Optimal Binary Search Tree
Last Updated :
23 Jul, 2025
Given a sorted array key [0.. n-1] of search keys and an array freq[0.. n-1] of frequency counts, where freq[i] is the number of searches for keys[i]. Construct a binary search tree of all keys such that the total cost of all the searches is as small as possible. The cost of a BST node is the level of that node multiplied by its frequency. The level of the root is 1.
Examples:
Input: keys[] = [10, 12], freq[]= [34, 50]
Output: 118
Explanation: There can be following two possible BSTs
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]
Output: 142
Explanation: There can be many possible BSTs
Among all possible BSTs, cost of the fifth BST is minimum.
Cost of this BST is 1*50 + 2*34 + 3*8 = 142.
Using Recursion - O(n*(2^n)) Time and O(n) Space
We can identify a recursive pattern in this problem.
Recurrence Relation:
- optCost(i, j) = fsum + min(optCost(i, r - 1) + optCost(r + 1, j)) for r in [i , j], where fsum is the sum of frequencies between the indices i and j.
Base Cases:
- If i > j, return 0 (no nodes).
- If i == j, return freq[i] (only one node at index i).
For each subproblem, we compute the cost of selecting each node r as the root of the subtree, recursively calculate the cost for the left and right subtrees, and then add the total frequency sum for the nodes between i and j.
C++
// C++ program for recursive implementation of
// optimal binary search tree
#include <bits/stdc++.h>
using namespace std;
// A utility function to get sum of
// array elements freq[i] to freq[j]
int sum(vector<int> &freq, int i, int j) {
int s = 0;
for (int k = i; k <= j; k++)
s += freq[k];
return s;
}
// A recursive function to calculate
// cost of optimal binary search tree
int optCost(vector<int> &freq, int i, int j) {
// no elements in this subarray
if (j < i)
return 0;
// one element in this subarray
if (j == i)
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;
}
int optimalSearchTree(vector<int> &keys, vector<int> &freq) {
int n = keys.size();
return optCost(freq, 0, n - 1);
}
int main() {
vector<int> keys = {10, 12, 20};
vector<int> freq = {34, 8, 50};
cout << optimalSearchTree(keys, freq);
return 0;
}
// Java program for recursive implementation of
// optimal binary search tree
import java.util.*;
class GfG {
// A utility function to get the sum of array elements
// freq[i] to freq[j]
static int sum(int[] freq, int i, int j) {
int s = 0;
for (int k = i; k <= j; k++)
s += freq[k];
return s;
}
// A recursive function to calculate the cost of an
// optimal binary search tree
static int optCost(int[] freq, int i, int j) {
// No elements in this subarray
if (j < i)
return 0;
// One element in this subarray
if (j == i)
return freq[i];
// Get the sum of freq[i], freq[i+1], ... freq[j]
int fsum = sum(freq, i, j);
// Initialize minimum value
int min = Integer.MAX_VALUE;
// 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;
}
static int optimalSearchTree(int[] keys, int[] freq) {
int n = keys.length;
return optCost(freq, 0, n - 1);
}
public static void main(String[] args) {
int[] keys = { 10, 12, 20 };
int[] freq = { 34, 8, 50 };
System.out.println(optimalSearchTree(keys, freq));
}
}
# Python program for recursive implementation of
# optimal binary search tree
def sum(freq, i, j):
s = 0
for k in range(i, j + 1):
s += freq[k]
return s
# A recursive function to calculate the
# cost of an optimal binary search tree
def optCost(freq, i, j):
# No elements in this subarray
if j < i:
return 0
# One element in this subarray
if j == i:
return freq[i]
# Get the sum of freq[i], freq[i+1], ... freq[j]
fsum = sum(freq, i, j)
