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📜  检查给定的数组是否可以从给定的子序列集中唯一地构造

📅  最后修改于: 2021-04-22 01:39:54             🧑  作者: Mango

给定不同的元件的阵列ARR和阵列的子序列seqs的列表,所述任务是检查是否与给定阵列可从给定的一组子序列被唯一构成。
例子:

方法:
为了解决此问题,我们需要找到所有数组元素的拓扑顺序,并检查是否仅存在一个元素的拓扑顺序,这可以通过在查找拓扑时的每个瞬间仅存在一个源来确认元素的排序。
下面是上述方法的实现:

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 originalSeq,
                vector > sequences)
{
    vector sortedOrder;
    if (originalSeq.size() <= 0) {
        return false;
    }
 
    // Count of incoming edges for every vertex
    unordered_map inDegree;
 
    // Adjacency list graph
    unordered_map > graph;
    for (auto seq : sequences) {
        for (int i = 0; i < seq.size(); i++) {
            inDegree[seq[i]] = 0;
            graph[seq[i]] = vector();
        }
    }
 
    // 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 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 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 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


输出:
No

时间复杂度:
上述算法的时间复杂度为O(N + E),其中“ N”是元素数,“ E”是规则总数。由于最多每个数字对可以给我们一个规则,因此我们可以得出结论,规则的上限为O(M),其中“ M”是所有序列中的数字计数。因此,可以说我们算法的时间复杂度为O(N + M)。
辅助空间: O(N + M),因为我们存储每个元素的所有可能规则。