给定一个长度为N的数组arr[] ，任务是检查它是否可以通过合并两个相同或不同长度的排列来形成。如果这种合并是可能的，则打印YES 。否则，打印NO 。
Permutations of length 3 are {1, 2, 3}, {2, 3, 1}, {1, 3, 2}, {3, 1, 2}, {3, 2, 1}, {2, 1, 3}.
例子：
Input: arr = [1, 3, 2, 4, 3, 1, 2]
Output: YES
Explanation:
The given array can be formed by merging a permutation of length 4 [1, 3, 2, 4] and permutation of length 3 [3, 1, 2]
Input: arr = [1, 2, 3, 2, 3, 2, 1]
Output: NO
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 ，而后缀a[l …… n]的 MEX 将为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，答案将是“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 offbeatPython3
# 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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