Find permutation of numbers 1 to N having X local maxima (peaks) and Y local minima (valleys)

Given three integers N, A and B, the task is to find a permutation of pairwise distinct numbers from 1 to N that has exactly ‘A’ local minima’s and ‘B’ local maxima’s. 

• A local minima is defined as the element which is less than both its neighbours.
• A local maxima is defined as the element which is greater than both its neighbours.
• The first and last element of the entire permutation can never be local minima or maxima.

If no such permutations exist print -1.

Example :

Input: N = 6 ,  A = 2 , B = 2
Output:  1, 3, 2, 5, 4, 6
Explanation : 
2 local minima’s: 2 and 5
2 local maxima’s: 3 and 5

Input: N = 5 , A = 2 , B = 2
Output: -1

Naive Approach (Brute Force): In this approach, generate all permutations of 1 to N numbers and check each one individually. Follow the below steps, to solve this problem:

• Generate all the permutations of numbers from 1 to N and store them in an array.
• Traverse through each permutation, if the following permutation has exactly A local minima’s and B local maxima’s, print the permutation.
• If no such permutation exists, then print -1.

Below is the implementation of the above approach : 

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to generate next permutation
void nextPermutation(vector<int>& nums)
{
    int n = nums.size(), k, l;
    for (k = n - 2; k >= 0; k--) {
        if (nums[k] < nums[k + 1]) {
            break;
        }
    }
    if (k < 0) {
        reverse(nums.begin(), nums.end());
    }
    else {
        for (l = n - 1; l > k; l--) {
            if (nums[l] > nums[k]) {
                break;
            }
        }
        swap(nums[k], nums[l]);
        reverse(nums.begin() + k + 1, nums.end());
    }
}

// Factorial function
int factorial(int n)
{
    return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
}

// Function to returns all the permutations of a given array
// or vector
vector<vector<int> > permute(vector<int>& nums)
{
    vector<vector<int> > permuted;
    int n = nums.size();
    int factn = factorial(n);
    for (int i = 0; i < factn; i++) {
        permuted.push_back(nums);
        nextPermutation(nums);
    }
    return permuted;
}

// Function to find the permutation of 1 to N numbers
// having A minimas and B maximas
void findPermutation(int n, int a, int b)
{

    // Generate the array containing one permutation
    vector<int> nums(n);
    for (int i = 0; i < n; i++) {
        nums[i] = i + 1;
    }

    // Generate all the permutations
    vector<vector<int> > allpermutations = permute(nums);

    int total = allpermutations.size();
    int ansindex = -1;

    for (int i = 0; i < total; i++) {
        // Count local minima and local maximas for each
        // permutation
        int minc = 0, maxc = 0;
        for (int j = 1; j < n - 1; j++) {
            if (allpermutations[i][j]
                    > allpermutations[i][j - 1]
                && allpermutations[i][j]
                       > allpermutations[i][j + 1]) {
                maxc++;
            }
            if (allpermutations[i][j]
                    < allpermutations[i][j - 1]
                && allpermutations[i][j]
                       < allpermutations[i][j + 1]) {
                minc++;
            }
        }
        if (minc == a && maxc == b) {

            // Store the index of a perfect permutation
            ansindex = i;
            break;
        }
    }

    // Print -1 if no such permutation exists
    if (ansindex == -1) {
        cout << -1;
    }
    else {
        // Print the perfect permutation if exists
        for (int i = 0; i < n; i++) {
            cout << allpermutations[ansindex][i] << " ";
        }
    }
}

int main()
{
    int N = 6, A = 2, B = 2;
    findPermutation(N, A, B);
    return 0;
}

