An Optimal Binary Search Tree (OBST), also known as a Weighted Binary Search Tree, is a binary search tree that minimizes the expected search cost. In a binary search tree, the search cost is the number of comparisons required to search for a given key.
In an OBST, each node is assigned a weight that represents the probability of the key being searched for. The sum of all the weights in the tree is 1.0. The expected search cost of a node is the sum of the product of its depth and weight, and the expected search cost of its children.
To construct an OBST, we start with a sorted list of keys and their probabilities. We then build a table that contains the expected search cost for all possible sub-trees of the original list. We can use dynamic programming to fill in this table efficiently. Finally, we use this table to construct the OBST.
The time complexity of constructing an OBST is O(n^3), where n is the number of keys. However, with some optimizations, we can reduce the time complexity to O(n^2). Once the OBST is constructed, the time complexity of searching for a key is O(log n), the same as for a regular binary search tree.
The OBST is a useful data structure in applications where the keys have different probabilities of being searched for. It can be used to improve the efficiency of searching and retrieval operations in databases, compilers, and other computer programs.
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.
Let us first define the cost of a BST. 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}
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
1) Optimal Substructure:
The optimal cost for freq[i..j] can be recursively calculated using the 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)
The reason for adding the sum of frequencies from i to j:
This can be divided into 2 parts one is the freq[r]+sum of frequencies of all elements from i to j except r. The term freq[r] is added because it is going to be root and that means level of 1, so freq[r]*1=freq[r]. Now the actual part comes, we are adding the frequencies of remaining elements because as we take r as root then all the elements other than that are going 1 level down than that is calculated in the subproblem. Let me put it in a more clear way, for calculating optcost(i,j) we assume that the r is taken as root and calculate min of opt(i,r-1)+opt(r+1,j) for all i<=r<=j. Here for every subproblem we are choosing one node as a root. But in reality the level of subproblem root and all its descendant nodes will be 1 greater than the level of the parent problem root. Therefore the frequency of all the nodes except r should be added which accounts to the descend in their level compared to level assumed in subproblem.
2) Overlapping Subproblems
Following is recursive implementation that simply follows the recursive structure mentioned above.
C++
// A naive recursive implementation of // optimal binary search tree problem #include <bits/stdc++.h>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 Codeint 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 |
Java
// A naive recursive implementation of optimal binary // search tree problempublic class GFG { // A recursive function to calculate cost of // optimal binary search tree static 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 = 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; } // 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) { // 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] static 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 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)); }}// This code is contributed by Sumit Ghosh |
Python3
# A naive recursive implementation of # optimal binary search tree problem # A recursive function to calculate # cost of optimal binary search tree def optCost(freq, i, 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] fsum = Sum(freq, i, j) # Initialize minimum value Min = 999999999999 # 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) + 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. def optimalSearchTree(keys, freq, 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] def Sum(freq, i, j): s = 0 for k in range(i, j + 1): s += freq[k] return s# Driver Codeif __name__ == '__main__': keys = [10, 12, 20] freq = [34, 8, 50] n = len(keys) print("Cost of Optimal BST is", optimalSearchTree(keys, freq, n)) # This code is contributed by PranchalK |
C#
// A naive recursive implementation of optimal binary // search tree problemusing System;class GFG{ // A recursive function to calculate cost of // optimal binary search tree static int optCost(int []freq, int i, int 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 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; } // 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) { // 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] static 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 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)); }}// This code is contributed by Sam007 |
