Minimize Cost to reduce the Array to a single element by given operations

Given an array a[] consisting of N integers and an integer K, the task is to find the minimum cost to reduce the given array to a single element by choosing any pair of consecutive array elements and replace them by (a[i] + a[i+1]) for a cost K * (a[i] + a[i+1]).

Examples:

Input: a[] = {1, 2, 3}, K = 2 
Output: 18 
Explanation:
Replacing {1, 2} by 3 modifies the array to {3, 3}. Cost 2 * 3 = 6 
Replacing {3, 3} by 6 modifies the array to {6}. Cost 2 * 6 = 12 
Therefore, the total cost is 18
Input: a[] = {4, 5, 6, 7}, K = 3 
Output: 132

Naive Approach: 
The simplest solution is to split the array into two halves, for every index and compute the cost of the two halves recursively and finally add their respective costs.
Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
#define inf 10000009

// Function to combine the sum of the two halves
int Combine(int a[], int i, int j)
{
    int sum = 0;

    // Calculate the sum from i to j
    for (int l = i; l <= j; l++)
        sum += a[l];

    return sum;
}

// Function to minimize the cost to
// reduce the array to a single element
int minCost(int a[], int i, int j, int k)
{
    if (i >= j)
    {

        // Base case
        // If n = 1 or n = 0
        return 0;
    }

    // Initialize cost to maximum value
    int best_cost = inf;

    // Iterate through all possible indices
    // and find the best index
    // to combine the subproblems
    for (int pos = i; pos < j; pos++)
    {

        // Compute left subproblem
        int left = minCost(a, i, pos, k);

        // Compute right subproblem
        int right = minCost(a, pos + 1, j, k);

        // Calculate the best cost
        best_cost = min(best_cost, left + right +
                                   k * Combine(a, i, j));
    }

    // Return the answer
    return best_cost;
}

// Driver code
int main()
{
    int n = 4;
    int a[] = { 4, 5, 6, 7 };
    int k = 3;

    cout << minCost(a, 0, n - 1, k) << endl;
    return 0;
}

// This code is contributed by PrinciRaj1992

Java

// Java Program to implement
// the above approach
import java.io.*;

class GFG {

    static int inf = 10000009;

    // Function to minimize the cost to
    // reduce the array to a single element
    public static int minCost(int a[], int i,
                              int j, int k)
    {
        if (i >= j) {

            // Base case
            // If n = 1 or n = 0
            return 0;
        }

        // Initialize cost to maximum value
        int best_cost = inf;

        // Iterate through all possible indices
        // and find the best index
        // to combine the subproblems
        for (int pos = i; pos < j; pos++) {

            // Compute left subproblem
            int left = minCost(a, i, pos, k);

            // Compute right subproblem
            int right = minCost(a, pos + 1, j, k);

            // Calculate the best  cost
            best_cost = Math.min(
                best_cost,
                left + right + k * Combine(a, i, j));
        }

        // Return the answer
        return best_cost;
    }

    // Function to combine the sum of the two halves
    public static int Combine(int a[], int i, int j)
    {
        int sum = 0;

        // Calculate the sum from i to j
        for (int l = i; l <= j; l++)
            sum += a[l];

        return sum;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 4;
        int a[] = { 4, 5, 6, 7 };
        int k = 3;

        System.out.println(minCost(a, 0, n - 1, k));
    }
}

Python3

# Python3 Program to implement
# the above approach
inf = 10000009;

# Function to minimize the cost to
# reduce the array to a single element
def minCost(a, i, j, k):
    if (i >= j):

        # Base case
        # If n = 1 or n = 0
        return 0;

    # Initialize cost to maximum value
    best_cost = inf;

    # Iterate through all possible indices
    # and find the best index
    # to combine the subproblems
    for pos in range(i, j):

        # Compute left subproblem
        left = minCost(a, i, pos, k);

        # Compute right subproblem
        right = minCost(a, pos + 1, j, k);

        # Calculate the best cost
        best_cost = min(best_cost,
                        left + right +
                        k * Combine(a, i, j));

    # Return the answer
    return best_cost;

# Function to combine
# the sum of the two halves
def Combine(a, i, j):
    sum = 0;

    # Calculate the sum from i to j
    for l in range(i, j + 1):
        sum += a[l];

    return sum;

