# Check if the given array can be constructed uniquely from the given set of subsequences

• Difficulty Level : Hard
• Last Updated : 15 Jun, 2022

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++

 `// C++ program to Check if ``// the given array can be constructed``// uniquely from the given set of subsequences` `#include ``using` `namespace` `std;` `bool` `canConstruct(vector<``int``> originalSeq,``                ``vector > 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 > 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

 `// 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 sortedOrder``            ``= ``new` `ArrayList();``        ``if` `(originalSeq.length <= ``0``) {``            ``return` `false``;``        ``}` `        ``// Count of incoming edges for every vertex``        ``Map inDegree``            ``= ``new` `HashMap();` `        ``// Adjacency list graph``        ``Map > graph``            ``= ``new` `HashMap >();``        ``for` `(``int``[] seq : sequences)``        ``{``            ``for` `(``int` `i = ``0``; i < seq.length; i++)``            ``{``                ``inDegree.put(seq[i], ``0``);``                ``graph.put(seq[i], ``new` `ArrayList());``            ``}``        ``}` `        ``// 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 sources = ``new` `ArrayList();``        ``for` `(Map.Entry 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 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`

## Python3

 `# 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`

## Javascript

 ``

Output:

`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.

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