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 <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
// Java program to Check if// the given array can be constructed// uniquely from the given set of subsequencesimport 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 |
Python3
# Python 3 program to Check if # the given array can be constructed # uniquely from the given set of subsequencesdef 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 |
C#
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"); }} |
Javascript
<script>// JavaScript program to Check if// the given array can be constructed// uniquely from the given set of subsequencesfunction 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 codelet 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> |
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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