Given an array arr of distinct elements and a list of subsequences seqs of the array, the task is to check whether the given array can be uniquely constructed from the given set of subsequences.
Examples:
Input : arr[] = {1, 2, 3, 4}, seqs[][] = {{1, 2}, {2, 3}, {3, 4}}
Output: Yes
Explanations: The sequences [1, 2], [2, 3], and [3, 4] can uniquely reconstruct
the original array {1, 2, 3, 4}.
Input: arr[] = {1, 2, 3, 4}, seqs[][] = {{1, 2}, {2, 3}, {2, 4}}
Output: No
Explanations : The sequences [1, 2], [2, 3], and [2, 4] cannot uniquely reconstruct
{1, 2, 3, 4}. There are two possible sequences that can be constructed from the given sequences:
1) {1, 2, 3, 4}
2) {1, 2, 4, 3}
Approach:
In order to solve this problem we need to find the Topological Ordering of all the array elements and check if only one topological ordering of the elements exists or not, which can be confirmed by the presence of only single source at every instant while finding the topological ordering of elements.
Below is the implementation of the above approach:
// C++ program to Check if // the given array can be constructed // uniquely from the given set of subsequences #include <bits/stdc++.h> using namespace std;
bool canConstruct(vector< int > originalSeq,
vector<vector< int > > sequences)
{ vector< int > sortedOrder;
if (originalSeq.size() <= 0) {
return false ;
}
// Count of incoming edges for every vertex
unordered_map< int , int > inDegree;
// Adjacency list graph
unordered_map< int , vector< int > > graph;
for ( auto seq : sequences) {
for ( int i = 0; i < seq.size(); i++) {
inDegree[seq[i]] = 0;
graph[seq[i]] = vector< int >();
}
}
// Build the graph
for ( auto seq : sequences) {
for ( int i = 1; i < seq.size(); i++) {
int parent = seq[i - 1], child = seq[i];
graph[parent].push_back(child);
inDegree[child]++;
}
}
// if ordering rules for all the numbers
// are not present
if (inDegree.size() != originalSeq.size()) {
return false ;
}
// Find all sources i.e., all vertices
// with 0 in-degrees
queue< int > sources;
for ( auto entry : inDegree) {
if (entry.second == 0) {
sources.push(entry.first);
}
}
// For each source, add it to the sortedOrder
// and subtract one from all of in-degrees
// if a child's in-degree becomes zero
// add it to the sources queue
while (!sources.empty()) {
// If there are more than one source
if (sources.size() > 1) {
// Multiple sequences exist
return false ;
}
// If the next source is different from the origin
if (originalSeq[sortedOrder.size()] !=
sources.front()) {
return false ;
}
int vertex = sources.front();
sources.pop();
sortedOrder.push_back(vertex);
vector< int > children = graph[vertex];
for ( auto child : children) {
// Decrement the node's in-degree
inDegree[child]--;
if (inDegree[child] == 0) {
sources.push(child);
}
}
}
// Compare the sizes of sortedOrder
// and the original sequence
return sortedOrder.size() == originalSeq.size();
} int main( int argc, char * argv[])
{ vector< int > arr = { 1, 2, 6, 7, 3, 5, 4 };
vector<vector< int > > seqs = { { 1, 2, 3 },
{ 7, 3, 5 },
{ 1, 6, 3, 4 },
{ 2, 6, 5, 4 } };
bool result = canConstruct(arr, seqs);
if (result)
cout << "Yes" << endl;
else
cout << "No" << endl;
} |
// Java program to Check if // the given array can be constructed // uniquely from the given set of subsequences import java.io.*;
import java.util.*;
class GFG {
static boolean canConstruct( int [] originalSeq,
int [][] sequences)
{
List<Integer> sortedOrder
= new ArrayList<Integer>();
if (originalSeq.length <= 0 ) {
return false ;
}
// Count of incoming edges for every vertex
Map<Integer, Integer> inDegree
= new HashMap<Integer, Integer>();
// Adjacency list graph
Map<Integer, ArrayList<Integer> > graph
= new HashMap<Integer, ArrayList<Integer> >();
for ( int [] seq : sequences)
{
for ( int i = 0 ; i < seq.length; i++)
{
inDegree.put(seq[i], 0 );
graph.put(seq[i], new ArrayList<Integer>());
}
}
// Build the graph
for ( int [] seq : sequences)
{
for ( int i = 1 ; i < seq.length; i++)
{
int parent = seq[i - 1 ], child = seq[i];
graph.get(parent).add(child);
inDegree.put(child,
inDegree.get(child) + 1 );
}
}
// if ordering rules for all the numbers
// are not present
if (inDegree.size() != originalSeq.length)
{
return false ;
}
// Find all sources i.e., all vertices
