# Make Binary Search Tree

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `using` `namespace` `std; ` ` `  `// Maxium number of vertices ` `#define N 705 ` ` `  `// To store is it possible at ` `// particular pace or not ` `int` `dp[N][N]; ` ` `  `// 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); ` ` `  `    ``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);  ` `}  ` ` `  `// Maxium 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 ` ` `  `# Maxium 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 `

Output:

```Yes
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.