# Driver code
if __name__ == '__main__':
    n = 4;
    a = [4, 5, 6, 7];
    k = 3;

    print(minCost(a, 0, n - 1, k));

# This code is contributed by Amit Katiyar

C#

// C# Program to implement
// the above approach
using System;
class GFG{

  static int inf = 10000009;

  // Function to minimize the cost to
  // reduce the array to a single element
  public static int minCost(int []a, int i,
                            int j, int k)
  {
    if (i >= j)
    {

      // Base case
      // If n = 1 or n = 0
      return 0;
    }

    // Initialize cost to maximum value
    int best_cost = inf;

    // Iterate through all possible indices
    // and find the best index
    // to combine the subproblems
    for (int pos = i; pos < j; pos++)
    {

      // Compute left subproblem
      int left = minCost(a, i, pos, k);

      // Compute right subproblem
      int right = minCost(a, pos + 1, j, k);

      // Calculate the best  cost
      best_cost = Math.Min(best_cost,
                           left + right +
                           k * Combine(a, i, j));
    }

    // Return the answer
    return best_cost;
  }

  // Function to combine the sum of the two halves
  public static int Combine(int []a, int i, int j)
  {
    int sum = 0;

    // Calculate the sum from i to j
    for (int l = i; l <= j; l++)
      sum += a[l];

    return sum;
  }

  // Driver code
  public static void Main(String[] args)
  {
    int n = 4;
    int []a = { 4, 5, 6, 7 };
    int k = 3;

    Console.WriteLine(minCost(a, 0, n - 1, k));
  }
}

// This code is contributed by Rohit_ranjan

Javascript

<script>

// JavaScript Program to implement
// the above approach
var inf = 10000009;

// Function to combine the sum of the two halves
function Combine(a, i, j)
{
    var sum = 0;

    // Calculate the sum from i to j
    for (var l = i; l <= j; l++)
        sum += a[l];

    return sum;
}

// Function to minimize the cost to
// reduce the array to a single element
function minCost(a, i, j, k)
{
    if (i >= j)
    {

        // Base case
        // If n = 1 or n = 0
        return 0;
    }

    // Initialize cost to maximum value
    var best_cost = inf;

    // Iterate through all possible indices
    // and find the best index
    // to combine the subproblems
    for (var pos = i; pos < j; pos++)
    {

        // Compute left subproblem
        var left = minCost(a, i, pos, k);

        // Compute right subproblem
        var right = minCost(a, pos + 1, j, k);

        // Calculate the best cost
        best_cost = Math.min(best_cost, left + right +
                                   k * Combine(a, i, j));
    }

    // Return the answer
    return best_cost;
}

// Driver code

var n = 4;
var a = [4, 5, 6, 7];
var k = 3;
document.write( minCost(a, 0, n - 1, k));


</script>

Output:

132

Time Complexity: O(2^N) 
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use the concept of Dynamic Programming. Follow the steps below to solve the problem:

• Initialize a matrix dp[][] and such that dp[i][j] stores the sum from index i to j.
• Calculate sum(i, j) using Prefix Sum technique.
• Compute the sum of the two subproblems and update the cost with the minimum value.
• Store in dp[][] and return.

Below is the implementation of the above approach:

C++

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

int inf = 10000000;

// Function to generate the cost using
// Prefix Sum Array technique
vector<int> preprocess(vector<int> a, int n)
{
    vector<int> p(n);
    p[0] = a[0];

    for(int i = 1; i < n; i++)
    {
        p[i] = p[i - 1] + a[i];
    }
    return p;
}

// Function to combine the sum of the
// two subproblems
int Combine(vector<int> p, int i, int j)
{
    if (i == 0)
        return p[j];
    else
        return p[j] - p[i - 1];
}

// Function to minimize the cost to
// add the array elements to a single element
int minCost(vector<int> a, int i, int j, int k,
            vector<int> prefix, vector<vector<int>> dp)
{
    if (i >= j)
        return 0;

    // Check if the value is
    // already stored in the array
    if (dp[i][j] != -1)
        return dp[i][j];

    int best_cost = inf;
    for(int pos = i; pos < j; pos++)
    {

        // Compute left subproblem
        int left = minCost(a, i, pos,
                           k, prefix, dp);

