Draw Optimal Binary Search Tree
Optimal Binary Search Tree | DP-24
Given a sorted array keys[0.. n-1] of search keys and an array freq[0.. n-1] of frequency counts, where freq[i] is the number of searches to keys[i]. Construct a binary search tree of all keys such that the total cost of all the searches is as small as possible.
Let us first define the cost of a BST. The cost of a BST node is level of that node multiplied by its frequency. Level of root is 1.
Examples:
Input: keys[] = {10, 12}, freq[] = {34, 50} There can be following two possible BSTs 10 12 / 12 10 I II Frequency of searches of 10 and 12 are 34 and 50 respectively. The cost of tree I is 34*1 + 50*2 = 134 The cost of tree II is 50*1 + 34*2 = 118 Input: keys[] = {10, 12, 20}, freq[] = {34, 8, 50} There can be following possible BSTs 10 12 20 10 20 / / / 12 10 20 12 20 10 / / 20 10 12 12 I II III IV V Among all possible BSTs, cost of the fifth BST is minimum. Cost of the fifth BST is 1*50 + 2*34 + 3*8 = 142
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
1) Optimal Substructure:
The optimal cost for freq[i..j] can be recursively calculated using following formula.
We need to calculate optCost(0, n-1) to find the result.
The idea of above formula is simple, we one by one try all nodes as root (r varies from i to j in second term). When we make rth node as root, we recursively calculate optimal cost from i to r-1 and r+1 to j.
We add sum of frequencies from i to j (see first term in the above formula), this is added because every search will go through root and one comparison will be done for every search.
2) Overlapping Subproblems
Following is recursive implementation that simply follows the recursive structure mentioned above.
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C++
// A naive recursive implementation of
// optimal binary search tree problem
#include
using namespace std;
// A utility function to get sum of
// array elements freq[i] to freq[j]
int sum(int freq[], int i, int j);
// A recursive function to calculate
// cost of optimal binary search tree
int optCost(int freq[], int i, int j)
{
// Base cases
if (j < i) // no elements in this subarray return 0; if (j == i) // one element in this subarray return freq[i]; // Get sum of freq[i], freq[i+1], ... freq[j] int fsum = sum(freq, i, j); // Initialize minimum value int min = INT_MAX; // One by one consider all elements // as root and recursively find cost // of the BST, compare the cost with // min and update min if needed for (int r = i; r <= j; ++r) { int cost = optCost(freq, i, r - 1) + optCost(freq, r + 1, j); if (cost < min) min = cost; } // Return minimum value return min + fsum; } // The main function that calculates // minimum cost of a Binary Search Tree. // It mainly uses optCost() to find // the optimal cost. int optimalSearchTree(int keys[], int freq[], int n) { // Here array keys[] is assumed to be // sorted in increasing order. If keys[] // is not sorted, then add code to sort // keys, and rearrange freq[] accordingly. return optCost(freq, 0, n - 1); } // A utility function to get sum of // array elements freq[i] to freq[j] int sum(int freq[], int i, int j) { int s = 0; for (int k = i; k <= j; k++) s += freq[k]; return s; } // Driver Code int main() { int keys[] = {10, 12, 20}; int freq[] = {34, 8, 50}; int n = sizeof(keys) / sizeof(keys[0]); cout << "Cost of Optimal BST is " << optimalSearchTree(keys, freq, n); return 0; } // This is code is contributed // by rathbhupendra [tabby title="C"]
#include <stdio.h>
#include <limits.h>
int sum( int freq[], int i, int j);
int optCost( int freq[], int i, int j)
{
if (j < i)
return 0;
if (j == i)
return freq[i];
int fsum = sum(freq, i, j);
int min = INT_MAX;
for ( int r = i; r <= j; ++r)
{
int cost = optCost(freq, i, r-1) +