# Initialize minimum value
min_val = float('inf')
# One by one, consider all elements as root and recursively find
# cost of the BST, compare the cost
# with min_val, and update min_val if needed
for r in range(i, j + 1):
cost = optCost(freq, i, r - 1) + optCost(freq, r + 1, j)
min_val = min(min_val, cost)
# Return minimum value
return min_val + fsum
def optimalSearchTree(keys, freq):
n = len(keys)
return optCost(freq, 0, n - 1)
keys = [10, 12, 20]
freq = [34, 8, 50]
print(optimalSearchTree(keys, freq))
// C# program for recursive implementation of
// optimal binary search tree
using System;
class GfG {
// A utility function to get the sum of array elements
// freq[i] to freq[j]
static int Sum(int[] freq, int i, int j) {
int s = 0;
for (int k = i; k <= j; k++)
s += freq[k];
return s;
}
// A recursive function to calculate the cost of an
// optimal binary search tree
static int OptCost(int[] freq, int i, int j) {
// No elements in this subarray
if (j < i)
return 0;
// One element in this subarray
if (j == i)
return freq[i];
// Get the sum of freq[i], freq[i+1], ... freq[j]
int fsum = Sum(freq, i, j);
// Initialize minimum value
int min = int.MaxValue;
// 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;
}
static int OptimalSearchTree(int[] keys, int[] freq) {
int n = keys.Length;
return OptCost(freq, 0, n - 1);
}
static void Main(string[] args) {
int[] keys = { 10, 12, 20 };
int[] freq = { 34, 8, 50 };
Console.WriteLine(OptimalSearchTree(keys, freq));
}
}
// JavaScript program for recursive implementation of
// optimal binary search tree
// A utility function to get the sum of array elements
// freq[i] to freq[j]
function sum(freq, i, j) {
let s = 0;
for (let k = i; k <= j; k++)
s += freq[k];
return s;
}
// A recursive function to calculate the cost of an optimal
// binary search tree
function optCost(freq, i, j) {
// Base cases
// No elements in this subarray
if (j < i)
return 0;
// One element in this subarray
if (j === i)
return freq[i];
// Get the sum of freq[i], freq[i+1], ... freq[j]
let fsum = sum(freq, i, j);
// Initialize minimum value
let min = Number.MAX_SAFE_INTEGER;
// 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 (let r = i; r <= j; r++) {
let cost = optCost(freq, i, r - 1)
+ optCost(freq, r + 1, j);
min = Math.min(min, cost);
}
// Return minimum value
return min + fsum;
}
function optimalSearchTree(keys, freq) {
let n = keys.length;
return optCost(freq, 0, n - 1);
}
// Driver Code
let keys = [ 10, 12, 20 ];
let freq = [ 34, 8, 50 ];
console.log(optimalSearchTree(keys, freq));
Using Top - Dowm Dp (Memoization) - O(n^3) Time and O(n^2) Space
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming:
1. Optimal Substructure: Minimum cost for a given i, j i.e. optCost(i, j), depends on the optimal solutions of the subproblems optCost(i, r-1) and optCost(r + 1, j). By choosing the total of these optimal substructures, we can efficiently calculate answer.
2. Overlapping Subproblems: While applying a recursive approach in this problem, we notice that certain subproblems are computed multiple times. For example optCost(1, 1) is computed multiple times from optCost(1, 2) and optCost(1, 3).
- There are two parameter that change in the recursive solution: i going from 0 to n-1, j ranges from i to n-1. So we create a 2D array of size n*n for memoization.
- We initialize array as -1 to indicate nothing is computed initially.
- Now we modify our recursive solution to first check if the value is -1, then only make recursive calls. This way, we avoid re-computations of the same subproblems.
