Number of steps to sort the array by changing order of three elements in each step
Last Updated :
12 Jul, 2025
Given an array arr[] of size N consisting of unique elements in the range [0, N-1], the task is to find K which is the number of steps required to sort the given array by selecting three distinct elements and rearranging them. And also, print the indices selected in those K steps in K lines.
For example, in the array {5, 4, 3, 2, 1, 0}, one possible way to sort the given array by selecting three distinct elements is to select the numbers {2, 1, 0} and sort them as {0, 1, 2} thereby making the array {5, 4, 3, 0, 1, 2}. Similarly, the remaining operations are performed and the indices selected ({3, 4, 5} in the above case) are printed in separate lines.
Examples:
Input: arr[] = {0, 5, 4, 3, 2, 1}
Output:
2
1 2 5
2 5 4
Explanation:
The above array can be sorted in 2 steps:
Step I: We change the order of elements at indices 1, 2, 5 then the array becomes {0, 1, 5, 3, 2, 4}.
Step II: We again change the order of elements at the indices 2, 5, 4 then the array becomes {0, 1, 2, 3, 4, 5} which is sorted.
Input: arr[] = {0, 3, 1, 6, 5, 2, 4}
Output: -1
Explanation:
The above array cannot be sorted in any number of steps.
Suppose we choose indices 1, 3, 2 then the array becomes {0, 1, 6, 3, 5, 2, 4}
After that, we choose indices 2, 6, 4 then the array becomes {0, 1, 5, 3, 4, 2, 6}.
Now only two elements are left unsorted so we cannot choose 3 elements so the above array cannot be sorted. We can try with any order of indices and we will always be left with 2 elements unsorted.
Approach: The idea is to first count the elements which are not sorted and insert them in an unordered set. If count is 0 then we don't need any number of steps for sorting the array so we print 0 and exit. Else, we first erase all the elements from the set for which i = A[A[i]] then we perform the following operation till the set becomes empty:
- We select all the possible combination of indices(if available) such that minimum two elements will get sorted.
- Now, change the order of the elements and erase them from the set if i = A[i].
- Then, we are left with only those elements such that i = A[A[i]] and the count of those must be a multiple of 4 otherwise it is not possible to sort the elements.
- Then, we choose any two pair and perform changing the order of elements two times. Then all the four chosen elements will get sorted.
- We store all the indices which are involved in the changing of orders of elements in a vector and print it as the answer.
Let's understand the above approach with an example. Let the array arr[] = {0, 8, 9, 10, 1, 7, 12, 4, 3, 2, 6, 5, 11}. Then:
- Initially, the set will contain all the 12 elements and there are no elements such that i = A[A[i]].
- Now, {11, 5, 7} are chosen and the order of the elements are changed. Then, arr[] = {0, 8, 9, 10, 1, 5, 12, 7, 3, 2, 6, 4, 11}.
- Now, {11, 4, 1} are chosen and the order of the elements are changed. Then, arr[] = {0, 1, 9, 10, 4, 5, 12, 7, 3, 2, 6, 8, 11}.
- Now, {11, 8, 3} are chosen and the order of the elements are changed. Then, arr[] = {0, 1, 9, 3, 4, 5, 12, 7, 8, 2, 6, 10, 11}.
- Now, {11, 10, 6} are chosen and the order of the elements are changed. Then, arr[] = {0, 1, 9, 3, 4, 5, 6, 7, 8, 2, 10, 12, 11}.
- After the above step, we are left with two pairs of unsorted elements such that i = A[A[i]].
- Finally, {2, 11, 9} and {11, 9, 5} are chosen and reordered. Then, arr[] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} which is sorted.