Javascript
<script>//Javascript Implementation // A recursive function to calculate// cost of optimal binary search treefunction optCost(freq, i, 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] var fsum = sum(freq, i, j); // Initialize minimum value var 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 (var r = i; r <= j; ++r) { var 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.function optimalSearchTree(keys, freq, 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]function sum(freq, i, j){ var s = 0; for (var k = i; k <= j; k++) s += freq[k]; return s;} // Driver Codevar keys = [10, 12, 20];var freq = [34, 8, 50];var n = keys.length;document.write("Cost of Optimal BST is " + optimalSearchTree(keys, freq, n)); // This code is contributed by shubhamsingh10</script> |
C
// A naive recursive implementation of optimal binary // search tree problem#include <stdio.h>#include <limits.h>// 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 treeint 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 program to test above functionsint 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;} |
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 subproblems are called again, this problem has Overlapping Subproblems 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++
// Dynamic Programming code for Optimal Binary Search // Tree Problem #include <bits/stdc++.h>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 Dynamic Programming based function that calculates minimum cost of a Binary Search Tree. */int optimalSearchTree(int keys[], int freq[], int n) { /* Create an auxiliary 2D matrix to store results of subproblems */ int cost[n][n]; /* cost[i][j] = Optimal cost of binary search tree that can be formed from keys[i] to keys[j]. cost[0][n-1] will store the resultant cost */ // For a single key, cost is equal to frequency of the key for (int i = 0; i < n; i++) cost[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 cost[][] for (int i = 0; i <= n-L+1; i++) { // Get column number j from row number i and // chain length L int j = i+L-1; cost[i][j] = INT_MAX; int off_set_sum = sum(freq, i, j); // Try making all keys in interval keys[i..j] as root for (int r = i; r <= j; r++) { // c = cost when keys[r] becomes root of this subtree int c = ((r > i)? cost[i][r-1]:0) + ((r < j)? cost[r+1][j]:0) + off_set_sum; if (c < cost[i][j]) cost[i][j] = c; } } } return cost[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 code is contributed by rathbhupendra |
C
// Dynamic Programming code for Optimal Binary Search // Tree Problem#include <stdio.h>#include <limits.h>// A utility function to get sum of array elements // freq[i] to freq[j]int sum(int freq[], int i, int j);/* A Dynamic Programming based function that calculates minimum cost of a Binary Search Tree. */int optimalSearchTree(int keys[], int freq[], int n){ /* Create an auxiliary 2D matrix to store results of subproblems */ int cost[n][n]; /* cost[i][j] = Optimal cost of binary search tree that can be formed from keys[i] to keys[j]. cost[0][n-1] will store the resultant cost */ // For a single key, cost is equal to frequency of the key for (int i = 0; i < n; i++) cost[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 cost[][] for (int i=0; i<=n-L+1; i++) { // Get column number j from row number i and // chain length L int j = i+L-1; int off_set_sum = sum(freq, i, j); cost[i][j] = INT_MAX; // Try making all keys in interval keys[i..j] as root for (int r=i; r<=j; r++) { // c = cost when keys[r] becomes root of this subtree int c = ((r > i)? cost[i][r-1]:0) + ((r < j)? cost[r+1][j]:0) + off_set_sum; if (c < cost[i][j]) cost[i][j] = c; } } } return cost[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 program to test above functionsint 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
// Dynamic Programming Java code for Optimal Binary Search// Tree Problempublic class Optimal_BST2 { /* A Dynamic Programming based function that calculates minimum cost of a Binary Search Tree. */ static int optimalSearchTree(int keys[], int freq[], int n) { /* Create an auxiliary 2D matrix to store results of subproblems */ int cost[][] = new int[n + 1][n + 1]; /* cost[i][j] = Optimal cost of binary search tree that can be formed from keys[i] to keys[j]. cost[0][n-1] will store the resultant cost */ // For a single key, cost is equal to frequency of the key for (int i = 0; i < n; i++) cost[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 cost[][] for (int i = 0; i <= n - L + 1; i++) { // Get column number j from row number i and // chain length L int j = i + L - 1; cost[i][j] = Integer.MAX_VALUE; int off_set_sum = sum(freq, i, j); // Try making all keys in interval keys[i..j] as root for (int r = i; r <= j; r++) { // c = cost when keys[r] becomes root of this subtree int c = ((r > i) ? cost[i][r - 1] : 0) + ((r < j) ? cost[r + 1][j] : 0) + off_set_sum; if (c < cost[i][j]) cost[i][j] = c; } } } return cost[0][n - 1]; } // 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++) { 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)); }}//This code is contributed by Sumit Ghosh |