Java

// Java program for the above approach

import java.io.*;
import java.util.*;

class GFG {

  // Function to generate next permutation
  public static void nextPermutation(List<Integer> nums)
  {
    int n = nums.size();
    int k, l;
    for (k = n - 2; k >= 0; k--) {
      if (nums.get(k) < nums.get(k + 1)) {
        break;
      }
    }
    if (k < 0) {
      Collections.reverse(nums);
    }
    else {
      for (l = n - 1; l > k; l--) {
        if (nums.get(l) > nums.get(k)) {
          break;
        }
      }
      Collections.swap(nums, k, l);
      List<Integer> subList = nums.subList(k + 1, n);
      Collections.reverse(subList);
    }
  }

  // Factorial function
  public static int factorial(int n)
  {
    return (n == 1 || n == 0) ? 1
      : factorial(n - 1) * n;
  }

  // Function to returns all the permutations of a given
  // array
  // or vector
  public static List<List<Integer> >
    permute(List<Integer> nums)
  {
    List<List<Integer> > permuted = new ArrayList<>();
    int n = nums.size();
    int factn = factorial(n);
    for (int i = 0; i < factn; i++) {
      permuted.add(new ArrayList<>(nums));
      nextPermutation(nums);
    }
    return permuted;
  }

  // Function to find the permutation of 1 to N numbers
  // having A minimas and B maximas
  public static void findPermutation(int n, int a, int b)
  {
    // Generate the array containing one permutation
    List<Integer> nums = new ArrayList<>(n);
    for (int i = 0; i < n; i++) {
      nums.add(i + 1);
    }

    // Generate all the permutations
    List<List<Integer> > allpermutations
      = permute(nums);
    int total = allpermutations.size();
    int ansindex = -1;
    for (int i = 0; i < total; i++) {
      // Count local minima and local maximas for each
      // permutation
      int minc = 0, maxc = 0;
      for (int j = 1; j < n - 1; j++) {
        if (allpermutations.get(i).get(j)
            > allpermutations.get(i).get(j - 1)
            && allpermutations.get(i).get(j)
            > allpermutations.get(i).get(
              j + 1)) {
          maxc++;
        }
        if (allpermutations.get(i).get(j)
            < allpermutations.get(i).get(j - 1)
            && allpermutations.get(i).get(j)
            < allpermutations.get(i).get(
              j + 1)) {
          minc++;
        }
      }
      if (minc == a && maxc == b) {
        // Store the index of a perfect permutation
        ansindex = i;
        break;
      }
    }

    // Print -1 if no such permutation exists
    if (ansindex == -1) {
      System.out.println(-1);
    }
    else {
      // Print the perfect permutation if exists
      for (int i = 0; i < n; i++) {
        System.out.print(
          allpermutations.get(ansindex).get(i)
          + " ");
      }
    }
  }

  public static void main(String[] args)
  {
    int N = 6, A = 2, B = 2;
    findPermutation(N, A, B);
  }
}

// This code is contributed by lokeshmvs21.

C#

using System;
using System.Collections.Generic;

class GFG {
  // Function to generate next permutation
  static void NextPermutation(List<int> nums)
  {
    int n = nums.Count, k, l;
    for (k = n - 2; k >= 0; k--) {
      if (nums[k] < nums[k + 1]) {
        break;
      }
    }
    if (k < 0) {
      nums.Reverse();
    }
    else {
      for (l = n - 1; l > k; l--) {
        if (nums[l] > nums[k]) {
          break;
        }
      }
      int temp = nums[k];
      nums[k] = nums[l];
      nums[l] = temp;
      nums.Reverse(k + 1, n - k - 1);
    }
  }

  // Factorial function
  static int Factorial(int n)
  {
    return (n == 1 || n == 0) ? 1
      : Factorial(n - 1) * n;
  }

  // Function to returns all the permutations of a given
  // array or list
  static List<List<int> > Permute(List<int> nums)
  {
    List<List<int> > permuted = new List<List<int> >();
    int n = nums.Count;
    int factn = Factorial(n);
    for (int i = 0; i < factn; i++) {
      permuted.Add(new List<int>(nums));
      NextPermutation(nums);
    }
    return permuted;
  }