Below is the implementation of the above approach:
CPP
// C++ program to sort the array
// by changing the order of
// three elements
#include <bits/stdc++.h>
using namespace std;
// Function to change the order of
// the elements having a temporary
// vector and the required indices
// as the arguments
void cngorder(vector<int>& v, int i,
int j, int k)
{
int temp = v[k];
v[k] = v[j];
v[j] = v[i];
v[i] = temp;
}
// Function to sort the elements having
// the given array and its size.
void sortbyorder3(vector<int>& A, int n)
{
// Flag to check whether the sorting
// is possible or not
bool flag = 0;
int count = 0;
// Set that will contains unsorted
// elements
unordered_set<int> s;
// Iterating through the elements
for (int i = 0; i < n; i++) {
// Inserting the required elements
// in the set
if (i != A[i])
count++, s.insert(i);
}
// When the given array is
// already sorted
if (count == 0)
cout << "0" << endl;
else {
// Vector that will contain
// the answer
vector<vector<int> > ans;
// Temporary vector to store
// the indices
vector<int> vv;
int x, y, z;
count = 0;
// Loop that will execute till the
// set becomes empty
while (!s.empty()) {
auto it = s.begin();
int i = *it;
// Check for the condition
if (i == A[A[i]]) {
s.erase(i);
s.erase(A[i]);
continue;
}
// Case when the minimum two
// elements will get sorted
else {
x = A[i], y = A[A[i]], z = A[A[A[i]]];
vv.push_back(x), vv.push_back(y),
vv.push_back(z);
// Changing the order of elements
cngorder(A, x, y, z);
// Pushing the indices to the
// answer vector
ans.push_back(vv);
// If the third element also
// gets sorted
if (vv[0] == A[vv[0]])
s.erase(vv[0]);
// Erasing the two sorted elements
// from the set
s.erase(vv[1]), s.erase(vv[2]);
vv.clear();
}
}
count = 0;
// The count of the remaining
// unsorted elements
for (int i = 0; i < n; i++) {
if (i != A[i])
count++;
}
// If the count of the left
// unsorted elements is not
// a multiple of 4, then
// sorting is not possible
if (count % 4 != 0)
flag = 1;
// Only the elements such that
// i = A[A[i]] are left
// for sorting
else {
// Indices of any one element
// from the two pairs that
// will be sorted in 2 steps
int i1 = -1, i2 = -1;
for (int i = 0; i < n; i++) {
// Index of any element of
// the pair
if (A[i] != i && i1 == -1) {
i1 = i;
}
// When we find the second
// pair and the index of
// any one element is stored
else if (A[i] != i && i1 != -1
&& i2 == -1) {
if (i1 == A[i])
continue;
else
i2 = i;
}
// When we got both the pair
// of elements
if (i1 != -1 && i2 != -1) {
// Remaining two indices
// of the elements
int i3 = A[i1], i4 = A[i2];
// The first order of indices
vv.push_back(i1),
vv.push_back(i2),
vv.push_back(A[i1]);
// Pushing the indices to the
// answer vector
ans.push_back(vv);
vv.clear();
// The second order of indices
vv.push_back(i2),
vv.push_back(A[i1]),
vv.push_back(A[i2]);
// Pushing the indices to the
// answer vector
ans.push_back(vv);
vv.clear();
// Changing the order of the
// first combination of
// the indices
cngorder(A, i1, i2, i3);
// Changing the order of the
// second combination of
// the indices after which all
// the 4 elements will be sorted
cngorder(A, i2, i3, i4);
i1 = -1, i2 = -1;
}
}
}
// If the flag value is 1
// the sorting is not possible
if (flag == 1)
cout << "-1" << endl;
else {
// Printing the required number
// of steps
cout << ans.size() << endl;
// Printing the indices involved
// in the shifting
for (int i = 0; i < ans.size(); i++) {
cout << ans[i][0]
<< " " << ans[i][1]
<< " " << ans[i][2]
<< endl;
}
}
}
}
// Driver code
int main()
{
int n;
vector<int> A{ 0, 8, 9, 10, 1, 7, 12,
4, 3, 2, 6, 5, 11 };
n = A.size();
// Calling the sorting function
sortbyorder3(A, n);
return 0;
}
// java program to sort the array
// by changing the order of
// three elements
import java.util.*;
class Main {
// Function to change the order of
// the elements having a temporary
// vector and the required indices
// as the arguments
static void cngorder(List<Integer> v, int i, int j, int k) {
int temp = v.get(k);
v.set(k, v.get(j));
v.set(j, v.get(i));
v.set(i, temp);
}
// Function to sort the elements having
// the given array and its size.