C++
// C++ program for implementation of
// optimal binary search tree using memoization
#include <bits/stdc++.h>
using namespace std;
// A utility function to get sum of
// array elements freq[i] to freq[j]
int sum(vector<int> &freq, int i, int j) {
int s = 0;
for (int k = i; k <= j; k++)
s += freq[k];
return s;
}
// A recursive function to calculate
// cost of optimal binary search tree
int optCost(vector<int> &freq, int i, int j, vector<vector<int>> &memo) {
// Base cases
if (j < i)
return 0;
if (j == i)
return freq[i];
if (memo[i][j] != -1)
return memo[i][j];
// 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, memo) + optCost(freq, r + 1, j, memo);
if (cost < min)
min = cost;
}
// Return minimum value
return memo[i][j] = 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(vector<int> &keys, vector<int> &freq) {
int n = keys.size();
vector<vector<int>> memo(n, vector<int>(n, -1));
return optCost(freq, 0, n - 1, memo);
}
int main() {
vector<int> keys = {10, 12, 20};
vector<int> freq = {34, 8, 50};
cout << optimalSearchTree(keys, freq);
return 0;
}
// Java program for implementation of
// optimal binary search tree using memoization
import java.util.Arrays;
class GfG {
// A utility function to get sum of
// array elements freq[i] to freq[j]
static int sum(int[] freq, int i, int j) {
int s = 0;
for (int k = i; k <= j; k++)
s += freq[k];
return s;
}
// A recursive function to calculate
// cost of optimal binary search tree
static int optCost(int[] freq, int i, int j,
int[][] memo) {
// no elements in this subarray
if (j < i)
return 0;
// one element in this subarray
if (j == i)
return freq[i];
if (memo[i][j] != -1)
return memo[i][j];
// Get sum of freq[i], freq[i+1], ... freq[j]
int fsum = sum(freq, i, j);
// Initialize minimum value
int min = Integer.MAX_VALUE;
// 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, memo)
+ optCost(freq, r + 1, j, memo);
if (cost < min)
min = cost;
}
// Return minimum value
return memo[i][j] = min + fsum;
}
// The main function that calculates
// minimum cost of a Binary Search Tree.
// It mainly uses optCost() to find
// the optimal cost.
static int optimalSearchTree(int[] keys, int[] freq) {
int n = keys.length;
int[][] memo = new int[n][n];
for (int[] row : memo)
Arrays.fill(row, -1);
return optCost(freq, 0, n - 1, memo);
}
public static void main(String[] args) {
int[] keys = { 10, 12, 20 };
int[] freq = { 34, 8, 50 };
System.out.println(optimalSearchTree(keys, freq));
}
}
# Python program for implementation of
# optimal binary search tree using memoization
def sum(freq, i, j):
s = 0
for k in range(i, j + 1):
s += freq[k]
return s
# A recursive function to calculate
# cost of optimal binary search tree
def optCost(freq, i, j, memo):
# Base cases
# no elements in this subarray
if j < i:
return 0
# one element in this subarray
if j == i:
return freq[i]
if memo[i][j] != -1:
return memo[i][j]
# Get sum of freq[i], freq[i+1], ... freq[j]
fsum = sum(freq, i, j)
# Initialize minimum value
min_cost = float('inf')
# 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 r in range(i, j + 1):
cost = optCost(freq, i, r - 1, memo) + \
optCost(freq, r + 1, j, memo)
if cost < min_cost:
min_cost = cost
# Return minimum value
memo[i][j] = min_cost + fsum
return memo[i][j]