Python3
# Dynamic Programming code for Optimal Binary Search # Tree Problem INT_MAX = 2147483647""" A Dynamic Programming based function thatcalculates minimum cost of a Binary Search Tree. """def optimalSearchTree(keys, freq, n): """ Create an auxiliary 2D matrix to store results of subproblems """ cost = [[0 for x in range(n)] for y in range(n)] """ cost[i][j] = Optimal cost of binary search tree that can be formed from keys[i] to keys[j]. cost[0][n-1] will store the resultant cost """ # For a single key, cost is equal to # frequency of the key for i in range(n): cost[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 cost for i in range(n - L + 2): # Get column number j from row number # i and chain length L j = i + L - 1 off_set_sum = sum(freq, i, j) if i >= n or j >= n: break cost[i][j] = INT_MAX # Try making all keys in interval # keys[i..j] as root for r in range(i, j + 1): # c = cost when keys[r] becomes root # of this subtree c = 0 if (r > i): c += cost[i][r - 1] if (r < j): c += cost[r + 1][j] c += off_set_sum if (c < cost[i][j]): cost[i][j] = c return cost[0][n - 1] # A utility function to get sum of # array elements freq[i] to freq[j] def sum(freq, i, j): s = 0 for k in range(i, j + 1): s += freq[k] return s # Driver Codeif __name__ == '__main__': keys = [10, 12, 20] freq = [34, 8, 50] n = len(keys) print("Cost of Optimal BST is", optimalSearchTree(keys, freq, n)) # This code is contributed by SHUBHAMSINGH10 |
C#
// Dynamic Programming C# code for Optimal Binary Search// Tree Problemusing System;class GFG{ /* A Dynamic Programming based function that calculates minimum cost of a Binary Search Tree. */ static int optimalSearchTree(int []keys, int []freq, int n) { /* Create an auxiliary 2D matrix to store results of subproblems */ int [,]cost = new int[n + 1,n + 1]; /* cost[i][j] = Optimal cost of binary search tree that can be formed from keys[i] to keys[j]. cost[0][n-1] will store the resultant cost */ // For a single key, cost is equal to frequency of the key for (int i = 0; i < n; i++) cost[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 cost[][] for (int i = 0; i <= n - L + 1; i++) { // Get column number j from row number i and // chain length L int j = i + L - 1; int off_set_sum = sum(freq, i, j); cost[i,j] = int.MaxValue; // Try making all keys in interval keys[i..j] as root for (int r = i; r <= j; r++) { // c = cost when keys[r] becomes root of this subtree int c = ((r > i) ? cost[i,r - 1] : 0) + ((r < j) ? cost[r + 1,j] : 0) + off_set_sum; if (c < cost[i,j]) cost[i,j] = c; } } } return cost[0,n - 1]; } // 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++) { 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)); }}// This code is contributed by Sam007 |
Javascript
<script>// Dynamic Programming code for Optimal Binary Search// Tree Problem/* A Dynamic Programming based function that calculatesminimum cost of a Binary Search Tree. */function optimalSearchTree(keys, freq, n){ /* Create an auxiliary 2D matrix to store results of subproblems */ var cost = new Array(n); for (var i = 0; i < n; i++) cost[i] = new Array(n); /* cost[i][j] = Optimal cost of binary search tree that can be formed from keys[i] to keys[j]. cost[0][n-1] will store the resultant cost */ // For a single key, cost is equal to frequency of the key for (var i = 0; i < n; i++) cost[i][i] = freq[i]; // Now we need to consider chains of length 2, 3, ... . // L is chain length. for (var L = 2; L <= n; L++) { // i is row number in cost[][] for (var i = 0; i <= n-L+1; i++) { // Get column number j from row number i and // chain length L var j = i+L-1; var off_set_sum = sum(freq, i, j); if ( i >= n || j >= n) break cost[i][j] = Number. MAX_SAFE_INTEGER; // Try making all keys in interval keys[i..j] as root for (var r = i; r <= j; r++) { // c = cost when keys[r] becomes root of this subtree var c = 0; if (r > i) c += cost[i][r-1] if (r < j) c += cost[r+1][j] c += off_set_sum; if (c < cost[i][j]) cost[i][j] = c; } } } return cost[0][n-1];} // A utility function to get sum of array elements// freq[i] to freq[j]function sum(freq, i, j){ var s = 0; for (var k = i; k <= j; k++) s += freq[k]; return s;}var keys = [10, 12, 20];var freq = [34, 8, 50];var n = keys.length;document.write("Cost of Optimal BST is " + optimalSearchTree(keys, freq, n)); // This code contributed by shubhamsingh10</script> |
Output
Cost of Optimal BST is 142
Notes
1) The time complexity of the above solution is O(n^3). We have optimized the implementation by calculating the sum of the subarray freq[i…j] only once.
