原文:
给定一个包含 n 整数的数组 arr[] ，任务是打印 k 不同的索引排列，使得这些索引处的值形成一个非递减序列。如果不可能，打印 -1 。
示例:
输入: arr[] = {1，3，3，1}，k = 3 输出: 0 3 1 2 3 0 1 2 3 0 2 1 对于每个排列，索引处的值形成以下序列{1，1，3，3} 输入: arr[] = {1，2，3，4}，k = 3 输出:-1【t11
方法:对给定的数组进行排序，并跟踪每个元素的原始索引。这给出了一个必要的排列。如果任意两个连续的元素相等，那么它们可以交换，得到另一个排列。类似地，可以生成第三置换。
下面是上述方法的实现:
c
// c implementation of the approach
#include
using namespace std;
int next_pos = 1;
// utility function to print the original indices
// of the elements of the array
void printindices(int n, pair a[])
{
for (int i = 0; i < n; i )
cout << a[i].second << " ";
cout << endl;
}
// function to print the required permutations
void printpermutations(int n, int a[], int k)
{
// to keep track of original indices
pair arr[n];
for (int i = 0; i < n; i )
{
arr[i].first = a[i];
arr[i].second = i;
}
// sort the array
sort(arr, arr n);
// count the number of swaps that can
// be made
int count = 1;
for (int i = 1; i < n; i )
if (arr[i].first == arr[i - 1].first)
count ;
// cannot generate 3 permutations
if (count < k) {
cout << "-1";
return;
}
for (int i = 0; i < k - 1; i )
{
// print the first permutation
printindices(n, arr);
// find an index to swap and create
// second permutation
for (int j = next_pos; j < n; j )
{
if (arr[j].first == arr[j - 1].first)
{
swap(arr[j], arr[j - 1]);
next_pos = j 1;
break;
}
}
}
// print the last permutation
printindices(n, arr);
}
// driver code
int main()
{
int a[] = { 1, 3, 3, 1 };
int n = sizeof(a) / sizeof(a[0]);
int k = 3;
// function call
printpermutations(n, a, k);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java implementation of the approach
import java.util.*;
class gfg {
static int next_pos = 1;
static class pair {
int first, second;
pair()
{
first = 0;
second = 0;
}
}
// utility function to print the original indices
// of the elements of the array
static void printindices(int n, pair a[])
{
for (int i = 0; i < n; i )
system.out.print(a[i].second " ");
system.out.println();
}
static class sort implements comparator
{
// used for sorting in ascending order
public int compare(pair a, pair b)
{
return a.first < b.first ? -1 : 1;
}
}
// function to print the required permutations
static void printpermutations(int n, int a[], int k)
{
// to keep track of original indices
pair arr[] = new pair[n];
for (int i = 0; i < n; i )
{
arr[i] = new pair();
arr[i].first = a[i];
arr[i].second = i;
}
// sort the array
arrays.sort(arr, new sort());
// count the number of swaps that can
// be made
int count = 1;
for (int i = 1; i < n; i )
if (arr[i].first == arr[i - 1].first)
count ;
// cannot generate 3 permutations
if (count < k)
{
system.out.print("-1");
return;
}
for (int i = 0; i < k - 1; i )
{
// print the first permutation
printindices(n, arr);
// find an index to swap and create
// second permutation
for (int j = next_pos; j < n; j )
{
if (arr[j].first == arr[j - 1].first)
{
pair t = arr[j];
arr[j] = arr[j - 1];
arr[j - 1] = t;
next_pos = j 1;
break;
}
}
}
// print the last permutation
printindices(n, arr);
}
// driver code
public static void main(string arsg[])
{
int a[] = { 1, 3, 3, 1 };
int n = a.length;
int k = 3;
// function call
printpermutations(n, a, k);
}
}
// this code is contributed by arnab kundu
python 3
# python 3 implementation of the approach
# utility function to print the original
# indices of the elements of the array
def printindices(n, a):
for i in range(n):
print(a[i][1], end=" ")
print("\n", end="")
# function to print the required
# permutations
def printpermutations(n, a, k):
# to keep track of original indices
arr = [[0, 0] for i in range(n)]
for i in range(n):
arr[i][0] = a[i]
arr[i][1] = i
# sort the array
arr.sort(reverse=false)
# count the number of swaps that
# can be made
count = 1
for i in range(1, n):
if (arr[i][0] == arr[i - 1][0]):
count = 1
# cannot generate 3 permutations
if (count < k):
print("-1", end="")
return
next_pos = 1
for i in range(k - 1):
# print the first permutation
printindices(n, arr)
# find an index to swap and create
# second permutation
for j in range(next_pos, n):
if (arr[j][0] == arr[j - 1][0]):
temp = arr[j]
arr[j] = arr[j - 1]
arr[j - 1] = temp
next_pos = j 1
break
# print the last permutation
printindices(n, arr)
# driver code
if __name__ == '__main__':
a = [1, 3, 3, 1]
n = len(a)
k = 3
# function call
printpermutations(n, a, k)
# this code is contributed by
# surendra_gangwar
c
// c# implementation of the approach
using system;
using system.collections;
using system.collections.generic;
class gfg {
static int next_pos = 1;
public class pair {
public int first, second;
public pair()
{
first = 0;
second = 0;
}
}
class sorthelper : icomparer
{
int icomparer.compare(object a, object b)
{
pair first = (pair)a;
pair second = (pair)b;
return first.first < second.first ? -1 : 1;
}
}
// utility function to print the original indices
// of the elements of the array
static void printindices(int n, pair []a)
{
for (int i = 0; i < n; i )
console.write(a[i].second " ");
console.writeline();
}
// function to print the required permutations
static void printpermutations(int n, int []a, int k)
{
// to keep track of original indices
pair []arr = new pair[n];
for (int i = 0; i < n; i )
{
arr[i] = new pair();
arr[i].first = a[i];
arr[i].second = i;
}
// sort the array
array.sort(arr, new sorthelper());
// count the number of swaps that can
// be made
int count = 1;
for (int i = 1; i < n; i )
if (arr[i].first == arr[i - 1].first)
count ;
// cannot generate 3 permutations
if (count < k)
{
console.write("-1");
return;
}
for (int i = 0; i < k - 1; i )
{
// print the first permutation
printindices(n, arr);
// find an index to swap and create
// second permutation
for (int j = next_pos; j < n; j )
{
if (arr[j].first == arr[j - 1].first)
{
pair t = arr[j];
arr[j] = arr[j - 1];
arr[j - 1] = t;
next_pos = j 1;
break;
}
}
}
// print the last permutation
printindices(n, arr);
}
// driver code
public static void main(string []args)
{
int []a = { 1, 3, 3, 1 };
int n = a.length;
int k = 3;
// function call
printpermutations(n, a, k);
}
}
// this code is contributed by rutvik_56.
java 描述语言
output
0 3 1 2
3 0 1 2
3 0 2 1
时间复杂度: o(n log n k n)
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