Given an integer N, the task is to generate a permutation of elements in the range [1, N] in such order that the length of LIS from starting is equal to LIS from the end of the array. Print -1 if no such array exists.
Examples:
Input: N = 5
Output: [1, 3, 5, 4, 2]
Explanation:
LIS from left side: [1, 3, 5]
LIS from right side: [2, 4, 5]
Length of LIS from left and right side is same that is 3.Input: N = 7
Output: [1, 3, 4, 7, 6, 5, 2]
Explanation:
LIS from left side: [1, 3, 4, 7]
LIS from right side: [2, 5, 6, 7]
Length of LIS from left and right side is same that is 4.
Approach: Use the below observation to solve this problem:
- If N = 2, then such an array doesn't exist.
- Else if N is odd,
- Then start by adding 1 at index 0 and adding 2 at index N-1.
- Now, add N at index N/2. After that, if N>3 then add remaining numbers 3, 4, 5, ... so on indices 1, 2, ..., (N/2-1).
- Now add remaining numbers in decreasing order from index N/2+1 to N-2.
- If N is even,
- say 4 then it has only 1 possible solution: {2, 1, 4, 3}.
- After 4, suppose N = 6, then just add 6 at starting and 5 at ending to the previous answer for N = 4 which forms array = [6, 2, 1, 4, 3, 5].
- Similarly, for next even N just add N at 0th index and N-1 at last index..
Follow the steps mentioned below to solve the problem:
- If the size mentioned is 2, then return -1.
- Otherwise, add the elements as per the observation mentioned above.
- Return the formed list and print the list elements.
Below is the implementation of the above approach:
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to create array having
// same LIS from both sides
vector<int> equalLIS(int N)
{
// If N = 2 we can't create array
if (N == 2) {
vector<int> al;
al.push_back(-1);
return al;
}
// If N is odd it is possible
// to make an array
else if (N % 2 != 0) {
// Hash Map contains index
// with its value
map<int, int> lis;
// indices 0, N/2, and N-1
lis.insert({ 0, 1 });
lis.insert({ N - 1, 2 });
int mid = N / 2;
lis.insert({ mid, N });
if (N > 3) {
// Adding numbers 3, 4, ...
// and so on at index from
// 1 to mid-1 respectively
int val = 3;
for (int i = 1; i < mid; i++) {
lis.insert({ i, val });
val++;
}
// Adding remaining integers
// in decreasing order
// from mid+1 to N-2
val = N - 1;
for (int i = mid + 1; i < N - 1; i++) {
lis.insert({ i, val });
val--;
}
}
// Creating Array List
// which will use to form
// required array
vector<int> al;
// Traversing from smallest key
// to largest key in Hash Map
// and adding its value in list
for (int i = 0; i < N; i++) {
al.push_back(lis[i]);
}
// Returning formed array
return al;
}
else {
// Hash Map which contains
// index with its value
map<int, int> lis;
// Adding value for N=4 in Hash Map
lis.insert({ 0, 2 });
lis.insert({ 1, 1 });
lis.insert({ 2, 4 });
lis.insert({ 3, 3 });
int i = 3;
int idx = 0;
if (N >= 6) {
i++;
idx--;
// Adding new N at starting index
// and N-1 at last index
int val = 5;
while (val <= N) {
lis.insert({ i, val });
lis.insert({ idx, val + 1 });
idx--;
i++;
val += 2;
}
}
// Creating Array List
// which will use to form
// required array
vector<int> al;
// Traversing from minimum key
// to maximum key add
// adding its value in Array List
for (int j = idx + 1; j < i; j++) {
al.push_back(lis[j]);
}
// Returning formed array
return al;
}
}
// Driver Code
int main()
{
int N = 7;
vector<int> lis = equalLIS(N);
for (auto x : lis)
cout << x << " ";
return 0;
}
// This code is contributed by rakeshsahni
// Java program for the above approach
import java.util.*;
class GFG {
// Driver Code
public static void main(String[] args)
{
int N = 7;
List<Integer> lis = equalLIS(N);
for (int x : lis)
System.out.print(x + " ");
}
// Function to create array having
// same LIS from both sides
public static List<Integer> equalLIS(int N)
{
// If N = 2 we can't create array
if (N == 2) {
List<Integer> al = new ArrayList<>();
al.add(-1);
return al;
}
// If N is odd it is possible
// to make an array
else if (N % 2 != 0) {
// Hash Map contains index
// with its value
Map<Integer, Integer> lis
= new HashMap<>();
// Adding 1, N, and 2 at
// indices 0, N/2, and N-1
lis.put(0, 1);
lis.put(N - 1, 2);
int mid = N / 2;
lis.put(mid, N);
if (N > 3) {
// Adding numbers 3, 4, ...