  // Function to find the permutation of 1 to N numbers
  // having A minimas and B maximas
  static void FindPermutation(int n, int a, int b)
  {

    // Generate the array containing one permutation
    List<int> nums = new List<int>(n);
    for (int i = 0; i < n; i++) {
      nums.Add(i + 1);
    }

    // Generate all the permutations
    List<List<int> > allpermutations = Permute(nums);

    int total = allpermutations.Count;
    int ansindex = -1;

    for (int i = 0; i < total; i++) {
      // Count local minima and local maximas for each
      // permutation
      int minc = 0, maxc = 0;
      for (int j = 1; j < n - 1; j++) {
        if (allpermutations[i][j]
            > allpermutations[i][j - 1]
            && allpermutations[i][j]
            > allpermutations[i][j + 1]) {
          maxc++;
        }
        if (allpermutations[i][j]
            < allpermutations[i][j - 1]
            && allpermutations[i][j]
            < allpermutations[i][j + 1]) {
          minc++;
        }
      }
      if (minc == a && maxc == b) {

        // Store the index of a perfect permutation
        ansindex = i;
        break;
      }
    }

    // Print -1 if no such permutation exists
    if (ansindex == -1) {
      Console.WriteLine("-1");
    }
    else {
      // Print the perfect permutation if exists
      for (int i = 0; i < n; i++) {
        Console.Write(allpermutations[ansindex][i]);
        Console.Write(" ");
      }
    }
  }

  static public void Main()
  {

    int N = 6, A = 2, B = 2;
    FindPermutation(N, A, B);
  }
}

// This code is contributed by akashish__

Javascript

// Function to generate next permutation
function nextPermutation(nums) {
    let n = nums.length, k, l;
    for (k = n - 2; k >= 0; k--) {
        if (nums[k] < nums[k + 1]) {
            break;
        }
    }
    if (k < 0) {
        nums.reverse();
    }
    else {
        for (l = n - 1; l > k; l--) {
            if (nums[l] > nums[k]) {
                break;
            }
        }
        let temp = nums[k];
        nums[k] = nums[l];
        nums[l] = temp;
        nums.splice(k + 1, n - k - 1, ...nums.slice(k + 1, n).reverse());
    }
}

// Factorial function
function factorial(n) {
    return (n === 1 || n === 0) ? 1 : factorial(n - 1) * n;
}

// Function to returns all the permutations of a given array
function permute(nums) {
    let permuted = [];
    let n = nums.length;
    let factn = factorial(n);
    for (let i = 0; i < factn; i++) {
        permuted.push([...nums]);
        nextPermutation(nums);
    }
    return permuted;
}

// Function to find the permutation of 1 to N numbers
// having A minimas and B maximas
function findPermutation(n, a, b) {

    // Generate the array containing one permutation
    let nums = [];
    for (let i = 0; i < n; i++) {
        nums.push(i + 1);
    }

    // Generate all the permutations
    let allpermutations = permute(nums);

    let total = allpermutations.length;
    let ansindex = -1;

    for (let i = 0; i < total; i++) {
        // Count local minima and local maximas for each
        // permutation
        let minc = 0, maxc = 0;
        for (let j = 1; j < n - 1; j++) {
            if (allpermutations[i][j] > allpermutations[i][j - 1] && allpermutations[i][j] > allpermutations[i][j + 1]) {
                maxc++;
            }
            if (allpermutations[i][j] < allpermutations[i][j - 1] && allpermutations[i][j] < allpermutations[i][j + 1]) {
                minc++;
            }
        }
        if (minc == a && maxc == b) {

            // Store the index of a perfect permutation
            ansindex = i;
            break;
        }
    }

    // Print -1 if no such permutation exists
    if (ansindex === -1) {
        console.log(-1);
    }
    else {
        // Print the perfect permutation if exists
        console.log(allpermutations[ansindex]);
    }
}


let N = 6, A = 2, B = 2;
findPermutation(N, A, B);