// with 0 in-degrees
List<Integer> sources = new ArrayList<Integer>();
for (Map.Entry<Integer, Integer> entry :
inDegree.entrySet())
{
if (entry.getValue() == 0 )
{
sources.add(entry.getKey());
}
}
// For each source, add it to the sortedOrder
// and subtract one from all of in-degrees
// if a child's in-degree becomes zero
// add it to the sources queue
while (!sources.isEmpty())
{
// If there are more than one source
if (sources.size() > 1 )
{
// Multiple sequences exist
return false ;
}
// If the next source is different from the
// origin
if (originalSeq[sortedOrder.size()]
!= sources.get( 0 ))
{
return false ;
}
int vertex = sources.get( 0 );
sources.remove( 0 );
sortedOrder.add(vertex);
List<Integer> children = graph.get(vertex);
for ( int child : children)
{
// Decrement the node's in-degree
inDegree.put(child,
inDegree.get(child) - 1 );
if (inDegree.get(child) == 0 )
{
sources.add(child);
}
}
}
// Compare the sizes of sortedOrder
// and the original sequence
return sortedOrder.size() == originalSeq.length;
}
// Driver code
public static void main(String[] args)
{
int [] arr = { 1 , 2 , 6 , 7 , 3 , 5 , 4 };
int [][] seqs = { { 1 , 2 , 3 },
{ 7 , 3 , 5 },
{ 1 , 6 , 3 , 4 },
{ 2 , 6 , 5 , 4 } };
boolean result = canConstruct(arr, seqs);
if (result)
System.out.println( "Yes" );
else
System.out.println( "No" );
}
} // This code is contributed by jitin |
# Python 3 program to Check if # the given array can be constructed # uniquely from the given set of subsequences def canConstruct(originalSeq, sequences):
sortedOrder = []
if ( len (originalSeq) < = 0 ):
return False
# Count of incoming edges for every vertex
inDegree = {i : 0 for i in range ( 100 )}
# Adjacency list graph
graph = {i : [] for i in range ( 100 )}
for seq in sequences:
for i in range ( len (seq)):
inDegree[seq[i]] = 0
graph[seq[i]] = []
# Build the graph
for seq in sequences:
for i in range ( 1 , len (seq)):
parent = seq[i - 1 ]
child = seq[i]
graph[parent].append(child)
inDegree[child] + = 1
# If ordering rules for all the numbers
# are not present
if ( len (inDegree) ! = len (originalSeq)):
return False
# Find all sources i.e., all vertices
# with 0 in-degrees
sources = []
for entry in inDegree:
if (entry[ 1 ] = = 0 ):
sources.append(entry[ 0 ])
# For each source, add it to the sortedOrder
# and subtract one from all of in-degrees
# if a child's in-degree becomes zero
# add it to the sources queue
while ( len (sources) > 0 ):
# If there are more than one source
if ( len (sources) > 1 ):
# Multiple sequences exist
return False
# If the next source is different from the origin
if (originalSeq[ len (sortedOrder)] ! = sources[ 0 ]):
return False
vertex = sources[ 0 ]
sources.remove(sources[ 0 ])
sortedOrder.append(vertex)
children = graph[vertex]
for child in children:
# Decrement the node's in-degree
inDegree[child] - = 1
if (inDegree[child] = = 0 ):
sources.append(child)
# Compare the sizes of sortedOrder
# and the original sequence
return len (sortedOrder) = = len (originalSeq)
if __name__ = = '__main__' :
arr = [ 1 , 2 , 6 , 7 , 3 , 5 , 4 ]
seqs = [[ 1 , 2 , 3 ],
[ 7 , 3 , 5 ],
[ 1 , 6 , 3 , 4 ],
[ 2 , 6 , 5 , 4 ]]
result = canConstruct(arr, seqs)
if (result):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by Bhupendra_Singh |
using System;
using System.Collections.Generic;
class GFG {
// Function to check if the given array can be
// constructed uniquely from the given set of
// subsequences
static bool CanConstruct( int [] originalSeq,
int [][] sequences)
{
// List to store the sorted order of elements in the
// original sequence
List< int > sortedOrder = new List< int >();
// If the original sequence is empty, return false
if (originalSeq.Length <= 0) {
return false ;
}
// Count of incoming edges for every vertex
Dictionary< int , int > inDegree
= new Dictionary< int , int >();
// Adjacency list graph
Dictionary< int , List< int > > graph
= new Dictionary< int , List< int > >();
foreach ( int [] seq in sequences)
{
// Initialize the in-degree and graph data
// structures for each element in the sequence
for ( int i = 0; i < seq.Length; i++) {
inDegree[seq[i]] = 0;
graph[seq[i]] = new List< int >();
}
}
// Build the graph
foreach ( int [] seq in sequences)
{
for ( int i = 1; i < seq.Length; i++) {