static void sortbyorder3(List<Integer> A, int n) {
// Flag to check whether the sorting
// is possible or not
boolean flag = false;
int count = 0;
// Set that will contains unsorted
// elements
HashSet<Integer> s = new HashSet<>();
// Iterating through the elements
for (int i = 0; i < n; i++) {
// Inserting the required elements
// in the set
if (i != A.get(i)) {
count++;
s.add(i);
}
}
// When the given array is
// already sorted
if (count == 0) {
System.out.println("0");
} else {
// Vector that will contain
// the answer
List<List<Integer>> ans = new ArrayList<>();
// Temporary vector to store
// the indices
List<Integer> vv = new ArrayList<>();
int x, y, z;
count = 0;
// Loop that will execute till the
// set becomes empty
while (!s.isEmpty()) {
Iterator<Integer> it = s.iterator();
// Check for the condition
int i = it.next();
if (i == A.get(A.get(i))) {
s.remove(i);
s.remove(A.get(i));
continue;
}
// Case when the minimum two
else {
x = A.get(i);
y = A.get(A.get(i));
z = A.get(A.get(A.get(i)));
vv.add(x);
vv.add(y);
vv.add(z);
// Changing the order of elements
cngorder(A, x, y, z);
// Pushing the indices to the
// answer vector
ans.add(new ArrayList<>(vv));
// If the third element also
// gets sorted
if (vv.get(0).equals(A.get(vv.get(0)))){
s.remove(vv.get(0));
}
// Erasing the two sorted elements
// from the set
s.remove(vv.get(1));
s.remove(vv.get(2));
vv.clear();
}
}
count = 0;
// The count of the remaining
// unsorted elements
for (int i = 0; i < n; i++) {
if (i != A.get(i)) {
count++;
}
}
// If the count of the left
// unsorted elements is not
// a multiple of 4, then
// sorting is not possible
if (count % 4 != 0) {
flag = true;
}
// Only the elements such that
// i = A[A[i]] are left
// for sorting
else {
// Indices of any one element
// from the two pairs that
// will be sorted in 2 steps
int i1 = -1, i2 = -1;
for (int i = 0; i < n; i++) {
// Index of any element of
// the pair
if (!A.get(i).equals(i) && i1 == -1) {
i1 = i;
}
// When we find the second
// pair and the index of
// any one element is stored
else if (!A.get(i).equals(i) && i1 != -1 && i2 == -1) {
if (i1 == A.get(i))
continue;
else
i2 = i;
}
if (i1 != -1 && i2 != -1) {
// Remaining two indices
// of the elements
int i3 = A.get(i1), i4 = A.get(i2);
// The first order of indices
vv.add(i1);
vv.add(i2);
vv.add(A.get(i1));
// Pushing the indices to the
// answer vector
ans.add(new ArrayList<>(vv));
vv.clear();
// The second order of indices
vv.add(i2);
vv.add(A.get(i1));
vv.add(A.get(i2));
// Pushing the indices to the
// answer vector
ans.add(new ArrayList<>(vv));
vv.clear();
// Changing the order of the
// first combination of
// the indices
cngorder(A, i1, i2, i3);
// Changing the order of the
// second combination of
// the indices after which all
// the 4 elements will be sorted
cngorder(A, i2, i3, i4);
i1 = -1;
i2 = -1;
}
}
}
// If the flag value is 1
// the sorting is not possible
if (flag) {
System.out.println("-1");
} else {
// Printing the required number
// of steps
System.out.println(ans.size());
// Printing the indices involved
// in the shifting
for (int i = 0; i < ans.size(); i++) {
System.out.println(ans.get(i).get(0) + " " + ans.get(i).get(1) + " " + ans.get(i).get(2));
