text: Optimal Binary Search Tree | DP | Dynamic Programming code for Optimal Binary Search Tree Problem ; A utility function to get sum of array elements freq [ i ] to freq [ j ] ; A Dynamic Programming based function that calculates minimum cost of a Binary Search Tree . ; Create an auxiliary 2D matrix to store results of subproblems ; For a single key , cost is equal to frequency of the key ; Now we need to consider chains of length 2 , 3 , ... . L is chain length . ; i is row number in cost [ ] [ ] ; Get column number j from row number i and chain length L ; Try making all keys in interval keys [ i . . j ] as root ; c = cost when keys [ r ] becomes root of this subtree ; Driver program to test above functions
code: #include <stdio.h> NEW_LINE #include <limits.h> NEW_LINE int sum ( int freq [ ] , int i , int j ) { int s = 0 ; for ( int k = i ; k <= j ; k ++ ) s += freq [ k ] ; return s ; } 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 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 ; }