# The main function that calculates
# minimum cost of a Binary Search Tree.
# It mainly uses optCost() to find
# the optimal cost.
def optimalSearchTree(keys, freq):
n = len(keys)
memo = [[-1 for _ in range(n)] for _ in range(n)]
return optCost(freq, 0, n - 1, memo)
keys = [10, 12, 20]
freq = [34, 8, 50]
print(optimalSearchTree(keys, freq))
// C# program for implementation of
// optimal binary search tree using memoization
using System;
class GfG {
// A utility function to get sum of
// array elements freq[i] to freq[j]
static int Sum(int[] freq, int i, int j) {
int s = 0;
for (int k = i; k <= j; k++)
s += freq[k];
return s;
}
// A recursive function to calculate
// cost of optimal binary search tree
static int OptCost(int[] freq, int i, int j,
int[, ] memo) {
// Base cases
// no elements in this subarray
if (j < i)
return 0;
// one element in this subarray
if (j == i)
return freq[i];
if (memo[i, j] != -1)
return memo[i, j];
// Get sum of freq[i], freq[i+1], ... freq[j]
int fsum = Sum(freq, i, j);
// Initialize minimum value
int min = int.MaxValue;
// 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, memo)
+ OptCost(freq, r + 1, j, memo);
if (cost < min)
min = cost;
}
// Return minimum value
memo[i, j] = min + fsum;
return memo[i, j];
}
// The main function that calculates
// minimum cost of a Binary Search Tree.
// It mainly uses OptCost() to find
// the optimal cost.
static int OptimalSearchTree(int[] keys, int[] freq) {
int n = keys.Length;
int[, ] memo = new int[n, n];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
memo[i, j] = -1;
return OptCost(freq, 0, n - 1, memo);
}
static void Main(string[] args) {
int[] keys = { 10, 12, 20 };
int[] freq = { 34, 8, 50 };
Console.WriteLine(OptimalSearchTree(keys, freq));
}
}
// JavaScript program for implementation of
// optimal binary search tree using memoization
function sum(freq, i, j) {
let s = 0;
for (let k = i; k <= j; k++)
s += freq[k];
return s;
}
// A recursive function to calculate
// cost of optimal binary search tree
function optCost(freq, i, j, memo) {
// no elements in this subarray
if (j < i)
return 0;
// one element in this subarray
if (j === i)
return freq[i];
if (memo[i][j] !== -1)
return memo[i][j];
// Get sum of freq[i], freq[i+1], ... freq[j]
let fsum = sum(freq, i, j);
// Initialize minimum value
let min = Number.MAX_VALUE;
// 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 (let r = i; r <= j; r++) {
let cost = optCost(freq, i, r - 1, memo)
+ optCost(freq, r + 1, j, memo);
if (cost < min)
min = cost;
}
// Return minimum value
memo[i][j] = min + fsum;
return memo[i][j];
}
// The main function that calculates
// minimum cost of a Binary Search Tree.
// It mainly uses optCost() to find
// the optimal cost.
function optimalSearchTree(keys, freq) {
let n = keys.length;
let memo
= Array.from({length : n}, () => Array(n).fill(-1));
return optCost(freq, 0, n - 1, memo);
}
let keys = [ 10, 12, 20 ];
let freq = [ 34, 8, 50 ];
console.log(optimalSearchTree(keys, freq));
Using Bottom - Up Dp (Tabulation) - O(n^3) Time and O(n^2) Space
We use an auxiliary array dp[n][n] to store the solutions of subproblems. dp[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 dp[i][i] values, then all dp[i][i+1] values, then all dp[i][i+2] values. 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++ program for implementation of
// optimal binary search tree using tabulation
#include <bits/stdc++.h>
using namespace std;
// A utility function to get sum of
// array elements freq[i] to freq[j]
int sum(vector<int> &freq, int i, int j) {
int s = 0;
for (int k = i; k <= j; k++)
s += freq[k];
return s;
}
// A Dynamic Programming based function that calculates the
// minimum cost of a Binary Search Tree.
int optimalSearchTree(vector<int> &keys, vector<int> &freq) {
int n = keys.size();
// Create an auxiliary 2D matrix to store
// results of subproblems
vector<vector<int>> dp(n, vector<int>(n, 0));
// dp[i][j] = Optimal cost of binary search tree
// that can be formed from keys[i] to keys[j].
// dp[0][n-1] will store the resultant dp
// For a single key, dp is equal to frequency of the key
for (int i = 0; i < n; i++) {
dp[i][i] = freq[i];
}
// Now we need to consider chains of length 2, 3, ... .