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 the 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.
Recursive Memoized Solution
We can use the recursive solution with a dynamic programming approach to have a more optimized code, reducing the complexity from O(n^3) from the pure dynamic programming to O(n). To do that, we have to store the subproblems calculations in a matrix of NxN and use that in the recursions, avoiding calculating all over again for every recursive call.
C++
#include <bits/stdc++.h>using namespace std;#define MAX 1000// Declare global cost matrixint cost[MAX][MAX];// Helper function to calculate the sum of frequencies from index i to jint Sum(int freq[], int i, int j) { int s = 0; for (int k = i; k <= j; k++) s += freq[k]; return s;}// Recursive function to find the optimal cost of a BST using memoizationint optCost_memoized(int freq[], int i, int j) { // Reuse cost already calculated for the subproblems. // Since we initialize cost matrix with 0 and frequency for a tree of one node, // it can be used as a stop condition if (cost[i][j]) return cost[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 c = optCost_memoized(freq, i, r - 1) + optCost_memoized(freq, r + 1, j) + fsum; if (c < Min) { Min = c; // replace cost with new optimal calc cost[i][j] = c; } } // Return minimum value return cost[i][j];}// Main function to calculate the minimum cost of a BSTint 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_memoized(freq, 0, n - 1);}int main() { int keys[] = {10, 12, 20}; int freq[] = {34, 8, 50}; int n = sizeof(keys) / sizeof(keys[0]); // cost[i][j] = Optimal cost of binary search // tree that can be formed from keys[i] to keys[j]. // cost[0][n-1] will store the resultant cost memset(cost, 0, sizeof(cost)); // For a single key, cost is equal to // frequency of the key for (int i = 0; i < n; i++) cost[i][i] = freq[i]; cout << "Cost of Optimal BST is " << optimalSearchTree(keys, freq, n) << endl; return 0;} |
Java
// Java Code for the approachimport java.util.*;public class Main { static final int MAX = 1000; static int cost[][] = new int[MAX][MAX]; // Helper function to calculate the sum of frequencies // from index i to 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; } // Recursive function to find the optimal cost of a BST // using memoization static int optCost_memoized(int freq[], int i, int j) { if (i < 0 || j < 0) return 0; // Reuse cost already calculated for the // subproblems. Since we initialize cost matrix with // 0 and frequency for a tree of one node, it can be // used as a stop condition if (cost[i][j] != 0) return cost[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 c = optCost_memoized(freq, i, r - 1) + optCost_memoized(freq, r + 1, j) + fsum; if (c < Min) { Min = c; // replace cost with new optimal calc cost[i][j] = c; } } // Return minimum value return cost[i][j]; } // Main function to calculate the minimum cost of a BST static 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_memoized(freq, 0, n - 1); } public static void main(String[] args) { int keys[] = { 10, 12, 20 }; int freq[] = { 34, 8, 50 }; int n = keys.length; // cost[i][j] = Optimal cost of binary search // tree that can be formed from keys[i] to keys[j]. // cost[0][n-1] will store the resultant cost for (int i = 0; i < n; i++) Arrays.fill(cost[i], 0); // For a single key, cost is equal to // frequency of the key for (int i = 0; i < n; i++) cost[i][i] = freq[i]; System.out.println( "Cost of Optimal BST is " + optimalSearchTree(keys, freq, n)); }} |
Python3
