create Matrix Chain Multiplication

Author: MayankUjawane

Dynamic Programming/43) Matrix Chain Multiplication.md

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+# Matrix Chain Multiplication
+Question -> [Matrix Chain Multiplication](https://www.codingninjas.com/codestudio/problems/matrix-chain-multiplication_975344)    
+
+### Recursion
+```java
+public class Solution {
+    public static int matrixMultiplication(int[] arr , int N) {
+        return func(1, N-1, arr);
+    }
+    private static int func(int i, int j, int[] arr) {
+        if(i==j) return 0;
+        int min = Integer.MAX_VALUE;
+        for(int k=i; k<j; k++) {
+            int value = (arr[i-1]*arr[k]*arr[j]) + func(i,k,arr) + func(k+1,j,arr);
+            if(value<min) min = value;
+        }
+        return min;
+    }
+}
+```           
+> `Time Complexity` : **Exponential**    
+> `Space Complexity` : **O(N)** 
+---
+### Memoization
+```java
+import java.util.*;
+public class Solution {
+    public static int matrixMultiplication(int[] arr , int N) {
+        int[][] dp = new int[N][N];
+        for(int[] row : dp) Arrays.fill(row,-1);
+        return func(1, N-1, arr, dp);
+    }
+    private static int func(int i, int j, int[] arr, int[][] dp) {
+        if(i==j) return 0;
+        if(dp[i][j] != -1) return dp[i][j];
+        int min = Integer.MAX_VALUE;
+        for(int k=i; k<j; k++) {
+            int value = (arr[i-1]*arr[k]*arr[j]) + func(i,k,arr,dp) + func(k+1,j,arr,dp);
+            if(value<min) min = value;
+        }
+        return dp[i][j] = min;
+    }
+}
+```
+> `Time Complexity` : **O(N<sup>3</sup>)**           
+> `Space Complexity` : **O(N)+O(N\*N)**, for Recursion Stack and dp array
+---
+### Tabulation
+```java
+public class Solution {
+    public static int matrixMultiplication(int[] arr , int N) {
+        int[][] dp = new int[N][N];
+        //base case
+        for(int i=1; i<N; i++) dp[i][i] = 0;
+        for(int i=N-1; i>=1; i--) {
+            for(int j=i+1; j<N; j++) {
+                int min = Integer.MAX_VALUE;
+                for(int k=i; k<j; k++) {
+                    int value = (arr[i-1]*arr[k]*arr[j]) + dp[i][k] + dp[k+1][j];
+                    if(value<min) min = value;
+                }
+                dp[i][j] = min;
+            }
+        }
+        return dp[1][N-1];
+    }
+}
+```
+> `Time Complexity` : **O(N<sup>3</sup>)**             
+> `Space Complexity` : **O(N\*N)** 
+---
+Video Explanations -> [Memoization](https://youtu.be/vRVfmbCFW7Y?list=PLgUwDviBIf0qUlt5H_kiKYaNSqJ81PMMY), [Tabulation](https://youtu.be/pDCXsbAw5Cg?list=PLgUwDviBIf0qUlt5H_kiKYaNSqJ81PMMY)   
+<hr>