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Make Binary Search Tree

  • Difficulty Level : Hard
  • Last Updated : 30 Jun, 2021

Given an array arr[] of size N. The task is to find whether it is possible to make Binary Search Tree with the given array of elements such that greatest common divisor of any two vertices connected by a common edge is > 1. If possible then print Yes else print No.
Examples: 
 

Input: arr[] = {3, 6, 9, 18, 36, 108} 
Output: Yes 
 

This is one of the possible Binary Search Tree with given array.
Input: arr[] = {2, 17} 
Output: No 
 

 



Approach: Let DP(l, r, root) be a DP determining whether it’s possible to assemble a tree rooted at root from the sub-segment [l..r]. 
It’s easy to see that calculating it requires extracting such rootleft from [l..root – 1] and rootright from [root + 1..right] such that: 
 

  • gcd(aroot, arootleft) > 1
  • gcd(aroot, arootright) > 1
  • DP(l, root-1, rootleft) = 1
  • DP(root+1, r, rootright) = 1

This can be done in O(r – l) provided we are given all DP(x, y, z) values for all sub-segments of [l..r]. Considering a total of O(n3) DP states, the final complexity is O(n4) and that’s too much.
Let’s turn our DP into DPnew(l, r, state) where the state can be either 0 or 1. It immediately turns out that DP(l, r, root) is inherited from DPnew(l, root-1, 1) and DPnew(root+1, r, 0). Now we have O(n2) states, but at the same time, all transitions are performed in linear time. Thus final complexity is O(n3) which is sufficient to pass.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Maximum number of vertices
#define N 705
 
// To store is it possible at
// particular pace or not
int dp[N][N][2];
 
// Return 1 if from l to r, it is possible with
// the given state
int possibleWithState(int l, int r, int state, int a[])
{
    // Base condition
    if (l > r)
        return 1;
 
    // If it is already calculated
    if (dp[l][r][state] != -1)
        return dp[l][r][state];
 
    // Choose the root
    int root;
    if (state == 1)
        root = a[r + 1];
    else
        root = a[l - 1];
 
    // Traverse in range l to r
    for (int i = l; i <= r; i++) {
 
        // If gcd is greater than one
        // check for both sides
        if (__gcd(a[i], root) > 1) {
            int x = possibleWithState(l, i - 1, 1, a);
            if (x != 1)
                continue;
            int y = possibleWithState(i + 1, r, 0, a);
            if (x == 1 && y == 1)
                return dp[l][r][state] = 1;
        }
    }
 
    // If not possible
    return dp[l][r][state] = 0;
}
 
// Function that return true if it is possible
// to make Binary Search Tree
bool isPossible(int a[], int n)
{
    memset(dp, -1, sizeof dp);
 
    // Sort the given array
    sort(a, a + n);
 
    // Check it is possible rooted at i
    for (int i = 0; i < n; i++)
 
        // Check at both sides
        if (possibleWithState(0, i - 1, 1, a)
            && possibleWithState(i + 1, n - 1, 0, a)) {
            return true;
        }
 
    return false;
}
 
// Driver code
int main()
{
    int a[] = { 3, 6, 9, 18, 36, 108 };
    int n = sizeof(a) / sizeof(a[0]);
 
    if (isPossible(a, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
class GFG
{
     
static int __gcd(int a, int b)
{
     
    // Everything divides 0
    if (a == 0)
        return b;
    if (b == 0)
        return a;
     
    // base case
    if (a == b)
        return a;
     
    // a is greater
    if (a > b)
        return __gcd(a - b, b);
    return __gcd(a, b-a);
}
 
// Maximum number of vertices
static final int N = 705;
 
// To store is it possible at
// particular pace or not
static int dp[][][] = new int[N][N][2];
 
// Return 1 if from l to r, it is
// possible with the given state
static int possibleWithState(int l, int r,
                        int state, int a[])
{
    // Base condition
    if (l > r)
        return 1;
 