optCost(freq, r+1, j);
if (cost < min)
min = cost;
}
return min + fsum;
}
int optimalSearchTree( int keys[], int freq[], int n)
{
return optCost(freq, 0, n-1);
}
int sum( int freq[], int i, int j)
{
int s = 0;
for ( int k = i; k <=j; k++)
s += freq[k];
return s;
}
int main()
{
int keys[] = {10, 12, 20};
int freq[] = {34, 8, 50};
int n = sizeof (keys)/ sizeof (keys[0]);
printf ( "Cost of Optimal BST is %d " ,
optimalSearchTree(keys, freq, n));
return 0;
}
Java
public class GFG
{
static int optCost( int freq[], int i, int j)
{
if (j < i)
return 0 ;
if (j == i)
return freq[i];
int fsum = sum(freq, i, j);
int min = Integer.MAX_VALUE;
for ( int r = i; r <= j; ++r)
{
int cost = optCost(freq, i, r- 1 ) +
optCost(freq, r+ 1 , j);
if (cost < min)
min = cost;
}
return min + fsum;
}
static int optimalSearchTree( int keys[], int freq[], int n)
{
return optCost(freq, 0 , n- 1 );
}
static int sum( int freq[], int i, int j)
{
int s = 0 ;
for ( int k = i; k <=j; k++)
s += freq[k];
return s;
}
public static void main(String[] args) {
int keys[] = { 10 , 12 , 20 };
int freq[] = { 34 , 8 , 50 };
int n = keys.length;
System.out.println( "Cost of Optimal BST is " +
optimalSearchTree(keys, freq, n));
}
}
Python3
def optCost(freq, i, j):
if j < i:
return 0
if j = = i:
return freq[i]
fsum = Sum (freq, i, j)
Min = 999999999999
for r in range (i, j + 1 ):
cost = (optCost(freq, i, r - 1 ) +
optCost(freq, r + 1 , j))
if cost < Min :
Min = cost
return Min + fsum
def optimalSearchTree(keys, freq, n):
return optCost(freq, 0 , n - 1 )
def Sum (freq, i, j):
s = 0
for k in range (i, j + 1 ):
s + = freq[k]
return s
if __name__ = = '__main__' :
keys = [ 10 , 12 , 20 ]
freq = [ 34 , 8 , 50 ]
n = len (keys)
print ( "Cost of Optimal BST is" ,
optimalSearchTree(keys, freq, n))
C#
using System;
class GFG
{
static int optCost( int []freq, int i, int j)
{
if (j < i)
return 0;
if (j == i)
return freq[i];
int fsum = sum(freq, i, j);
int min = int .MaxValue;
for ( int r = i; r <= j; ++r)
{
int cost = optCost(freq, i, r-1) +
optCost(freq, r+1, j);
if (cost < min)
min = cost;
}
return min + fsum;
}
static int optimalSearchTree( int []keys, int []freq, int n)
{
return optCost(freq, 0, n-1);
}
static int sum( int []freq, int i, int j)
{
int s = 0;
for ( int k = i; k <=j; k++)
s += freq[k];
return s;
}
public static void Main()
{
int []keys = {10, 12, 20};
int []freq = {34, 8, 50};
int n = keys.Length;
Console.Write( "Cost of Optimal BST is " +
optimalSearchTree(keys, freq, n));
}
}
Output:
Cost of Optimal BST is 142
Time complexity of the above naive recursive approach is exponential. It should be noted that the above function computes the same subproblems again and again. We can see many subproblems being repeated in the following recursion tree for freq[1..4].
Since same suproblems are called again, this problem has Overlapping Subprolems property. So optimal BST problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of same subproblems can be avoided by constructing a temporary array cost[][] in bottom up manner.
Dynamic Programming Solution
Following is C/C++ implementation for optimal BST problem using Dynamic Programming. We use an auxiliary array cost[n][n] to store the solutions of subproblems. cost[0][n-1] will hold the final result. The challenge in implementation is, all diagonal values must be filled first, then the values which lie on the line just above the diagonal. In other words, we must first fill all cost[i][i] values, then all cost[i][i+1] values, then all cost[i][i+2] values. So how to fill the 2D array in such manner> The idea used in the implementation is same as Matrix Chain Multiplication problem, we use a variable 'L' for chain length and increment 'L', one by one. We calculate column number 'j' using the values of 'i' and 'L'.