// This code is contributed by akashish__

Python3

# Python program for the above approach

from typing import List, Tuple

# Function to generate next permutation
def nextPermutation(nums: List[int]) -> None:
    n = len(nums)
    k, l = n - 2, n - 1
    while k >= 0:
        if nums[k] < nums[k + 1]:
            break
        k -= 1
    if k < 0:
        nums.reverse()
    else:
        while l > k:
            if nums[l] > nums[k]:
                break
            l -= 1
        nums[k], nums[l] = nums[l], nums[k]
        nums[k+1:] = reversed(nums[k+1:])

# Factorial function
def factorial(n: int) -> int:
    return 1 if n == 1 or n == 0 else factorial(n - 1) * n

# Function to returns all the permutations of a given array or vector
def permute(nums: List[int]) -> List[List[int]]:
    permuted = []
    factn = factorial(len(nums))
    for i in range(factn):
        permuted.append(nums.copy())
        nextPermutation(nums)
    return permuted

# Function to find the permutation of 1 to N numbers having A minimas and B maximas
def findPermutation(n: int, a: int, b: int) -> Tuple[int, List[int]]:
    # Generate the array containing one permutation
    nums = [i + 1 for i in range(n)]
    # Generate all the permutations
    allpermutations = permute(nums)
    ansindex = -1
    for i in range(len(allpermutations)):
        # Count local minima and local maximas for each permutation
        minc, maxc = 0, 0
        for j in range(1, n - 1):
            if allpermutations[i][j] > allpermutations[i][j-1] and allpermutations[i][j] > allpermutations[i][j+1]:
                maxc += 1
            if allpermutations[i][j] < allpermutations[i][j-1] and allpermutations[i][j] < allpermutations[i][j+1]:
                minc += 1
        if minc == a and maxc == b:
            # Store the index of a perfect permutation
            ansindex = i
            break
    # Print -1 if no such permutation exists
    if ansindex == -1:
        return -1;
    else:
        # Print the perfect permutation if exists
        return allpermutations[ansindex]

N, A, B = 6, 2, 2
print(findPermutation(N, A, B))

Output

1 3 2 5 4 6 

Time Complexity: O(N!)
Auxiliary Space: O(N!)

Efficient Approach (Greedy Method):

The above brute force method can be optimized using the Greedy Algorithm. Greedy is an algorithmic paradigm that builds up a solution piece by piece, always choosing the next piece that offers the most obvious and immediate benefit. So, break the problem into different usable pieces according to the values of N, A, B. Follow the below steps to solve this problem:

• As the first and the last element of an entire permutation cannot be maxima or minima so the total number of maxima’s and minima’s must be less than or equal to N-2. So if (A + B > N -2), print -1.
• Also, there cannot be two consecutive minima’s or maxima’s. There must be a minima in between two consecutive maxima’s and there must be a maxima in between two consecutive minima’s. So the absolute difference between the total number of minima’s and maxima’s must be less than or equal to 1. So if the absolute difference between A and B exceeds 1 then print -1. We can easily visualize that from the image below.

• After the two corner cases are done, create two variables, minValue to store the minimum value of the permutation, and maxValue to store the maxValue. Now divide the problem into three different cases that are (A > B), (A < B) and (A=B). Now solve each case individually:
  • If (A > B): Initialize the minValue as 1 and fill the array with the minValue starting from index 2 for A times. After each insertion increase the minValue by 1 as the values should be distinct. While filling, make sure to leave one index empty after each insertion as it is not possible for two minima’s to reside consecutively. After creating A minima’s fill the rest of the array in increasing order so that no new minima is created.
  • If (B > A): Initialize the maxValue as N and fill the array with the maxValue starting from index 2 for B times. After each insertion decrease the maxValue by 1 as the values should be distinct. While filling, make sure to leave one index empty after each insertion as it is not possible for two maxima’s to reside consecutively. After creating B maxima’s fill the rest of the array in decreasing order so that no new maxima is created.
  • If (A = B): then initialize two values minValue as 1 and maxValue as N. Fill the first element of the array with minValue and increase the minValue by 1. Then fill the even indices with maxValue and decrease maxValue by 1 and fill the odd indices with minValue and increase the minValue by 1 for A times. Now, fill the rest of the positions in increasing order.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the permutation of 1 to N numbers
// having A minimas and B maximas
void findPermutation(int N, int A, int B)
{