        // Compute left subproblem
        int right = minCost(a, pos + 1, j,
                            k, prefix, dp);

        // Calculate minimum cost
        best_cost = min(best_cost, left + right +
                       (k * Combine(prefix, i, j)));
    }

    // Store the answer to
    // avoid recalculation
    return dp[i][j] = best_cost;
}

// Driver code
int main()
{
    int n = 4;

    vector<int> a = { 4, 5, 6, 7 };

    int k = 3;

    // Initialise dp array
    vector<vector<int>> dp;
    dp.resize(n + 1, vector<int>(n + 1));
    for(int i = 0; i < n + 1; i++)
    {
        for(int j = 0; j < n + 1; j++)
        {
            dp[i][j] = -1;
        }
    }

    // Preprocessing the array
    vector<int> prefix = preprocess(a, n);

    cout << minCost(a, 0, n - 1, k, prefix, dp)
         << endl;

    return 0;
}

// This code is contributed by divyeshrabadiya07

Java

// Java Program for the above approach
import java.util.*;
public class Main {

    static int inf = 10000000;

    // Function to minimize the cost to
    // add the array elements to a single element
    public static int minCost(int a[], int i, int j, int k,
                              int[] prefix, int[][] dp)
    {
        if (i >= j)
            return 0;

        // Check if the value is
        // already stored in the array
        if (dp[i][j] != -1)
            return dp[i][j];

        int best_cost = inf;
        for (int pos = i; pos < j; pos++) {

            // Compute left subproblem
            int left = minCost(a, i, pos, k, prefix, dp);

            // Compute left subproblem
            int right
                = minCost(a, pos + 1, j, k, prefix, dp);

            // Calculate minimum cost
            best_cost = Math.min(
                best_cost,
                left + right + (k * Combine(prefix, i, j)));
        }

        // Store the answer to
        // avoid recalculation
        return dp[i][j] = best_cost;
    }

    // Function to generate the cost using
    // Prefix Sum Array technique
    public static int[] preprocess(int[] a, int n)
    {
        int p[] = new int[n];
        p[0] = a[0];
        for (int i = 1; i < n; i++)
            p[i] = p[i - 1] + a[i];
        return p;
    }

    // Function to combine the sum of the two subproblems
    public static int Combine(int[] p, int i, int j)
    {
        if (i == 0)
            return p[j];
        else
            return p[j] - p[i - 1];
    }

    // Driver Code
    public static void main(String args[])
    {
        int n = 4;

        int a[] = { 4, 5, 6, 7 };

        int k = 3;

        // Initialise dp array
        int dp[][] = new int[n + 1][n + 1];
        for (int i[] : dp)
            Arrays.fill(i, -1);

        // Preprocessing the array
        int prefix[] = preprocess(a, n);

        System.out.println(
            minCost(a, 0, n - 1, k, prefix, dp));
    }
}

Python3

# Python3 program for the above approach
inf = 10000000

# Function to minimize the cost to
# add the array elements to a single element
def minCost(a, i, j, k, prefix, dp):

    if (i >= j):
        return 0

    # Check if the value is
    # already stored in the array
    if (dp[i][j] != -1):
        return dp[i][j]

    best_cost = inf
    for pos in range(i, j):

        # Compute left subproblem
        left = minCost(a, i, pos,
                       k, prefix, dp)

        # Compute left subproblem
        right = minCost(a, pos + 1, j,
                        k, prefix, dp)

        # Calculate minimum cost
        best_cost = min(best_cost,
                        left + right +
                          (k * Combine(prefix, i, j)))

    # Store the answer to
    # avoid recalculation
    dp[i][j] = best_cost

    return dp[i][j]

# Function to generate the cost using
# Prefix Sum Array technique
def preprocess(a, n):

    p = [0] * n
    p[0] = a[0]

    for i in range(1, n):
        p[i] = p[i - 1] + a[i]

    return p

# Function to combine the sum
# of the two subproblems
def Combine(p, i, j):

    if (i == 0):
        return p[j]
    else:
        return p[j] - p[i - 1]