// l is chain length.
for (int l = 2; l <= n; l++) {
// i is row number in dp[][]
for (int i = 0; i <= n - l; i++) {
// Get column number j from row number i
// and chain length l
int j = i + l - 1;
dp[i][j] = INT_MAX;
int fsum = sum(freq, i, j);
// Try making all keys in interval keys[i..j] as root
for (int r = i; r <= j; r++) {
// c = dp when keys[r] becomes root of this subtree
int c = ((r > i) ? dp[i][r - 1] : 0) + ((r < j) ? dp[r + 1][j] : 0) + fsum;
if (c < dp[i][j]) {
dp[i][j] = c;
}
}
}
}
return dp[0][n - 1];
}
int main() {
vector<int> keys = {10, 12, 20};
vector<int> freq = {34, 8, 50};
cout << optimalSearchTree(keys, freq);
return 0;
}
// Java program for implementation of
// optimal binary search tree using tabulation
import java.util.Arrays;
class GfG {
// A utility function to get sum of
// array elements freq[i] to freq[j]
static int sum(int[] freq, int i, int j) {
int s = 0;
for (int k = i; k <= j; k++)
s += freq[k];
return s;
}
// A Dynamic Programming based function that calculates
// minimum cost of a Binary Search Tree.
static int optimalSearchTree(int[] keys, int[] freq) {
int n = keys.length;
// Create an auxiliary 2D matrix to store results
// of subproblems
int[][] dp = new int[n][n];
// dp[i][j] = Optimal cost of binary search tree
// that can be formed from keys[i] to keys[j].
// dp[0][n-1] will store the resultant dp
// For a single key, dp is equal to frequency of the
// key
for (int i = 0; i < n; i++)
dp[i][i] = freq[i];
// Now we need to consider chains of length 2, 3,
// ... l is chain length.
for (int l = 2; l <= n; l++) {
// i is row number in dp[][]
for (int i = 0; i <= n - l; i++) {
// Get column number j from row number i and
// chain length l
int j = i + l - 1;
dp[i][j] = Integer.MAX_VALUE;
int fsum = sum(freq, i, j);
// Try making all keys in interval
// keys[i..j] as root
for (int r = i; r <= j; r++) {
// c = dp when keys[r] becomes root of
// this subtree
int c = ((r > i) ? dp[i][r - 1] : 0)
+ ((r < j) ? dp[r + 1][j] : 0)
+ fsum;
if (c < dp[i][j])
dp[i][j] = c;
}
}
}
return dp[0][n - 1];
}
public static void main(String[] args) {
int[] keys = { 10, 12, 20 };
int[] freq = { 34, 8, 50 };
System.out.println(optimalSearchTree(keys, freq));
}
}
# Python program for implementation of
# optimal binary search tree using tabulation
def sum(freq, i, j):
s = 0
for k in range(i, j + 1):
s += freq[k]
return s
# A Dynamic Programming based function that calculates
# minimum cost of a Binary Search Tree.
def optimalSearchTree(keys, freq):
n = len(keys)
# Create an auxiliary 2D matrix to store results
# of subproblems
dp = [[0] * n for _ in range(n)]
# dp[i][j] = Optimal cost of binary search tree
# that can be formed from keys[i] to keys[j].
# dp[0][n-1] will store the resultant dp
# For a single key, dp is equal to frequency of the key
for i in range(n):
dp[i][i] = freq[i]
