> For the complete documentation index, see [llms.txt](https://coding-9.gitbook.io/untitled/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://coding-9.gitbook.io/untitled/dynamic-programming/mcm-dp-or-partition-dp/matrix-chain-multiplication.md).
# Matrix Chain Multiplication
## Problem Statement
\
Given a sequence of matrices, find the most efficient way to multiply these matrices together. The efficient way is the one that involves the least number of multiplications.
The dimensions of the matrices are given in an array **arr\[]** of size **N** (such that N = number of matrices + 1) where the **ith** matrix has the dimensions **(arr\[i-1] x arr\[i])**.
**Example 1:**
Input: N = 5
arr = {40, 20, 30, 10, 30}
Output: 26000
Explaination: There are 4 matrices of dimension
40x20, 20x30, 30x10, 10x30. Say the matrices are
named as A, B, C, D. Out of all possible combinations,
the most efficient way is (A*(B*C))*D.
The number of operations are -
20*30*10 + 40*20*10 + 40*10*30 = 26000.
\
**Example 2:**
Input: N = 4
arr = {10, 30, 5, 60}
Output: 4500
Explaination: The matrices have dimensions
10*30, 30*5, 5*60. Say the matrices are A, B
and C. Out of all possible combinations,the
most efficient way is (A*B)*C. The
number of multiplications are -
10*30*5 + 10*5*60 = 4500.
\
**Your Task:**\
You do not need to take input or print anything. Your task is to complete the function **matrixMultiplication()** which takes the value **N** and the array **arr\[]** as input parameters and returns the minimum number of multiplication operations needed to be performed.
\
**Expected Time Complexity:** O(N3)\
**Expected Auxiliary Space:** O(N2)
\
**Constraints:** \
2 ≤ N ≤ 100\
1 ≤ arr\[i] ≤ 500\
## Intuition
```
Recurse through the array
Partition at each point and find the answer
Basically
i * k * j + Left partition + right Partition
eg- ABC
A = 10*20
B = 20*30
C = 30*40
ABC Total size = 10 * 40
Multiplicaltion is 10*20*40 + let (AB) + let(C)
or 10*20*40 + let (A) + let(BC)
So on
```
### Links
### Video Links
\
\
### Approach 1:
```
Memoization
dp i j -> refers to value of multiplication count between index i and j
```
{% code title="C++" lineNumbers="true" %}
```cpp
class Solution{
public:
int find_ans(int i, int j, int arr[], vector> &dp){
if(i == j)
return 0;
if(dp[i][j] != -1)
return dp[i][j];
int mini = 1e9;
for(int k=i; k ans)
mini = ans;
}
return dp[i][j] = mini;
}
int matrixMultiplication(int N, int arr[]){
vector> dp(N, vector (N,-1));
return find_ans(1, N-1, arr, dp);
}
};
```
{% endcode %}
### Approach 2:
```
Tabulation
```
{% code title="C++" lineNumbers="true" %}
```cpp
class Solution{
public:
int matrixMultiplication(int n, int arr[]){
vector> dp(n, vector (n,0));
for(int i=n-1; i>=1; i--) {
for(int j=i+1; j ans)
mini = ans;
}
dp[i][j] = mini;
}
}
return dp[1][n-1];
}
};
```
{% endcode %}
### Approach 3:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Approach 4:
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Similar Problems
###
---
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