给定一个由N个字符组成的数组arr [] ，任务是生成最多X个元素(1≤X≤N)的所有可能组合。
例子：
Input: N = 3, X = 2, arr[] = {‘a’, ‘b’, ‘a’}
Output: a b c bc ca ab cb ac ba
Explanation: All possible combinations using 1 character is 3 {‘a’, ‘b’, ‘c’}. All possible combinations using 2 characters are {“bc” “ca” “ab” “cb” “ac” “ba”}.
Input: N = 3, X = 3, arr[] = {‘d’, ‘a’, ‘b’}
Output: d a b da ab bd ad ba db dab dba abd adb bda bad
方法：可以使用动态编程方法解决给定的问题。请按照以下步骤解决问题：
- 生成可以使用1个字符(即给定数组arr [])创建的所有可能的排列。
- 存储所有排列。
- 存储后，生成2个字符的所有可能排列并存储它们。
- 完成最后一步后，丢弃单个字符的所有排列。
- 以相同的方式迭代地计算置换，直到达到X。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to generate permutations of
// at most X elements from array arr[]
void differentFlagPermutations(int X,
vector arr)
{
vector ml;
ml = arr;
for(int i = 0; i < ml.size(); i++)
{
cout << ml[i] << " ";
}
int count = ml.size();
// Traverse all possible lengths
for(int z = 0; z < X - 1; z++)
{
// Stores all combinations
// of length z
vector tmp;
// Traverse the array
for(int i = 0; i < arr.size(); i++)
{
for(int k = 0; k < ml.size(); k++)
{
if (arr[i] != ml[k])
{
// Generate all
// combinations of length z
tmp.push_back(ml[k] + arr[i]);
count += 1;
}
}
}
// Print all combinations of length z
for(int i = 0; i < tmp.size(); i++)
{
cout << tmp[i] << " ";
}
// Replace all combinations of length z - 1
// with all combinations of length z
ml = tmp;
}
}
// Driver Code
int main()
{
// Given array
vector arr{ "c", "a", "b" };
// Given X
int X = 2;
differentFlagPermutations(X, arr);
return 0;
}
// This code is contributed by divyeshrabadiya07 Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to generate permutations of
// at most X elements from array arr[]
static void differentFlagPermutations(int X,
String[] arr)
{
String[] ml = arr;
for(int i = 0; i < ml.length; i++)
{
System.out.print(ml[i] + " ");
}
int count = ml.length;
// Traverse all possible lengths
for(int z = 0; z < X - 1; z++)
{
// Stores all combinations
// of length z
Vector tmp = new Vector();
// Traverse the array
for(int i = 0; i < arr.length; i++)
{
for(int k = 0; k < ml.length; k++)
{
if (arr[i] != ml[k])
{
// Generate all
// combinations of length z
tmp.add(ml[k] + arr[i]);
count += 1;
}
}
}
// Print all combinations of length z
for(int i = 0; i < tmp.size(); i++)
{
System.out.print(tmp.get(i) + " ");
}
// Replace all combinations of length z - 1
// with all combinations of length z
ml = tmp.toArray(new String[tmp.size()]);;
}
}
// Driver Code
public static void main(String[] args)
{
// Given array
String []arr = { "c", "a", "b" };
// Given X
int X = 2;
differentFlagPermutations(X, arr);
}
}
// This code is contributed by shikhasingrajput Python3
# Python3 program for the above approach
# Function to generate permutations of
# at most X elements from array arr[]
def differentFlagPermutations(X, arr):
ml = arr.copy()
print(" ".join(ml), end =" ")
count = len(ml)
# Traverse all possible lengths
for z in range(X-1):
# Stores all combinations
# of length z
tmp = []
# Traverse the array
for i in arr:
for k in ml:
if i not in k:
# Generate all
# combinations of length z
tmp.append(k + i)
count += 1
# Print all combinations of length z
print(" ".join(tmp), end =" ")
# Replace all combinations of length z - 1
# with all combinations of length z
ml = tmp
# Given array
arr = ['c', 'a', 'b']
# Given X
X = 2
differentFlagPermutations(X, arr)C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to generate permutations of
// at most X elements from array arr[]
static void differentFlagPermutations(int X, List arr)
{
List ml = new List();
ml = arr;
for(int i = 0; i < ml.Count; i++)
{
Console.Write(ml[i] + " ");
}
int count = ml.Count;
// Traverse all possible lengths
for(int z = 0; z < X - 1; z++)
{
// Stores all combinations
// of length z
List tmp = new List();
// Traverse the array
for(int i = 0; i < arr.Count; i++)
{
for(int k = 0; k < ml.Count; k++)
{
if (arr[i] != ml[k])
{
// Generate all
// combinations of length z
tmp.Add(ml[k] + arr[i]);
count += 1;
}
}
}
// Print all combinations of length z
for(int i = 0; i < tmp.Count; i++)
{
Console.Write(tmp[i] + " ");
}
// Replace all combinations of length z - 1
// with all combinations of length z
ml = tmp;
}
}
// Driver code
static void Main()
{
// Given array
List arr = new List(new string[] { "c", "a", "b" });
// Given X
int X = 2;
differentFlagPermutations(X, arr);
}
}
// This code is contributed by divyesh072019 输出：
c a b ac bc ca ba cb ab时间复杂度： O(X * N 2 )
辅助空间： O(N 2 )