    // Create the result array
    vector<int> arr(N);
    for (int i = 0; i < N; i++) {
        arr[i] = -1;
    }

    // If the absolute difference between A and B is
    // greater 1 or A+B is greater than N-2, then return -1
    if (abs(A - B) > 1 || A + B > N - 2) {
        cout << -1;
    }
    else {
        if (A > B) {

            // Initialize maxValue with N
            int maxValue = N;

            // Create a maxima's
            for (int i = 1; i < N - 1 && A > 0; i += 2) {
                arr[i] = maxValue;
                maxValue--;
                A--;
            }

            // Fill other elements in decreasing order
            for (int i = 0; i < N; i++) {
                if (arr[i] == -1) {
                    arr[i] = maxValue;
                    maxValue--;
                }
            }
        }
        else if (A < B) {
            // Initialize minValue with 1
            int minValue = 1;

            // Create B minima's
            for (int i = 1; i < N - 1 && B > 0; i += 2) {
                arr[i] = minValue;
                minValue++;
                B--;
            }

            // Fill other elements in increasing order
            for (int i = 0; i < N; i++) {
                if (arr[i] == -1) {
                    arr[i] = minValue;
                    minValue++;
                }
            }
        }
        else if (A == B) {

            // Initialize maxValue with n and minValue with
            // 1
            int minValue = 1, maxValue = N;
            arr[0] = minValue;
            minValue++;

            // Initialize fill equal number of minima and
            // maximas
            for (int i = 1; i < N - 1 && A > 0; i += 2) {
                arr[i] = maxValue;
                arr[i + 1] = minValue;
                A--;
                maxValue--;
                minValue++;
            }

            // Fill the rest in increasing order
            for (int i = 0; i < N; i++) {
                if (arr[i] == -1) {
                    arr[i] = minValue;
                    minValue++;
                }
            }
        }

        // Print the output
        for (int i = 0; i < N; i++) {
            cout << arr[i] << " ";
        }
    }
    cout << endl;
}

// Driver Code
int main()
{
    int N = 6, A = 2, B = 1;
    findPermutation(N, A, B);

    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function to find the permutation of 1 to N numbers
// having A minimas and B maximas
static void findPermutation(int N, int A, int B)
{

    // Create the result array
    int []arr = new int[N];
    for (int i = 0; i < N; i++) {
        arr[i] = -1;
    }

    // If the absolute difference between A and B is
    // greater 1 or A+B is greater than N-2, then return -1
    if (Math.abs(A - B) > 1 || A + B > N - 2) {
        System.out.print(-1);
    }
    else {
        if (A > B) {

            // Initialize maxValue with N
            int maxValue = N;

            // Create a maxima's
            for (int i = 1; i < N - 1 && A > 0; i += 2) {
                arr[i] = maxValue;
                maxValue--;
                A--;
            }

            // Fill other elements in decreasing order
            for (int i = 0; i < N; i++) {
                if (arr[i] == -1) {
                    arr[i] = maxValue;
                    maxValue--;
                }
            }
        }
        else if (A < B) {
            // Initialize minValue with 1
            int minValue = 1;

            // Create B minima's
            for (int i = 1; i < N - 1 && B > 0; i += 2) {
                arr[i] = minValue;
                minValue++;
                B--;
            }