def optCost_memoized(freq, i, j): # Reuse cost already calculated for the subproblems. # Since we initialize cost matrix with 0 and fredquency for a tree of one node, # it can be used as a stop condition if cost[i][j]: return cost[i][j] # Get sum of freq[i], freq[i+1], ... freq[j] fsum = Sum(freq, i, j) # Initialize minimum value Min = 999999999999 # 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): c = (optCost_memoized(freq, i, r - 1) + optCost_memoized(freq, r + 1, j)) c += fsum if c < Min: Min = c # replace cost with new optimal calc cost[i][j] = c # Return minimum value return cost[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): # 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_memoized(freq, 0, n - 1)# A utility function to get sum of# array elements freq[i] to freq[j]def Sum(freq, i, j): s = 0 for k in range(i, j + 1): s += freq[k] return sif __name__ == '__main__': keys = [10, 12, 20] freq = [34, 8, 50] n = len(keys) # cost[i][j] = Optimal cost of binary search # tree that can be formed from keys[i] to keys[j]. # cost[0][n-1] will store the resultant cost cost = [[0 for x in range(n+1)] for y in range(n+1)] # For a single key, cost is equal to # frequency of the key for i in range(n): cost[i][i] = freq[i] print("Cost of Optimal BST is", optimalSearchTree(keys, freq, n)) |
C#
// C# Code for the approachusing System;public class GFG { const int MAX = 1000; // Declare global cost matrix static int[, ] cost = new int[MAX, MAX]; // Helper function to calculate the sum of frequencies // from index i to 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; } // Recursive function to find the optimal cost of a BST // using memoization static int optCost_memoized(int[] freq, int i, int j) { if (i < 0 || j < 0) return 0; // Reuse cost already calculated for the // subproblems. Since we initialize cost matrix with // 0 and frequency for a tree of one node, it can be // used as a stop condition if (cost[i, j] != 0) return cost[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 c = optCost_memoized(freq, i, r - 1) + optCost_memoized(freq, r + 1, j) + fsum; if (c < Min) { Min = c; // replace cost with new optimal calc cost[i, j] = c; } } // Return minimum value return cost[i, j]; } // Main function to calculate the minimum cost of a BST static 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_memoized(freq, 0, n - 1); } public static void Main() { int[] keys = { 10, 12, 20 }; int[] freq = { 34, 8, 50 }; int n = keys.Length; // cost[i][j] = Optimal cost of binary search // tree that can be formed from keys[i] to keys[j]. // cost[0][n-1] will store the resultant cost for (int i = 0; i < MAX; i++) { for (int j = 0; j < MAX; j++) { cost[i, j] = 0; } } // For a single key, cost is equal to // frequency of the key for (int i = 0; i < n; i++) cost[i, i] = freq[i]; Console.WriteLine( "Cost of Optimal BST is " + optimalSearchTree(keys, freq, n)); }} |
Javascript
// JavaScript code for the approach// Declare global cost matrixconst MAX = 1000;let cost = new Array(MAX).fill(null).map( () => new Array(MAX).fill(0));// Helper function to calculate the sum of frequencies from// index i to jfunction Sum(freq, i, j){ let s = 0; for (let k = i; k <= j; k++) s += freq[k]; return s;}// Recursive function to find the optimal cost of a BST// using memoizationfunction optCost_memoized(freq, i, j){ if (i < 0 || j < 0) return 0; // Reuse cost already calculated for the subproblems. // Since we initialize cost matrix with 0 and frequency // for a tree of one node, it can be used as a stop // condition if (cost[i][j]) return cost[i][j]; // Get sum of freq[i], freq[i+1], ... freq[j] let fsum = Sum(freq, i, j); // Initialize minimum value let Min = Infinity; // 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 c = optCost_memoized(freq, i, r - 1) + optCost_memoized(freq, r + 1, j) + fsum; if (c < Min) { Min = c; // replace cost with new optimal calc cost[i][j] = c; } } // Return minimum value return cost[i][j];}// Main function to calculate the minimum cost of a BSTfunction optimalSearchTree(keys, freq, 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_memoized(freq, 0, n - 1);}// Usage examplelet keys = [ 10, 12, 20 ];let freq = [ 34, 8, 50 ];let n = keys.length;// cost[i][j] = Optimal cost of binary search// tree that can be formed from keys[i] to keys[j].// cost[0][n-1] will store the resultant costcost = new Array(MAX).fill(null).map( () => new Array(MAX).fill(0));// For a single key, cost is equal to// frequency of the keyfor (let i = 0; i < n; i++) cost[i][i] = freq[i];console.log("Cost of Optimal BST is " + optimalSearchTree(keys, freq, n)); |
Output
Cost of Optimal BST is 142
The time complexity of the above solution is O(n^3)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