    // If it is already calculated
    if (dp[l][r][state] != -1)
        return dp[l][r][state];
 
    // Choose the root
    int root;
    if (state == 1)
        root = a[r + 1];
    else
        root = a[l - 1];
 
    // Traverse in range l to r
    for (int i = l; i <= r; i++)
    {
 
        // If gcd is greater than one
        // check for both sides
        if (__gcd(a[i], root) > 1)
        {
            int x = possibleWithState(l, i - 1, 1, a);
            if (x != 1)
                continue;
                 
            int y = possibleWithState(i + 1, r, 0, a);
             
            if (x == 1 && y == 1)
                return dp[l][r][state] = 1;
        }
    }
 
    // If not possible
    return dp[l][r][state] = 0;
}
 
// Function that return true if it is possible
// to make Binary Search Tree
static boolean isPossible(int a[], int n)
{
    for(int i = 0; i < dp.length; i++)
        for(int j = 0; j < dp[i].length; j++)
            for(int k = 0; k < dp[i][j].length; k++)
                dp[i][j][k]=-1;
 
    // Sort the given array
    Arrays.sort(a);
 
    // Check it is possible rooted at i
    for (int i = 0; i < n; i++)
 
        // Check at both sides
        if (possibleWithState(0, i - 1, 1, a) != 0 &&
            possibleWithState(i + 1, n - 1, 0, a) != 0)
        {
            return true;
        }
 
    return false;
}
 
// Driver code
public static void main(String args[])
{
    int a[] = { 3, 6, 9, 18, 36, 108 };
    int n = a.length;
 
    if (isPossible(a, n))
        System.out.println("Yes");
    else
        System.out.println("No");
 
}
}
 
// This code is contributed by
// Arnab Kundu

Python3




# Python3 implementation of the approach
import math
 
# Maximum number of vertices
N = 705
 
# To store is it possible at
# particular pace or not
dp = [[[-1 for z in range(2)]
           for x in range(N)]
           for y in range(N)]
 
# Return 1 if from l to r, it is
# possible with the given state
def possibleWithState(l, r, state, a):
 
    # Base condition
    if (l > r):
        return 1
 
    # If it is already calculated
    if (dp[l][r][state] != -1):
        return dp[l][r][state]
 
    # Choose the root
    root = 0
    if (state == 1) :
        root = a[r + 1]
    else:
        root = a[l - 1]
 
    # Traverse in range l to r
    for i in range(l, r + 1):
 
        # If gcd is greater than one
        # check for both sides
        if (math.gcd(a[i], root) > 1):
            x = possibleWithState(l, i - 1, 1, a)
            if (x != 1):
                continue
            y = possibleWithState(i + 1, r, 0, a)
            if (x == 1 and y == 1) :
                return 1
 
    # If not possible
    return 0
 
# Function that return true if it is
# possible to make Binary Search Tree
def isPossible(a, n):
     
    # Sort the given array
    a.sort()
 
    # Check it is possible rooted at i
    for i in range(n):
         
        # Check at both sides
        if (possibleWithState(0, i - 1, 1, a) and
            possibleWithState(i + 1, n - 1, 0, a)):
            return True
             
    return False
 
# Driver Code
if __name__ == '__main__':
    a = [3, 6, 9, 18, 36, 108]
    n = len(a)
    if (isPossible(a, n)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
static int __gcd(int a, int b)
{
     
    // Everything divides 0
    if (a == 0)
        return b;
    if (b == 0)
        return a;
     
    // base case
    if (a == b)
        return a;
     
    // a is greater
    if (a > b)
        return __gcd(a - b, b);
    return __gcd(a, b-a);
}
 
// Maximum number of vertices
static int N = 705;
 
// To store is it possible at
// particular pace or not
static int [,,]dp = new int[N, N, 2];
 
// Return 1 if from l to r, it is
// possible with the given state
static int possibleWithState(int l, int r,
                        int state, int []a)
{
    // Base condition
    if (l > r)
        return 1;
 