// and so on at index from
// 1 to mid-1 respectively
int val = 3;
for (int i = 1; i < mid; i++) {
lis.put(i, val);
val++;
}
// Adding remaining integers
// in decreasing order
// from mid+1 to N-2
val = N - 1;
for (int i = mid + 1;
i < N - 1; i++) {
lis.put(i, val);
val--;
}
}
// Creating Array List
// which will use to form
// required array
List<Integer> al
= new ArrayList<>();
// Traversing from smallest key
// to largest key in Hash Map
// and adding its value in list
for (int i = 0; i < N; i++) {
al.add(lis.get(i));
}
// Returning formed array
return al;
}
else {
// Hash Map which contains
// index with its value
Map<Integer, Integer> lis
= new HashMap<>();
// Adding value for N=4 in Hash Map
lis.put(0, 2);
lis.put(1, 1);
lis.put(2, 4);
lis.put(3, 3);
int i = 3;
int idx = 0;
if (N >= 6) {
i++;
idx--;
// Adding new N at starting index
// and N-1 at last index
int val = 5;
while (val <= N) {
lis.put(i, val);
lis.put(idx, val + 1);
idx--;
i++;
val += 2;
}
}
// Creating Array List
// which will use to form
// required array
List<Integer> al
= new ArrayList<>();
// Traversing from minimum key
// to maximum key add
// adding its value in Array List
for (int j = idx + 1; j < i; j++) {
al.add(lis.get(j));
}
// Returning formed array
return al;
}
}
}
# Python3 program for the above approach
# Function to create array having
# same LIS from both sides
def equalLIS(N):
# If N = 2 we can't create array
if (N == 2):
al = []
al.append(-1)
return al
# If N is odd it is possible
# to make an array
elif (N % 2 != 0):
# Hash Map(dictionary) contains index
# with its value
lis = {}
# indices 0, N/2, and N-1
lis[0] = 1
lis[N - 1] = 2
mid = N // 2
lis[mid] = N
if (N > 3):
# Adding numbers 3, 4, ...
# and so on at index from
# 1 to mid-1 respectively
val = 3
for i in range(1, mid):
lis[i] = val
val += 1
# Adding remaining integers
# in decreasing order
# from mid+1 to N-2
val = N - 1
for i in range(mid+1, N-1):
lis[i] = val
val -= 1
# Creating Array List
# which will use to form
# required array
al = []
# Traversing from smallest key
# to largest key in Hash Map
# and adding its value in list
for i in range(N):
al.append(lis[i])
# Returning formed array
return al
else:
# Hash Map which contains
# index with its value
lis = {}
# Adding value for N=4 in Hash Map
lis[0] = 2
lis[1] = 1
lis[2] = 4
lis[3] = 3
i = 3
idx = 0
if (N >= 6):
i += 1
idx -= 1
# Adding new N at starting index
# and N-1 at last index
val = 5
while (val <= N):
lis[i] = val
lis[idx] = val + 1
idx -= 1
i += 1
val += 2
# Creating Array List
# which will use to form
# required array
al = []
# Traversing from minimum key
# to maximum key add
# adding its value in Array List
for j in range(idx+1, i):
al.append(lis[j])
# Returning formed array
return al
# Driver Code
N = 7
lis = equalLIS(N)
for x in lis:
print(x, end=" ")
# This code is contributed by vikkycirus
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Driver Code
public static void Main(string[] args)
{
int N = 7;
List<int> lis = equalLIS(N);
for (int i = 0; i < lis.Count; i++)
Console.Write(lis[i] + " ");
}
// Function to create array having
// same LIS from both sides
public static List<int> equalLIS(int N)
{
// If N = 2 we can't create array
if (N == 2) {
List<int> al = new List<int>();
al.Add(-1);
return al;
}
// If N is odd it is possible
// to make an array
else if (N % 2 != 0) {
// Hash Map contains index
// with its value
Dictionary<int, int> lis
= new Dictionary<int, int>();
// Adding 1, N, and 2 at
// indices 0, N/2, and N-1
lis[0] = 1;
lis[N - 1] = 2;
int mid = N / 2;
lis[mid] = N;
if (N > 3) {
// Adding numbers 3, 4, ...