C/C++
#include <stdio.h>
#include <limits.h>
int sum( int freq[], int i, int j);
int optimalSearchTree( int keys[], int freq[], int n)
{
int cost[n][n];
for ( int i = 0; i < n; i++)
cost[i][i] = freq[i];
for ( int L=2; L<=n; L++)
{
for ( int i=0; i<=n-L+1; i++)
{
int j = i+L-1;
cost[i][j] = INT_MAX;
for ( int r=i; r<=j; r++)
{
int c = ((r > i)? cost[i][r-1]:0) +
((r < j)? cost[r+1][j]:0) +
sum(freq, i, j);
if (c < cost[i][j])
cost[i][j] = c;
}
}
}
return cost[0][n-1];
}
int sum( int freq[], int i, int j)
{
int s = 0;
for ( int k = i; k <=j; k++)
s += freq[k];
return s;
}
int main()
{
int keys[] = {10, 12, 20};
int freq[] = {34, 8, 50};
int n = sizeof (keys)/ sizeof (keys[0]);
printf ( "Cost of Optimal BST is %d " ,
optimalSearchTree(keys, freq, n));
return 0;
}
Java
public class Optimal_BST2 {
static int optimalSearchTree( int keys[], int freq[], int n) {
int cost[][] = new int [n + 1 ][n + 1 ];
for ( int i = 0 ; i < n; i++)
cost[i][i] = freq[i];
for ( int L = 2 ; L <= n; L++) {
for ( int i = 0 ; i <= n - L + 1 ; i++) {
int j = i + L - 1 ;
cost[i][j] = Integer.MAX_VALUE;
for ( int r = i; r <= j; r++) {
int c = ((r > i) ? cost[i][r - 1 ] : 0 )
+ ((r < j) ? cost[r + 1 ][j] : 0 ) + sum(freq, i, j);
if (c < cost[i][j])
cost[i][j] = c;
}
}
}
return cost[ 0 ][n - 1 ];
}
static int sum( int freq[], int i, int j) {
int s = 0 ;
for ( int k = i; k <= j; k++) {
if (k >= freq.length)
continue ;
s += freq[k];
}
return s;
}
public static void main(String[] args) {
int keys[] = { 10 , 12 , 20 };
int freq[] = { 34 , 8 , 50 };
int n = keys.length;
System.out.println( "Cost of Optimal BST is "
+ optimalSearchTree(keys, freq, n));
}
}
Python3
INT_MAX = 2147483647
def optimalSearchTree(keys, freq, n):
cost = [[ 0 for x in range (n)]
for y in range (n)]
for i in range (n):
cost[i][i] = freq[i]
for L in range ( 2 , n + 1 ):
for i in range (n - L + 2 ):
j = i + L - 1
if i > = n or j > = n:
break
cost[i][j] = INT_MAX
for r in range (i, j + 1 ):
c = 0
if (r > i):
c + = cost[i][r - 1 ]
if (r < j):
c + = cost[r + 1 ][j]
c + = sum (freq, i, j)
if (c < cost[i][j]):
cost[i][j] = c
return cost[ 0 ][n - 1 ]
def sum (freq, i, j):
s = 0
for k in range (i, j + 1 ):
s + = freq[k]
return s
if __name__ = = '__main__' :
keys = [ 10 , 12 , 20 ]
freq = [ 34 , 8 , 50 ]
n = len (keys)
print ( "Cost of Optimal BST is" ,
optimalSearchTree(keys, freq, n))
C#
using System;
class GFG
{
static int optimalSearchTree( int []keys, int []freq, int n) {
int [,]cost = new int [n + 1,n + 1];
for ( int i = 0; i < n; i++)
cost[i,i] = freq[i];
for ( int L = 2; L <= n; L++) {
for ( int i = 0; i <= n - L + 1; i++) {
int j = i + L - 1;
cost[i,j] = int .MaxValue;
for ( int r = i; r <= j; r++) {
int c = ((r > i) ? cost[i,r - 1] : 0)
+ ((r < j) ? cost[r + 1,j] : 0) + sum(freq, i, j);
if (c < cost[i,j])
cost[i,j] = c;
}
}
}
return cost[0,n - 1];
}
static int sum( int []freq, int i, int j) {
int s = 0;
for ( int k = i; k <= j; k++) {
if (k >= freq.Length)
continue ;
s += freq[k];
}
return s;
}
public static void Main() {
int []keys = { 10, 12, 20 };
int []freq = { 34, 8, 50 };
int n = keys.Length;
Console.Write( "Cost of Optimal BST is "
+ optimalSearchTree(keys, freq, n));
}
}
Output:
Cost of Optimal BST is 142
Notes
1) The time complexity of the above solution is O(n^4). The time complexity can be easily reduced to O(n^3) by pre-calculating sum of frequencies instead of calling sum() again and again.
2) In the above solutions, we have computed optimal cost only. The solutions can be easily modified to store the structure of BSTs also. We can create another auxiliary array of size n to store the structure of tree. All we need to do is, store the chosen 'r' in the innermost loop.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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Source: https://tutorialspoint.dev/language/c-and-cpp-programs/dynamic-programming-set-24-optimal-binary-search-tree
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