
     
    // Stores the index of element in
    // the given array arr[]
    Dictionary<int,
               int> m = new Dictionary<int,
                                       int>();
 
    for(int i = 0; i < arr.Count; i++)
        m.Add(arr[i], i);
 
    // Traverse over the list pieces
    for(int i = 0; i < pieces.Count; i++)
    {
         
        // If item size is 1 and
        // exists in map
        if (pieces[i].Count == 1 &&
            m.ContainsKey(pieces[i][0]))
        {
            continue;
        }
 
        // If item contains > 1 element
        // then check order of element
        else if (pieces[i].Count > 1 &&
                 m.ContainsKey(pieces[i][0]))
        {
            int idx = m[pieces[i][0]];
 
            idx++;
 
            // If end of the array
            if (idx >= arr.Count)
                return false;
 
            // Check the order of elements
            for(int j = 1; j < pieces[i].Count; j++)
            {
                 
                // If order is same as
                // the array elements
                if (arr[idx] == pieces[i][j])
                {
                     
                    // Increment idx
                    idx++;
 
                    // If order breaks
                    if (idx >= arr.Count &&
                     j < pieces[i].Count - 1)
                        return false;
                }
 
                // Otherwise
                else
                {
                    return false;
                }
            }
        }
         
        // Return false if the first
        // element doesn't exist in m
        else
        {
            return false;
        }
    }
     
    // Return true
    return true;
}
 
// Driver Code
static public void Main()
{
     
    // Given target list
    List<int> arr = new List<int>(){ 1, 2, 4, 3 };
 
    List<List<int> > pieces = new List<List<int> >();
 
    // Given array of list
    pieces.Add(new List<int>(){ 1 });
    pieces.Add(new List<int>(){ 4, 3 });
    pieces.Add(new List<int>(){ 2 });
 
    // Function call
    if (check(arr, pieces))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}
 
// This code is contributed by akhilsaini

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to check if it is possible
// to obtain array by concatenating
// the arrays in list pieces[]
function check(arr, pieces)
{
     
    // Stores the index of element in
    // the given array arr[]
    var m = new Map();
 
    for(var i = 0; i < arr.length; i++)
        m.set(arr[i], i+1)
 
    // Traverse over the list pieces
    for(var i = 0; i < pieces.length; i++)
    {
         
        // If item size is 1 and
        // exists in map
        if (pieces[i].length == 1 &&
          m.get(pieces[i][0]) != 0)
        {
            continue;
        }
 
        // If item contains > 1 element
        // then check order of element
        else if (pieces[i].length > 1 &&
               m.get(pieces[i][0]) != 0)
        {
            var idx = m.get(pieces[i][0]) - 1;
 
            idx++;
 
            // If end of the array
            if (idx >= arr.length)
                return false;
 
            // Check the order of elements
            for(var j = 1; j < pieces[i].length; j++)
            {
                 
                // If order is same as
                // the array elements
                if (arr[idx] == pieces[i][j])
                {
                     
                    // Increment idx
                    idx++;
 
                    // If order breaks
                    if (idx >= arr.length &&
                     j < pieces[i].length - 1)
                        return false;
                }
 
                // Otherwise
                else
                {
                    return false;
                }
            }
        }
 
        // Return false if the first
        // element doesn't exist in m
        else
        {
            return false;
        }
    }
 
    // Return true
    return true;
}
 
// Driver Code
// Given target list
var arr = [ 1, 2, 4, 3 ];
 
// Given array of list
var pieces = [ [ 1 ], [ 4, 3 ], [ 2 ] ];
 
// Function call
if (check(arr, pieces))
{
    document.write( "Yes");
}
else
{
    document.write( "No");
}
 
// This code is contributed by itsok.
</script>
Output: 
Yes

 

Time Complexity: O(N*log N) where N is the length of the given array.
Auxiliary Space: O(N)




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