// and so on at index from
// 1 to mid-1 respectively
int val = 3;
for (int i = 1; i < mid; i++) {
lis[i] = val;
val++;
}
// Adding remaining integers
// in decreasing order
// from mid+1 to N-2
val = N - 1;
for (int i = mid + 1; i < N - 1; i++) {
lis[i] = val;
val--;
}
}
// Creating Array List
// which will use to form
// required array
List<int> al = new List<int>();
// Traversing from smallest key
// to largest key in Hash Map
// and adding its value in list
for (int i = 0; i < N; i++) {
al.Add(lis[i]);
}
// Returning formed array
return al;
}
else {
// Hash Map which contains
// index with its value
Dictionary<int, int> lis
= new Dictionary<int, int>();
// Adding value for N=4 in Hash Map
lis[0] = 2;
lis[1] = 1;
lis[2] = 4;
lis[3] = 3;
int i = 3;
int idx = 0;
if (N >= 6) {
i++;
idx--;
// Adding new N at starting index
// and N-1 at last index
int val = 5;
while (val <= N) {
lis[i] = val;
lis[idx] = val + 1;
idx -= 1;
i += 1;
val += 2;
}
}
// Creating Array List
// which will use to form
// required array
List<int> al = new List<int>();
// Traversing from minimum key
// to maximum key add
// adding its value in Array List
for (int j = idx + 1; j < i; j++) {
al.Add(lis[j]);
}
// Returning formed array
return al;
}
}
}
// This code is contributed by ukasp.
// Function to create array having same LIS from both sides
function equalLIS(N) {
// If N = 2 we can't create array
if (N === 2) {
let al = [];
al.push(-1);
return al;
}
// If N is odd it is possible to make an array
else if (N % 2 !== 0) {
// Object that contains index with its value
let lis = {};
// indices 0, N/2, and N-1
lis[0] = 1;
lis[N - 1] = 2;
let mid = Math.floor(N / 2);
lis[mid] = N;
if (N > 3) {
// Adding numbers 3, 4, ... and so on at index from 1 to mid-1
// respectively
let val = 3;
for (let i = 1; i < mid; i++) {
lis[i] = val;
val++;
}
// Adding remaining integers in decreasing order from mid+1 to N-2
val = N - 1;
for (let i = mid + 1; i < N - 1; i++) {
lis[i] = val;
val--;
}
}
// Array that will be used to form the required array
let al = [];
// Traversing from smallest key to largest key in Object
// and adding its value in the array
for (let i = 0; i < N; i++) {
al.push(lis[i]);
}
// Returning the formed array
return al;
} else {
// Object that contains index with its value
let lis = {};
// Adding value for N = 4 in Object
lis[0] = 2;
lis[1] = 1;
lis[2] = 4;
lis[3] = 3;
let i = 3;
let idx = 0;
if (N >= 6) {
i++;
idx--;
// Adding new N at starting index and N-1 at last index
let val = 5;
while (val <= N) {
lis[i] = val;
lis[idx] = val + 1;
idx--;
i++;
val += 2;
}
}
// Array that will be used to form the required array
let al = [];
// Traversing from minimum key to maximum key and adding
// its value in the array
for (let j = idx + 1; j < i; j++) {
al.push(lis[j]);
}
// Returning the formed array
return al;
}
}
// Driver code
let N = 7;
let lis = equalLIS(N);
console.log(lis.join(" "));
//This code is contributed by Edula Vinay Kumar Reddy
Output
1 3 4 7 6 5 2
Time Complexity: O(N)
Auxiliary Space: O(N)