# Driver Code
if __name__ == "__main__":

    n = 4
    a = [ 4, 5, 6, 7 ]
    k = 3

    # Initialise dp array
    dp = [[-1 for x in range (n + 1)]
              for y in range (n + 1)]

    # Preprocessing the array
    prefix = preprocess(a, n)

    print(minCost(a, 0, n - 1, k, prefix, dp))

# This code is contributed by chitranayal

C#

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

  static int inf = 10000000;

  // Function to minimize the cost to
  // add the array elements to a single element
  public static int minCost(int []a, int i, int j, int k,
                            int[] prefix, int[,] dp)
  {
    if (i >= j)
      return 0;

    // Check if the value is
    // already stored in the array
    if (dp[i, j] != -1)
      return dp[i, j];

    int best_cost = inf;
    for (int pos = i; pos < j; pos++)
    {

      // Compute left subproblem
      int left = minCost(a, i, pos, k, prefix, dp);

      // Compute left subproblem
      int right = minCost(a, pos + 1, j,
                          k, prefix, dp);

      // Calculate minimum cost
      best_cost = Math.Min(best_cost, left + right +
                          (k * Combine(prefix, i, j)));
    }

    // Store the answer to
    // avoid recalculation
    return dp[i, j] = best_cost;
  }

  // Function to generate the cost using
  // Prefix Sum Array technique
  public static int[] preprocess(int[] a, int n)
  {
    int []p = new int[n];
    p[0] = a[0];
    for (int i = 1; i < n; i++)
      p[i] = p[i - 1] + a[i];
    return p;
  }

  // Function to combine the sum of the two subproblems
  public static int Combine(int[] p, int i, int j)
  {
    if (i == 0)
      return p[j];
    else
      return p[j] - p[i - 1];
  }

  // Driver Code
  public static void Main(String []args)
  {
    int n = 4;

    int []a = { 4, 5, 6, 7 };

    int k = 3;

    // Initialise dp array
    int [,]dp = new int[n + 1, n + 1];
    for(int i = 0; i < n + 1; i++)
    {
      for (int j = 0; j < n + 1; j++)
      {
        dp[i, j] = -1;
      }
    }
    // Preprocessing the array
    int []prefix = preprocess(a, n);

    Console.WriteLine(minCost(a, 0, n - 1, k,
                              prefix, dp));
  }
}

// This code is contributed by sapnasingh4991

Javascript

<script>

// JavaScript program for the above approach

var inf = 10000000;

// Function to generate the cost using
// Prefix Sum Array technique
function preprocess(a, n)
{
    var p = Array(n);
    p[0] = a[0];

    for(var i = 1; i < n; i++)
    {
        p[i] = p[i - 1] + a[i];
    }
    return p;
}

// Function to combine the sum of the
// two subproblems
function Combine(p, i, j)
{
    if (i == 0)
        return p[j];
    else
        return p[j] - p[i - 1];
}

// Function to minimize the cost to
// add the array elements to a single element
function minCost(a, i, j, k, prefix, dp)
{
    if (i >= j)
        return 0;

    // Check if the value is
    // already stored in the array
    if (dp[i][j] != -1)
        return dp[i][j];

    var best_cost = inf;
    for(var pos = i; pos < j; pos++)
    {

        // Compute left subproblem
        var left = minCost(a, i, pos,
                           k, prefix, dp);

        // Compute left subproblem
        var right = minCost(a, pos + 1, j,
                            k, prefix, dp);

        // Calculate minimum cost
        best_cost = Math.min(best_cost, left + right +
                       (k * Combine(prefix, i, j)));
    }

    // Store the answer to
    // avoid recalculation
    return dp[i][j] = best_cost;
}

// Driver code
var n = 4;
var a = [4, 5, 6, 7];
var k = 3;
// Initialise dp array
var dp = Array.from(Array(n+1), ()=>Array(n+1).fill(-1));

// Preprocessing the array
var prefix = preprocess(a, n);

document.write( minCost(a, 0, n - 1, k, prefix, dp))

</script>

Output:

132

Time Complexity: O(N²) 
Auxiliary Space: O(N²)

Another approach : Using DP Tabulation method ( Iterative approach )

In this approach we use Dp to store computation of subproblems and get the desired output without the help of recursion.