}
}
}
}
// Driver code
public static void main(String[] args) {
List<Integer> A = new ArrayList<>(Arrays.asList(0, 8, 9, 10, 1, 7, 12, 4, 3, 2, 6, 5, 11));
int n = A.size();
sortbyorder3(A, n);
}
}
//This code is contributed by shivregkec
def cngorder(arr, i, j, k):
"""
Function to change the order of elements
"""
temp = arr[k]
arr[k] = arr[j]
arr[j] = arr[i]
arr[i] = temp
def sortbyorder3(A, n):
"""
Function to sort the array by changing the order of three elements
"""
# Flag to check whether the sorting is possible or not
flag = 0
count = 0
# Set that will contain unsorted elements
s = set()
# Iterating through the elements
for i in range(n):
# Inserting the required elements in the set
if i != A[i]:
count += 1
s.add(i)
# When the given array is already sorted
if count == 0:
print("0")
else:
# Vector that will contain the answer
ans = []
# Temporary list to store the indices
vv = []
x, y, z = 0, 0, 0
count = 0
# Loop that will execute till the set becomes empty
while s:
i = next(iter(s))
# Check for the condition
if i == A[A[i]]:
s.remove(i)
s.remove(A[i])
continue
else:
x, y, z = A[i], A[A[i]], A[A[A[i]]]
vv.extend([x, y, z])
# Changing the order of elements
cngorder(A, x, y, z)
# Pushing the indices to the answer list
ans.append(list(vv))
# If the third element also gets sorted
if vv[0] == A[vv[0]]:
s.remove(vv[0])
# Erasing the two sorted elements from the set
s.remove(vv[1])
s.remove(vv[2])
vv.clear()
count = 0
# The count of the remaining unsorted elements
for i in range(n):
if i != A[i]:
count += 1
# If the count of the left unsorted elements is not a multiple of 4, then sorting is not possible
if count % 4 != 0:
flag = 1
else:
i1, i2 = -1, -1
for i in range(n):
if A[i] != i and i1 == -1:
i1 = i
elif A[i] != i and i1 != -1 and i2 == -1:
if i1 == A[i]:
continue
else:
i2 = i
if i1 != -1 and i2 != -1:
i3, i4 = A[i1], A[i2]
# The first order of indices
vv.extend([i1, i2, A[i1]])
# Pushing the indices to the answer list
ans.append(list(vv))
vv.clear()
# The second order of indices
vv.extend([i2, A[i1], A[i2]])
# Pushing the indices to the answer list
ans.append(list(vv))
vv.clear()
# Changing the order of the first combination of the indices
cngorder(A, i1, i2, i3)
# Changing the order of the second combination of the indices after which all the 4 elements will be sorted
cngorder(A, i2, i3, i4)
i1, i2 = -1, -1
# If the flag value is 1, the sorting is not possible
if flag == 1:
print("-1")
else:
# Printing the required number of steps
print(len(ans))
# Printing the indices involved in the shifting
for a in ans:
print(*a)
# Driver code
if __name__ == "__main__":
A = [0, 8, 9, 10, 1, 7, 12, 4, 3, 2, 6, 5, 11]
n = len(A)
# Calling the sorting function
sortbyorder3(A, n)
// C# program to sort the array
// by changing the order of
// three elements
using System;
using System.Collections.Generic;
class GFG {
// Function to change the order of
// the elements having a temporary
// vector and the required indices
// as the arguments
static void cngorder(List<int> v, int i, int j, int k)
{
int temp = v[k];
v[k] = v[j];
v[j] = v[i];
v[i] = temp;
}
// Function to sort the elements having
// the given array and its size.
