Print k different sorted permutations of a given array
Last Updated :
22 Jul, 2021
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Given an array arr[] containing N integers, the task is to print k different permutations of indices such that the values at those indices form a non-decreasing sequence. Print -1 if it is not possible.
Examples:
Input: arr[] = {1, 3, 3, 1}, k = 3
Output:
0 3 1 2
3 0 1 2
3 0 2 1
For every permutation, the values at the indices form the following sequence {1, 1, 3, 3}
Input: arr[] = {1, 2, 3, 4}, k = 3
Output: -1
There is only 1 non decreasing sequence possible {1, 2, 3, 4}.
Approach: Sort the given array and keep track of the original indices of each element. That gives one required permutation. Now if any 2 continuous elements are equal then they can be swapped to get another permutation. Similarly, the third permutation can be generated.
Below is the implementation of the above approach:
// C++ implementation of the approach
#include <bits/stdc++.h>
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<int, int> 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<int, int> 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 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<pair>
{
// 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 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# 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.
<script>
// Javascript implementation of the approach
var next_pos = 1;
// Utility function to print the original indices
// of the elements of the array
function printIndices(n, a)
{
for (var i = 0; i < n; i++)
document.write( a[i][1] + " ");
document.write("<br>");
}
// Function to print the required permutations
function printPermutations(n, a, k)
{
// To keep track of original indices
var arr = Array.from(Array(n), ()=>Array(2));
for (var i = 0; i < n; i++)
{
arr[i][0] = a[i];
arr[i][1] = i;
}
// Sort the array
arr.sort();
// Count the number of swaps that can
// be made
var count = 1;
for (var i = 1; i < n; i++)
if (arr[i][0] == arr[i - 1][0])
count++;
// Cannot generate 3 permutations
if (count < k) {
document.write( "-1");
return;
}
for (var i = 0; i < k - 1; i++)
{
// Print the first permutation
printIndices(n, arr);
// Find an index to swap and create
// second permutation
for (var j = next_pos; j < n; j++)
{
if (arr[j][0] == arr[j - 1][0])
{
[arr[j], arr[j - 1]] = [arr[j - 1], arr[j]];
next_pos = j + 1;
break;
}
}
}
// Print the last permutation
printIndices(n, arr);
}
// Driver code
var a = [1, 3, 3, 1];
var n = a.length;
var k = 3;
// Function call
printPermutations(n, a, k);
// This code is contributed by famously.
</script>
Output
0 3 1 2 3 0 1 2 3 0 2 1
Time Complexity: O(N log N + K N)