            // Fill other elements in increasing order
            for (int i = 0; i < N; i++) {
                if (arr[i] == -1) {
                    arr[i] = minValue;
                    minValue++;
                }
            }
        }
        else if (A == B) {

            // Initialize maxValue with n and minValue with
            // 1
            int minValue = 1, maxValue = N;
            arr[0] = minValue;
            minValue++;

            // Initialize fill equal number of minima and
            // maximas
            for (int i = 1; i < N - 1 && A > 0; i += 2) {
                arr[i] = maxValue;
                arr[i + 1] = minValue;
                A--;
                maxValue--;
                minValue++;
            }

            // Fill the rest in increasing order
            for (int i = 0; i < N; i++) {
                if (arr[i] == -1) {
                    arr[i] = minValue;
                    minValue++;
                }
            }
        }

        // Print the output
        for (int i = 0; i < N; i++) {
            System.out.print(arr[i]+ " ");
        }
    }
    System.out.println();
}

// Driver Code
public static void main(String[] args)
{
    int N = 6, A = 2, B = 1;
    findPermutation(N, A, B);

}
}

// This code is contributed by 29AjayKumar

Python3

# Python 3 program for the above approach

# Function to find the permutation of 1 to N numbers
# having A minimas and B maximas
def findPermutation(N, A, B):

    # Create the result array
    arr = [0]*(N)
    for i in range(N):
        arr[i] = -1

    # If the absolute difference between A and B is
    # greater 1 or A+B is greater than N-2, then return -1
    if (abs(A - B) > 1 or A + B > N - 2):
        print(-1)

    else:
        if (A > B):

            # Initialize maxValue with N
            maxValue = N

            # Create a maxima's
            i = 1
            while i < N - 1 and A > 0:
                arr[i] = maxValue
                maxValue -= 1
                A -= 1
                i += 2

            # Fill other elements in decreasing order
            for i in range(N):
                if (arr[i] == -1):
                    arr[i] = maxValue
                    maxValue -= 1
        elif (A < B):
            # Initialize minValue with 1
            minValue = 1

            # Create B minima's
            i = 1
            while i < N - 1 and B > 0:
                arr[i] = minValue
                minValue += 1
                B -= 1
                i += 2

            # Fill other elements in increasing order
            for i in range(N):
                if (arr[i] == -1):
                    arr[i] = minValue
                    minValue += 1
        elif (A == B):

            # Initialize maxValue with n and minValue with
            # 1
            minValue = 1
            maxValue = N
            arr[0] = minValue
            minValue += 1

            # Initialize fill equal number of minima and
            # maximas
            i = 1
            while i < N - 1 and A > 0:
                arr[i] = maxValue
                arr[i + 1] = minValue
                A -= 1
                maxValue -= 1
                minValue += 1
                i += 2

            # Fill the rest in increasing order
            for i in range(N):
                if (arr[i] == -1):
                    arr[i] = minValue
                    minValue += 1

        # Print the output
        for i in range(N):
            print(arr[i], end=" ")

    print()

# Driver Code
if __name__ == "__main__":

    N = 6
    A = 2
    B = 1
    findPermutation(N, A, B)

    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
class GFG{

  // Function to find the permutation of 1 to N numbers
  // having A minimas and B maximas
  static void findPermutation(int N, int A, int B)
  {

    // Create the result array
    int []arr = new int[N];
    for (int i = 0; i < N; i++) {
      arr[i] = -1;
    }

    // If the absolute difference between A and B is
    // greater 1 or A+B is greater than N-2, then return -1
    if (Math.Abs(A - B) > 1 || A + B > N - 2) {
      Console.Write(-1);
    }
    else {
      if (A > B) {

        // Initialize maxValue with N
        int maxValue = N;

        // Create a maxima's
        for (int i = 1; i < N - 1 && A > 0; i += 2) {
          arr[i] = maxValue;
          maxValue--;
          A--;
        }