static void sortbyorder3(List<int> A, int n)
{
// Flag to check whether the sorting
// is possible or not
bool flag = false;
int count = 0;
// Set that will contains unsorted
// elements
HashSet<int> s = new HashSet<int>();
// Iterating through the elements
for (int i = 0; i < n; i++) {
// Inserting the required elements
// in the set
if (i != A[i]) {
count++;
s.Add(i);
}
}
// When the given array is
// already sorted
if (count == 0) {
Console.WriteLine("0");
}
else {
// Vector that will contain
// the answer
List<List<int>> ans = new List<List<int>>();
// Temporary vector to store
// the indices
List<int> vv = new List<int>();
int x, y, z;
count = 0;
// Loop that will execute till the
// set becomes empty
while (s.Count > 0) {
var it = s.GetEnumerator();
it.MoveNext();
// Check for the condition
int i = it.Current;
if (i == A[A[i]]) {
s.Remove(i);
s.Remove(A[i]);
continue;
}
// Case when the minimum two
else {
x = A[i];
y = A[A[i]];
z = A[A[A[i]]];
vv.Add(x);
vv.Add(y);
vv.Add(z);
// Changing the order of elements
cngorder(A, x, y, z);
// Pushing the indices to the
// answer vector
ans.Add(new List<int>(vv));
// If the third element also
// gets sorted
if (vv[0] == A[vv[0]]) {
s.Remove(vv[0]);
}
// Erasing the two sorted elements
// from the set
s.Remove(vv[1]);
s.Remove(vv[2]);
vv.Clear();
}
}
count = 0;
// The count of the remaining
// unsorted elements
for (int i = 0; i < n; i++) {
if (i != A[i]) {
count++;
}
}
// If the count of the left
// unsorted elements is not
// a multiple of 4, then
// sorting is not possible
if (count % 4 != 0) {
flag = true;
}
// Only the elements such that
// i = A[A[i]] are left
// for sorting
else {
// Indices of any one element
// from the two pairs that
// will be sorted in 2 steps
int i1 = -1, i2 = -1;
for (int i = 0; i < n; i++) {
// Index of any element of
// the pair
if (A[i] != i && i1 == -1) {
i1 = i;
}
// When we find the second
// pair and the index of
// any one element is stored
else if (A[i] != i && i1 != -1 && i2 == -1) {
if (i1 == A[i])
continue;
else
i2 = i;
}
if (i1 != -1 && i2 != -1) {
// Remaining two indices
// of the elements
int i3 = A[i1], i4 = A[i2];
// The first order of indices
vv.Add(i1);
vv.Add(i2);
vv.Add(A[i1]);
// Pushing the indices to the
// answer vector
ans.Add(new List<int>(vv));
vv.Clear();
// The second order of indices
vv.Add(i2);
vv.Add(A[i1]);
vv.Add(A[i2]);
// Pushing the indices to the
// answer vector
ans.Add(new List<int>(vv));
vv.Clear();
// Changing the order of the
// first combination of
// the indices
cngorder(A, i1, i2, i3);
// Changing the order of the
// second combination of
// the indices after which all
// the 4 elements will be sorted
cngorder(A, i2, i3, i4);
i1 = -1;
i2 = -1;
}
}
}
// If the flag value is 1
// the sorting is not possible
if (flag) {
Console.WriteLine("-1");
}
else {
// Printing the required number
// of steps
Console.WriteLine(ans.Count);
// Printing the indices involved
// in the shifting
for (int i = 0; i < ans.Count; i++) {
Console.WriteLine(ans[i][0] + " " + ans[i][1] + " " + ans[i][2]);
}
}
}
}
// Driver code
static void Main(string[] args)
{
List<int> A = new List<int> { 0, 8, 9, 10, 1, 7, 12, 4, 3, 2, 6, 5, 11 };
int n = A.Count;
sortbyorder3(A, n);
}
}
//This code is contributed by Shivhack999
// Function to change the order of
// the elements having a temporary
// array and the required indices
function cngorder(v, i, j, k) {
let temp = v[k];
v[k] = v[j];
v[j] = v[i];
v[i] = temp;
}
// Function to sort the elements having
// the given array and its size.