    // If it is already calculated
    if (dp[l, r, state] != -1)
        return dp[l, r, state];
 
    // Choose the root
    int root;
    if (state == 1)
        root = a[r + 1];
    else
        root = a[l - 1];
 
    // Traverse in range l to r
    for (int i = l; i <= r; i++)
    {
 
        // If gcd is greater than one
        // check for both sides
        if (__gcd(a[i], root) > 1)
        {
            int x = possibleWithState(l, i - 1, 1, a);
            if (x != 1)
                continue;
                 
            int y = possibleWithState(i + 1, r, 0, a);
             
            if (x == 1 && y == 1)
                return dp[l,r,state] = 1;
        }
    }
 
    // If not possible
    return dp[l,r,state] = 0;
}
 
// Function that return true
// if it is possible to make
// Binary Search Tree
static bool isPossible(int []a, int n)
{
    for(int i = 0; i < dp.GetLength(0); i++)
        for(int j = 0; j < dp.GetLength(1); j++)
            for(int k = 0; k < dp.GetLength(2); k++)
                dp[i, j, k]=-1;
 
    // Sort the given array
    Array.Sort(a);
 
    // Check it is possible rooted at i
    for (int i = 0; i < n; i++)
 
        // Check at both sides
        if (possibleWithState(0, i - 1, 1, a) != 0 &&
            possibleWithState(i + 1, n - 1, 0, a) != 0)
        {
            return true;
        }
 
    return false;
}
 
// Driver code
public static void Main(String []args)
{
    int []a = { 3, 6, 9, 18, 36, 108 };
    int n = a.Length;
 
    if (isPossible(a, n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
 
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
 
// Javascript implementation of the approach
     
function __gcd(a, b)
{
     
    // Everything divides 0
    if (a == 0)
        return b;
    if (b == 0)
        return a;
     
    // base case
    if (a == b)
        return a;
     
    // a is greater
    if (a > b)
        return __gcd(a - b, b);
    return __gcd(a, b-a);
}
 
// Maximum number of vertices
var N = 705;
 
// To store is it possible at
// particular pace or not
var dp = Array.from(Array(N), ()=>Array(N));
for(var i = 0; i<N; i++)
{
    for(var j = 0; j<N; j++)
    {
        dp[i][j] = new Array(2).fill(-1);
    }
}
 
// Return 1 if from l to r, it is
// possible with the given state
function possibleWithState(l, r, state,a)
{
    // Base condition
    if (l > r)
        return 1;
 
    // If it is already calculated
    if (dp[l][r][state] != -1)
        return dp[l][r][state];
 
    // Choose the root
    var root;
    if (state == 1)
        root = a[r + 1];
    else
        root = a[l - 1];
 
    // Traverse in range l to r
    for (var i = l; i <= r; i++)
    {
 
        // If gcd is greater than one
        // check for both sides
        if (__gcd(a[i], root) > 1)
        {
            var x = possibleWithState(l, i - 1, 1, a);
            if (x != 1)
                continue;
                 
            var y = possibleWithState(i + 1, r, 0, a);
             
            if (x == 1 && y == 1)
                return dp[l][r][state] = 1;
        }
    }
 
    // If not possible
    return dp[l][r][state] = 0;
}
 
// Function that return true
// if it is possible to make
// Binary Search Tree
function isPossible(a, n)
{
 
    // Sort the given array
    a.sort();
 
    // Check it is possible rooted at i
    for (var i = 0; i < n; i++)
 
        // Check at both sides
        if (possibleWithState(0, i - 1, 1, a) != 0 &&
            possibleWithState(i + 1, n - 1, 0, a) != 0)
        {
            return true;
        }
 
    return false;
}
 
// Driver code
var a = [3, 6, 9, 18, 36, 108];
var n = a.length;
if (isPossible(a, n))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by itsok.
</script>
Output: 
Yes

 

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