        // Fill other elements in decreasing order
        for (int i = 0; i < N; i++) {
          if (arr[i] == -1) {
            arr[i] = maxValue;
            maxValue--;
          }
        }
      }
      else if (A < B) {
        // Initialize minValue with 1
        int minValue = 1;

        // Create B minima's
        for (int i = 1; i < N - 1 && B > 0; i += 2) {
          arr[i] = minValue;
          minValue++;
          B--;
        }

        // Fill other elements in increasing order
        for (int i = 0; i < N; i++) {
          if (arr[i] == -1) {
            arr[i] = minValue;
            minValue++;
          }
        }
      }
      else if (A == B) {

        // Initialize maxValue with n and minValue with
        // 1
        int minValue = 1, maxValue = N;
        arr[0] = minValue;
        minValue++;

        // Initialize fill equal number of minima and
        // maximas
        for (int i = 1; i < N - 1 && A > 0; i += 2) {
          arr[i] = maxValue;
          arr[i + 1] = minValue;
          A--;
          maxValue--;
          minValue++;
        }

        // Fill the rest in increasing order
        for (int i = 0; i < N; i++) {
          if (arr[i] == -1) {
            arr[i] = minValue;
            minValue++;
          }
        }
      }

      // Print the output
      for (int i = 0; i < N; i++) {
        Console.Write(arr[i]+ " ");
      }
    }
    Console.Write("\n");
  }

  // Driver Code
  public static void Main()
  {
    int N = 6, A = 2, B = 1;
    findPermutation(N, A, B);

  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
   // JavaScript code for the above approach

   // Function to find the permutation of 1 to N numbers
   // having A minimas and B maximas
   function findPermutation(N, A, B)
   {

     // Create the result array
     let arr = new Array(N);
     for (let i = 0; i < N; i++) {
       arr[i] = -1;
     }

     // If the absolute difference between A and B is
     // greater 1 or A+B is greater than N-2, then return -1
     if (Math.abs(A - B) > 1 || A + B > N - 2) {
       document.write(-1);
     }
     else {
       if (A > B) {

         // Initialize maxValue with N
         let maxValue = N;

         // Create a maxima's
         for (let i = 1; i < N - 1 && A > 0; i += 2) {
           arr[i] = maxValue;
           maxValue--;
           A--;
         }

         // Fill other elements in decreasing order
         for (let i = 0; i < N; i++) {
           if (arr[i] == -1) {
             arr[i] = maxValue;
             maxValue--;
           }
         }
       }
       else if (A < B)
       {

         // Initialize minValue with 1
         let minValue = 1;

         // Create B minima's
         for (let i = 1; i < N - 1 && B > 0; i += 2) {
           arr[i] = minValue;
           minValue++;
           B--;
         }

         // Fill other elements in increasing order
         for (let i = 0; i < N; i++) {
           if (arr[i] == -1) {
             arr[i] = minValue;
             minValue++;
           }
         }
       }
       else if (A == B) {

         // Initialize maxValue with n and minValue with
         // 1
         let minValue = 1, maxValue = N;
         arr[0] = minValue;
         minValue++;

         // Initialize fill equal number of minima and
         // maximas
         for (let i = 1; i < N - 1 && A > 0; i += 2) {
           arr[i] = maxValue;
           arr[i + 1] = minValue;
           A--;
           maxValue--;
           minValue++;
         }

         // Fill the rest in increasing order
         for (let i = 0; i < N; i++) {
           if (arr[i] == -1) {
             arr[i] = minValue;
             minValue++;
           }
         }
       }

       // Print the output
       for (let i = 0; i < N; i++) {
         document.write(arr[i] + " ");
       }
     }
     document.write('<br>')
   }

   // Driver Code
   let N = 6, A = 2, B = 1;
   findPermutation(N, A, B);

 // This code is contributed by Potta Lokesh
 </script>

Output

4 6 3 5 2 1 

Time Complexity: O(N)
Auxiliary Space: O(N)