Steps to solve this problem :

• Create a table DP to store the solution of the subproblems .
• Initialize the table with base cases
• Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
• Return the final solution stored in dp[0][n-1].

Implementation :

C++

// C++ code for above approach

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

int inf = 10000000;

// Function to generate the cost using
// Prefix Sum Array technique
vector<int> preprocess(vector<int> a, int n)
{
    vector<int> p(n);
    p[0] = a[0];

    for(int i = 1; i < n; i++)
    {
        p[i] = p[i - 1] + a[i];
    }
    return p;
}

// Function to combine the sum of the
// two subproblems
int Combine(vector<int> p, int i, int j)
{
    if (i == 0)
        return p[j];
    else
        return p[j] - p[i - 1];
}

// Function to minimize the cost to
// add the array elements to a single element
int minCost(vector<int> a, int n, int k)
{
    // Initialise dp array
    vector<vector<int>> dp(n + 1, vector<int>(n + 1));

    // Preprocessing the array
    vector<int> prefix = preprocess(a, n);

    // Base case
    for(int i = 0; i < n; i++)
    {
        dp[i][i] = 0;
    }

    // DP tabulation
    for(int len = 2; len <= n; len++)
    {
        for(int i = 0; i <= n - len; i++)
        {
            int j = i + len - 1;
            dp[i][j] = inf;

            for(int pos = i; pos < j; pos++)
            {
                int left = dp[i][pos];
                int right = dp[pos + 1][j];
                int cost = k * Combine(prefix, i, j);
                dp[i][j] = min(dp[i][j], left + right + cost);
            }
        }
    }

      // return answer
    return dp[0][n - 1];
}

// Driver code
int main()
{
    int n = 4;

    vector<int> a = { 4, 5, 6, 7 };

    int k = 3;

      // function call
    cout << minCost(a, n, k) << endl;

    return 0;
}

// this code is contributed by bhardwajji

Java

import java.util.*;

public class Main {
  static int inf = 10000000;

  // Function to generate the cost using
  // Prefix Sum Array technique
  static List<Integer> preprocess(List<Integer> a, int n)
  {
    List<Integer> p = new ArrayList<>();
    p.add(a.get(0));

    for (int i = 1; i < n; i++) {
      p.add(p.get(i - 1) + a.get(i));
    }
    return p;
  }

  // Function to combine the sum of the
  // two subproblems
  static int Combine(List<Integer> p, int i, int j)
  {
    if (i == 0)
      return p.get(j);
    else
      return p.get(j) - p.get(i - 1);
  }

  // Function to minimize the cost to
  // add the array elements to a single element
  static int minCost(List<Integer> a, int n, int k)
  {
    // Initialise dp array
    int[][] dp = new int[n + 1][n + 1];

    // Preprocessing the array
    List<Integer> prefix = preprocess(a, n);

    // Base case
    for (int i = 0; i < n; i++) {
      dp[i][i] = 0;
    }

    // DP tabulation
    for (int len = 2; len <= n; len++) {
      for (int i = 0; i <= n - len; i++) {
        int j = i + len - 1;
        dp[i][j] = inf;

        for (int pos = i; pos < j; pos++) {
          int left = dp[i][pos];
          int right = dp[pos + 1][j];
          int cost = k * Combine(prefix, i, j);
          dp[i][j] = Math.min(
            dp[i][j], left + right + cost);
        }
      }
    }

    // return answer
    return dp[0][n - 1];
  }

  // Driver code
  public static void main(String[] args)
  {
    int n = 4;
    List<Integer> a
      = new ArrayList<>(Arrays.asList(4, 5, 6, 7));
    int k = 3;

    // function call
    System.out.println(minCost(a, n, k));
  }
}

Python3

# code
INF = 10000000

# Function to generate the cost using
# Prefix Sum Array technique


def preprocess(a, n):
    p = [0] * n
    p[0] = a[0]

    for i in range(1, n):
        p[i] = p[i - 1] + a[i]

    return p

# Function to combine the sum of the
# two subproblems


def Combine(p, i, j):
    if i == 0:
        return p[j]
    else:
        return p[j] - p[i - 1]