function sortbyorder3(A, n) {
// Flag to check whether the sorting
// is possible or not
let flag = 0;
let count = 0;
// Set that will contain unsorted
// elements
let s = new Set();
// Iterating through the elements
for (let i = 0; i < n; i++) {
// Inserting the required elements
// in the set
if (i !== A[i])
count++, s.add(i);
}
// When the given array is
// already sorted
if (count === 0)
console.log("0");
else {
// Array that will contain
// the answer
let ans = [];
// Temporary array to store
// the indices
let vv = [];
let x, y, z;
count = 0;
// Loop that will execute till the
// set becomes empty
while (s.size !== 0) {
let it = s.values().next().value;
let i = it;
// Check for the condition
if (i === A[A[i]]) {
s.delete(i);
s.delete(A[i]);
continue;
}
// Case when the minimum two
// elements will get sorted
else {
x = A[i], y = A[A[i]], z = A[A[A[i]]];
vv.push(x), vv.push(y),
vv.push(z);
// Changing the order of elements
cngorder(A, x, y, z);
// Pushing the indices to the
// answer array
ans.push([...vv]);
// If the third element also
// gets sorted
if (vv[0] === A[vv[0]])
s.delete(vv[0]);
// Erasing the two sorted elements
// from the set
s.delete(vv[1]);
s.delete(vv[2]);
vv = [];
}
}
count = 0;
// The count of the remaining
// unsorted elements
for (let i = 0; i < n; i++) {
if (i !== A[i])
count++;
}
// If the count of the left
// unsorted elements is not
// a multiple of 4, then
// sorting is not possible
if (count % 4 !== 0)
flag = 1;
// Only the elements such that
// i = A[A[i]] are left
// for sorting
else {
// Indices of any one element
// from the two pairs that
// will be sorted in 2 steps
let i1 = -1, i2 = -1;
for (let i = 0; i < n; i++) {
// Index of any element of
// the pair
if (A[i] !== i && i1 === -1) {
i1 = i;
}
// When we find the second
// pair and the index of
// any one element is stored
else if (A[i] !== i && i1 !== -1
&& i2 === -1) {
if (i1 === A[i])
continue;
else
i2 = i;
}
// When we got both the pair
// of elements
if (i1 !== -1 && i2 !== -1) {
// Remaining two indices
// of the elements
let i3 = A[i1], i4 = A[i2];
// The first order of indices
vv.push(i1),
vv.push(i2),
vv.push(i3);
// Pushing the indices to the
// answer array
ans.push([...vv]);
vv = [];
// The second order of indices
vv.push(i2),
vv.push(i3),
vv.push(i4);
// Pushing the indices to the
// answer array
ans.push([...vv]);
vv = [];
// Changing the order of the
// first combination of
// the indices
cngorder(A, i1, i2, i3);
// Changing the order of the
// second combination of
// the indices after which all
// the 4 elements will be sorted
cngorder(A, i2, i3, i4);
i1 = -1, i2 = -1;
}
}
}
// If the flag value is 1
// the sorting is not possible
if (flag === 1)
console.log("-1");
else {
// Printing the required number
// of steps
console.log(ans.length);
// Printing the indices involved
// in the shifting
for (let i = 0; i < ans.length; i++) {
console.log(ans[i][0],
ans[i][1],
ans[i][2]);
}
}
}
}
// Driver code
function main() {
let n;
let A = [0, 8, 9, 10, 1, 7, 12,
4, 3, 2, 6, 5, 11];
n = A.length;
// Calling the sorting function
sortbyorder3(A, n);
}
// Invoke the main function
main();
Output6
11 5 7
11 4 1
11 8 3
11 10 6
2 11 9
11 9 12
Time Complexity: O(N), where N is the size of the array.
Auxiliary space: O(N)
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