62636465666768697071

package main

import (
	"fmt"
	"math"
)

// Recursive function to compute the minimum cost of matrix multiplication
func funcRec(arr []int, i, j int, dp [][]int) int {

	/* Base case: When there is only one
	matrix, no multiplication is needed */
	if i == j {
		return 0
	}

	// Check if the subproblem is already calculated
	if dp[i][j] != -1 {
		return dp[i][j]
	}

	mini := math.MaxInt32

	// Partition the matrices between i and j
	for k := i; k <= j-1; k++ {
		/* Compute the cost of multiplying matrices
		from i to k and from k+1 to j and add cost
		of multiplying the two resulting matrices */
		ans := funcRec(arr, i, k, dp) + funcRec(arr, k+1, j, dp) + arr[i-1]*arr[k]*arr[j]

		// Update the minimum cost
		if ans < mini {
			mini = ans
		}
	}

	// Store and return the minimum cost found
	dp[i][j] = mini
	return dp[i][j]
}

// Function to set up the parameters and call the recursive function
func matrixMultiplication(nums []int) int {
	N := len(nums)

	// Create a DP table initialized with -1
	dp := make([][]int, N)
	for i := range dp {
		dp[i] = make([]int, N)
		for j := range dp[i] {
			dp[i][j] = -1
		}
	}

	// Starting index of the matrix chain
	i := 1

	// Ending index of the matrix chain
	j := N - 1

	// Call the recursive function
	return funcRec(nums, i, j, dp)
}

func main() {
	arr := []int{10, 20, 30, 40, 50}

	// Print the result
	fmt.Println("The minimum number of operations is", matrixMultiplication(arr))
}

Complexity Analysis:

Time Complexity: O(N*N*N), There are N*N states and we explicitly run a loop inside the function which will run for N times, therefore at max ‘N*N*N’ new problems will be solved.

Space Complexity:O(N*N) + O(N). We are using an auxiliary recursion stack space O(N) and a 2D array O(N*N).1e:T38ba,

Intuition

The matrix chain multiplication problem aims to determine the optimal order of multiplying a given sequence of matrices to minimize the total number of scalar multiplications. The key insight is to utilize dynamic programming by breaking down the problem into smaller subproblems, solving each subproblem once, and storing the solutions. This method avoids redundant calculations and efficiently computes the minimum cost. The process involves considering every possible way to split the chain of matrices, calculating the cost for each split, and then choosing the split that results in the minimum cost.

Approach

Define a 2D array dp where dp[i][j] represents the minimum cost of multiplying the chain of matrices from index i to index j.
Initialize the diagonal elements of the dp array to 0, as a single matrix requires no multiplication.
Iterate over the length of the chain, starting from 2 up to the length of the matrix array.
For each length, iterate over the possible starting indices of the chain.
For each starting index, determine the ending index based on the current chain length.
For each pair of starting and ending indices, iterate over all possible split points.
Calculate the cost of multiplying the matrices at the current split point, which includes the cost of multiplying the left sub-chain, the right sub-chain, and the cost of multiplying the resulting two matrices.
Update the dp array with the minimum cost found for each pair of starting and ending indices.
The result is stored in dp[1][n-1], representing the minimum cost of multiplying the entire chain of matrices.

Solution