# Function to minimize the cost to
# add the array elements to a single element


def minCost(a, n, k):
    # Initialise dp array
    dp = [[0] * (n + 1) for i in range(n + 1)]

    # Preprocessing the array
    prefix = preprocess(a, n)

    # Base case
    for i in range(n):
        dp[i][i] = 0

    # DP tabulation
    for length in range(2, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            dp[i][j] = INF

            for pos in range(i, j):
                left = dp[i][pos]
                right = dp[pos + 1][j]
                cost = k * Combine(prefix, i, j)
                dp[i][j] = min(dp[i][j], left + right + cost)

    # return answer
    return dp[0][n - 1]


if __name__ == '__main__':
    n = 4
    a = [4, 5, 6, 7]
    k = 3

    # function call
    print(minCost(a, n, k))

C#

using System;
using System.Collections.Generic;

class Program
{
    static int inf = 10000000;

    // Function to generate the cost using
    // Prefix Sum Array technique
    static List<int> preprocess(List<int> a, int n)
    {
        List<int> p = new List<int>();
        p.Add(a[0]);

        for (int i = 1; i < n; i++)
        {
            p.Add(p[i - 1] + a[i]);
        }
        return p;
    }

    // Function to combine the sum of the
    // two subproblems
    static int Combine(List<int> p, int i, int j)
    {
        if (i == 0)
            return p[j];
        else
            return p[j] - p[i - 1];
    }

    // Function to minimize the cost to
    // add the array elements to a single element
    static int minCost(List<int> a, int n, int k)
    {
        // Initialise dp array
        int[,] dp = new int[n + 1, n + 1];

        // Preprocessing the array
        List<int> prefix = preprocess(a, n);

        // Base case
        for (int i = 0; i < n; i++)
        {
            dp[i, i] = 0;
        }

        // DP tabulation
        for (int len = 2; len <= n; len++)
        {
            for (int i = 0; i <= n - len; i++)
            {
                int j = i + len - 1;
                dp[i, j] = inf;

                for (int pos = i; pos < j; pos++)
                {
                    int left = dp[i, pos];
                    int right = dp[pos + 1, j];
                    int cost = k * Combine(prefix, i, j);
                    dp[i, j] = Math.Min(dp[i, j], left + right + cost);
                }
            }
        }

        // return answer
        return dp[0, n - 1];
    }

    // Driver code
    static void Main(string[] args)
    {
        int n = 4;
        List<int> a = new List<int>() { 4, 5, 6, 7 };
        int k = 3;

        // function call
        Console.WriteLine(minCost(a, n, k));
    }
}

Javascript

// JavaScript code for above approach
const inf = 10000000;

// Function to generate the cost using
// Prefix Sum Array technique
function preprocess(a, n) {
    let p = [];
    p[0] = a[0];
    for (let i = 1; i < n; i++) {
        p[i] = p[i - 1] + a[i];
    }
    return p;
}

// Function to combine the sum of the
// two subproblems
function Combine(p, i, j) {
    if (i == 0)
        return p[j];
    else
        return p[j] - p[i - 1];
}

// Function to minimize the cost to
// add the array elements to a single element
function minCost(a, n, k) {
    // Initialise dp array
    let dp = new Array(n + 1).fill().map(() => new Array(n + 1).fill(0));
    // Preprocessing the array
    let prefix = preprocess(a, n);

    // Base case
    for (let i = 0; i < n; i++) {
        dp[i][i] = 0;
    }

    // DP tabulation
    for (let len = 2; len <= n; len++) {
        for (let i = 0; i <= n - len; i++) {
            let j = i + len - 1;
            dp[i][j] = inf;

            for (let pos = i; pos < j; pos++) {
                let left = dp[i][pos];
                let right = dp[pos + 1][j];
                let cost = k * Combine(prefix, i, j);
                dp[i][j] = Math.min(dp[i][j], left + right + cost);
            }
        }
    }

    // return answer
    return dp[0][n - 1];
}

// Driver code
let n = 4;
let a = [4, 5, 6, 7];
let k = 3;

// function call
console.log(minCost(a, n, k));

// This code is contributed by user_dtewbxkn77n

Output

132

Time Complexity: O(N^3) 
Auxiliary Space: O(N^2)