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📜  检查是否可以通过合并2个非空排列形成数组

📅  最后修改于: 2021-06-25 11:44:37             🧑  作者: Mango

给定长度为N的数组arr [] ,任务是检查是否可以通过合并两个相同或不同长度的排列形成。如果可以合并,则打印是”。否则,打印NO

例子:

方法 :
我们可以观察到长度为N的排列的最小excludant(MEX)是N + 1。
因此,如果第一个排列的长度为l ,则前缀arr [0……l-1]的MEX为l + 1 ,后缀a [l……n]的MEX为N – l + 1
因此,我们可以计算前缀和后缀的MEX,如果满足上述条件,则答案为“是” 。否则,答案将为“否”

下面是上述方法的实现:

C++
// C++ program for the
// above approach
#include
using namespace std;
void if_merged_permutations(int a[],
                            int n)
{
  int pre_mex[n];
 
  // Calculate the mex of the
  // array a[0...i]
  int freq[n + 1];
   
  memset(freq, 0, sizeof(freq));
   
  for(int i = 0; i < n; i++)
  {
    pre_mex[i] = 1;
  }
 
  // Mex of empty
  // array is 1
  int mex = 1;
 
  // Calculating the frequency
  // of array elements
  for(int i = 0; i < n; i++)
  {
    freq[a[i]]++;
    if(freq[a[i]] > 1)
    {
      // In a permutation
      // each element is
      // present one time,
      // So there is no chance
      // of getting permutations
      // for the prefix of
      // length greater than i
      break;
    }
 
    // The current element
    // is the mex
    if(a[i] == mex)
    {
      // While mex is present
      // in the array
      while(freq[mex] != 0)
      {
        mex++;
      }
    }
    pre_mex[i] = mex;
  }
   
  int suf_mex[n];
   
  for(int i = 0; i < n; i++)
  {
    suf_mex[i] = 1;
  }
   
  // Calculate the mex of the
  // array a[i..n]
  memset(freq, 0, sizeof(freq));
 
  // Mex of empty
  // array is 1
  mex = 1;
 
  // Calculating the frequency
  // of array elements
  for(int i = n - 1; i > -1; i--)
  {
    freq[a[i]]++;
    if(freq[a[i]] > 1)
    {
      // In a permutation each
      // element is present
      // one time, So there is
      // no chance of getting
      // permutations for the
      // suffix of length lesser
      // than i
      continue;
    }
     
    // The current element is
    // the mex
    if(a[i] == mex)
    {
      // While mex is present
      // in the array
      while(freq[mex] != 0)
      {
        mex++;
      }
    }
    suf_mex[i] = mex;
  }
 
  // Now check if there is atleast
  // one index i such that mex of
  // the prefix a[0..i]= i +
  // 2(0 based indexing) and  mex
  // of the suffix a[i + 1..n]= n-i
  for(int i = 0; i < n - 1; i++)
  {
    if(pre_mex[i] == i + 2 &&
       suf_mex[i + 1] == n - i)
    {
      cout << "YES" << endl;
      return;
    }
  }
  cout << "NO" << endl;
}
 
// Driver code
int main()
{
    int a[] = {1, 3, 2,
               4, 3, 1, 2};
    int n = sizeof(a)/ sizeof(a[0]);
    if_merged_permutations(a, n);   
}
 
//This code is contributed by avanitrachhadiya2155


Java
// Java program for above approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG{
 
static void if_merged_permutations(int a[],
                                   int n)
{
  int[] pre_mex = new int[n];
 
  // Calculate the mex of the
  // array a[0...i]
  int[] freq = new int[n + 1];
   
  for(int i = 0; i < n; i++)
  {
    pre_mex[i] = 1;
  }
 
  // Mex of empty
  // array is 1
  int mex = 1;
 
  // Calculating the frequency
  // of array elements
  for(int i = 0; i < n; i++)
  {
    freq[a[i]]++;
     
    if (freq[a[i]] > 1)
    {
       
      // In a permutation
      // each element is
      // present one time,
      // So there is no chance
      // of getting permutations
      // for the prefix of
      // length greater than i
      break;
    }
 
    // The current element
    // is the mex
    if (a[i] == mex)
    {
       
      // While mex is present
      // in the array
      while (freq[mex] != 0)
      {
        mex++;
      }
    }
    pre_mex[i] = mex;
  }
   
  int[] suf_mex = new int[n];
   
  for(int i = 0; i < n; i++)
  {
    suf_mex[i] = 1;
  }
   
  // Calculate the mex of the
  // array a[i..n]
  Arrays.fill(freq, 0);
 
  // Mex of empty
  // array is 1
  mex = 1;
 
  // Calculating the frequency
  // of array elements
  for(int i = n - 1; i > -1; i--)
  {
    freq[a[i]]++;
     
    if (freq[a[i]] > 1)
    {
       
      // In a permutation each
      // element is present
      // one time, So there is
      // no chance of getting
      // permutations for the
      // suffix of length lesser
      // than i
      continue;
    }
     
    // The current element is
    // the mex
    if (a[i] == mex)
    {
       
      // While mex is present
      // in the array
      while (freq[mex] != 0)
      {
        mex++;
      }
    }
    suf_mex[i] = mex;
  }
 