# Now we need to consider chains of length 2, 3, ...
# l is chain length.
for l in range(2, n + 1):
# i is row number in dp[][]
for i in range(n - l + 1):
# Get column number j from row number i and
# chain length l
j = i + l - 1
dp[i][j] = float('inf')
fsum = sum(freq, i, j)
# Try making all keys in interval keys[i..j] as root
for r in range(i, j + 1):
# c = dp when keys[r] becomes root of this subtree
c = ((dp[i][r - 1] if r > i else 0) +
(dp[r + 1][j] if r < j else 0) +
fsum)
if c < dp[i][j]:
dp[i][j] = c
return dp[0][n - 1]
keys = [10, 12, 20]
freq = [34, 8, 50]
print(optimalSearchTree(keys, freq))
// C# program for implementation of
// optimal binary search tree using tabulation
using System;
class GfG {
// A utility function to get sum of
// array elements freq[i] to freq[j]
static int Sum(int[] freq, int i, int j) {
int s = 0;
for (int k = i; k <= j; k++)
s += freq[k];
return s;
}
// A Dynamic Programming based function that calculates
// minimum cost of a Binary Search Tree.
static int OptimalSearchTree(int[] keys, int[] freq) {
int n = keys.Length;
// Create an auxiliary 2D matrix to store results
// of subproblems
int[, ] dp = new int[n, n];
// dp[i][j] = Optimal cost of binary search tree
// that can be formed from keys[i] to keys[j].
// dp[0][n-1] will store the resultant dp
// For a single key, dp is equal to frequency of the
// key
for (int i = 0; i < n; i++)
dp[i, i] = freq[i];
// Now we need to consider chains of length 2, 3,
// ... l is chain length.
for (int l = 2; l <= n; l++) {
// i is row number in dp[][]
for (int i = 0; i <= n - l; i++) {
// Get column number j from row number i and
// chain length l
int j = i + l - 1;
dp[i, j] = int.MaxValue;
int fsum = Sum(freq, i, j);
// Try making all keys in interval
// keys[i..j] as root
for (int r = i; r <= j; r++) {
// c = dp when keys[r] becomes root of
// this subtree
int c = ((r > i ? dp[i, r - 1] : 0)
+ (r < j ? dp[r + 1, j] : 0)
+ fsum);
if (c < dp[i, j])
dp[i, j] = c;
}
}
}
return dp[0, n - 1];
}
static void Main(string[] args) {
int[] keys = { 10, 12, 20 };
int[] freq = { 34, 8, 50 };
Console.WriteLine(OptimalSearchTree(keys, freq));
}
}
// JavaScript program for implementation of
// optimal binary search tree using tabulation
// A utility function to get sum of
// array elements freq[i] to freq[j]
function sum(freq, i, j) {
let s = 0;
for (let k = i; k <= j; k++)
s += freq[k];
return s;
}
// A Dynamic Programming based function that calculates
// minimum cost of a Binary Search Tree.
function optimalSearchTree(keys, freq) {
let n = keys.length;
// Create an auxiliary 2D matrix to store results
// of subproblems
let dp
= Array.from({length : n}, () => Array(n).fill(0));
// dp[i][j] = Optimal cost of binary search tree
// that can be formed from keys[i] to keys[j].
// dp[0][n-1] will store the resultant dp
// For a single key, dp is equal to frequency of the key
for (let i = 0; i < n; i++)
dp[i][i] = freq[i];
// Now we need to consider chains of length 2, 3, ...
// l is chain length.
for (let l = 2; l <= n; l++) {
// i is row number in dp[][]
for (let i = 0; i <= n - l; i++) {
// Get column number j from row number i and
// chain length l
let j = i + l - 1;
dp[i][j] = Number.MAX_VALUE;
let fsum = sum(freq, i, j);
// Try making all keys in interval keys[i..j] as
// root
for (let r = i; r <= j; r++) {
// c = dp when keys[r] becomes root of this
// subtree
let c
= ((r > i ? dp[i][r - 1] : 0)
+ (r < j ? dp[r + 1][j] : 0) + fsum);
if (c < dp[i][j])
dp[i][j] = c;
}
}
}
return dp[0][n - 1];
}
// Driver Code
let keys = [ 10, 12, 20 ];
let freq = [ 34, 8, 50 ];
console.log(optimalSearchTree(keys, freq));
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