int parent = seq[i - 1], child = seq[i];
graph[parent].Add(child);
inDegree[child] = inDegree[child] + 1;
}
}
// If ordering rules for all the numbers are not
// present, return false
if (inDegree.Count != originalSeq.Length) {
return false ;
}
// Find all sources i.e., all vertices with 0
// in-degrees
List< int > sources = new List< int >();
foreach (KeyValuePair< int , int > entry in inDegree)
{
if (entry.Value == 0) {
sources.Add(entry.Key);
}
}
// For each source, add it to the sortedOrder and
// subtract one from all of in-degrees If a child's
// in-degree becomes zero, add it to the sources
// queue
while (sources.Count > 0) {
// If there are more than one source, multiple
// sequences exist and return false
if (sources.Count > 1) {
return false ;
}
// If the next source is different from the
// original sequence, return false
if (originalSeq[sortedOrder.Count]
!= sources[0]) {
return false ;
}
int vertex = sources[0];
sources.RemoveAt(0);
sortedOrder.Add(vertex);
List< int > children = graph[vertex];
foreach ( int child in children)
{
// Decrement the node's in-degree
inDegree[child] = inDegree[child] - 1;
if (inDegree[child] == 0) {
sources.Add(child);
}
}
}
// Compare the sizes of sortedOrder and the original
// sequence
return sortedOrder.Count == originalSeq.Length;
}
// Driver code
static void Main( string [] args)
{
int [] arr = { 1, 2, 6, 7, 3, 5, 4 };
int [][] seqs = { new int [] { 1, 2, 3 },
new int [] { 7, 3, 5 },
new int [] { 1, 6, 3, 4 },
new int [] { 2, 6, 5, 4 } };
bool result = CanConstruct(arr, seqs);
if (result)
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
} |
<script> // JavaScript program to Check if // the given array can be constructed // uniquely from the given set of subsequences function canConstruct(originalSeq, sequences)
{ let sortedOrder = [];
if (originalSeq.length <= 0) {
return false ;
}
// Count of incoming edges for every vertex
let inDegree = new Map();
// Adjacency list graph
let graph = new Array(100).fill(0).map(()=>[]);
for (let seq of sequences) {
for (let i = 0; i < seq.length; i++) {
inDegree.set(seq[i] , 0);
graph[seq[i]] = [];
}
}
// Build the graph
for (let seq of sequences) {
for (let i = 1; i < seq.length; i++) {
let parent = seq[i - 1], child = seq[i];
graph[parent].push(child);
if (inDegree.has(child))
inDegree.set(child,inDegree.get(child)+1);
else
inDegree.set(child,1);
}
}
// if ordering rules for all the numbers
// are not present
if (inDegree.length != originalSeq.length) {
return false ;
}
// Find all sources i.e., all vertices
// with 0 in-degrees
let sources = [];
for (let [entry,res] of inDegree) {
if (res== 0) {
sources.push(entry);
}
}
// For each source, add it to the sortedOrder
// and subtract one from all of in-degrees
// if a child's in-degree becomes zero
// add it to the sources queue
while (sources.length > 0) {
// If there are more than one source
if (sources.length > 1) {
// Multiple sequences exist
return false ;
}
// If the next source is different from the origin
if (originalSeq[sortedOrder.length] !=
sources[0]) {
return false ;
}
let vertex = sources.shift();
sortedOrder.push(vertex);
let children = graph[vertex];
for (let child of children) {
// Decrement the node's in-degree
if (inDegree.has(child)){
inDegree.set(child,inDegree.get(child)-1);
}
if (inDegree.get(child) == 0) {
sources.push(child);
}
}
}
// Compare the sizes of sortedOrder
// and the original sequence
return sortedOrder.length == originalSeq.length;
} // driver code let arr = [ 1, 2, 6, 7, 3, 5, 4 ]; let seqs = [ [ 1, 2, 3], [ 7, 3, 5],
[ 1, 6, 3, 4],
[ 2, 6, 5, 4]];
let result = canConstruct(arr, seqs); if (result)
document.write( "Yes" , "</br>" );
else document.write( "No" , "</br>" );
// This code is contributed by shinjanpatra </script> |
No
Time complexity :
The time complexity of the above algorithm will be O(N+E), where ‘N’ is the number of elements and ‘E’ is the total number of the rules. Since, at most, each pair of numbers can give us one rule, we can conclude that the upper bound for the rules is O(M) where ‘M’ is the count of numbers in all sequences. So, we can say that the time complexity of our algorithm is O(N + M).
Auxiliary Space : O(N+ M), since we are storing all possible rules for each element.