  // Now check if there is atleast
  // one index i such that mex of
  // the prefix a[0..i]= i +
  // 2(0 based indexing) and  mex
  // of the suffix a[i + 1..n]= n-i
  for(int i = 0; i < n - 1; i++)
  {
    if (pre_mex[i] == i + 2 &&
        suf_mex[i + 1] == n - i)
    {
      System.out.println("YES");
      return;
    }
  }
  System.out.println("NO");
}
   
// Driver code
public static void main(String[] args)
{
  int a[] = { 1, 3, 2, 4, 3, 1, 2 };
  int n = a.length;
   
  if_merged_permutations(a, n);   
}
}
 
// This code is contributed by offbeat


Python3
# Python3 program for above approach
def if_merged_permutations(a, n):
    pre_mex =[1 for i in range(n)]
     
    # Calculate the mex of the
    # array a[0...i]
    freq =[0 for i in range(n + 1)]
     
    # Mex of empty array is 1
    mex = 1
     
    # Calculating the frequency
    # of array elements
    for i in range(n):
        freq[a[i]]+= 1
        if freq[a[i]]>1:
            # In a permutation
            # each element is
            # present one time,
            # So there is no chance
            # of getting permutations
            # for the prefix of
            # length greater than i
            break
         
        # The current element
        # is the mex   
        if a[i]== mex:
       
            # While mex is present
            # in the array
            while freq[mex]!= 0 :
                mex+= 1
        pre_mex[i]= mex
                 
    suf_mex =[1 for i in range(n)]
     
    # Calculate the mex of the
    # array a[i..n]
    freq =[0 for i in range(n + 1)]
   
    # Mex of empty array is 1
    mex = 1
     
    # Calculating the frequency
    # of array elements
    for i in range(n-1, -1, -1):
        freq[a[i]]+= 1
        if freq[a[i]]>1:
 
            # In a permutation each
            # element is present
            # one time, So there is
            # no chance of getting
            # permutations for the
            # suffix of length lesser
            # than i
            break
 
        # The current element is
        # the mex
        if a[i]== mex:
            # While mex is present
            # in the array
            while freq[mex]!= 0 :
                mex+= 1
        suf_mex[i]= mex
 
    # Now check if there is atleast
    # one index i such that mex of
    # the prefix a[0..i]= i +
    # 2(0 based indexing) and  mex
    # of the suffix a[i + 1..n]= n-i
 
    for i in range(n-1):
        if pre_mex[i]== i + 2 and suf_mex[i + 1]== n-i:
            print("YES")
            return
    print("NO")
     
a =[1, 3, 2, 4, 3, 1, 2]   
n = len(a)       
if_merged_permutations(a, n)


C#
// C# program for above approach
using System;
 
class GFG{
     
static void if_merged_permutations(int[] a,
                                   int n)
{
    int[] pre_mex = new int[n];
     
    // Calculate the mex of the
    // array a[0...i]
    int[] freq = new int[n + 1];
     
    for(int i = 0; i < n; i++)
    {
        pre_mex[i] = 1;
    }
     
    // Mex of empty
    // array is 1
    int mex = 1;
     
    // Calculating the frequency
    // of array elements
    for(int i = 0; i < n; i++)
    {
        freq[a[i]]++;
     
        if (freq[a[i]] > 1)
        {
             
            // In a permutation
            // each element is
            // present one time,
            // So there is no chance
            // of getting permutations
            // for the prefix of
            // length greater than i
            break;
        }
     
        // The current element
        // is the mex
        if (a[i] == mex)
        {
             
            // While mex is present
            // in the array
            while (freq[mex] != 0)
            {
                mex++;
            }
        }
        pre_mex[i] = mex;
    }
     
    int[] suf_mex = new int[n];
     
    for(int i = 0; i < n; i++)
    {
        suf_mex[i] = 1;
    }
     
    // Calculate the mex of the
    // array a[i..n]
    Array.Fill(freq, 0);
     
    // Mex of empty
    // array is 1
    mex = 1;
     
    // Calculating the frequency
    // of array elements
    for(int i = n - 1; i > -1; i--)
    {
        freq[a[i]]++;
         
        if (freq[a[i]] > 1)
        {
             
            // In a permutation each
            // element is present
            // one time, So there is
            // no chance of getting
            // permutations for the
            // suffix of length lesser
            // than i
            continue;
        }
         
        // The current element is
        // the mex
        if (a[i] == mex)
        {
             
            // While mex is present
            // in the array
            while (freq[mex] != 0)
            {
                mex++;
            }
        }
        suf_mex[i] = mex;
    }
     
    // Now check if there is atleast
    // one index i such that mex of
    // the prefix a[0..i]= i +
    // 2(0 based indexing) and  mex
    // of the suffix a[i + 1..n]= n-i
    for(int i = 0; i < n - 1; i++)
    {
        if (pre_mex[i] == i + 2 &&
            suf_mex[i + 1] == n - i)
        {
            Console.WriteLine("YES");
            return;
        }
    }
    Console.WriteLine("NO");
}
 
// Driver Code
static void Main()
{
    int[] a = { 1, 3, 2, 4, 3, 1, 2 };
    int n = a.Length;
     
    if_merged_permutations(a, n);
}
}
 
// This code is contributed by divyeshrabadiya07


输出:
YES

时间复杂度: O(N)

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