added matrix-chain-multiplication and egg-dropping puzzle -2 approaches

C++/dynamic programming/eggDropping.cpp

@@ -0,0 +1,41 @@
+#include<bits/stdc++.h>
+using namespace std;
+
+// recursive solution
+int eggDrop(int e, int f){
+	if(f==0 || f==1) return f;
+	if(e == 1) return f;
+
+	int res = INT_MAX;
+	for(int k=1; k<=f; k++){
+		int tempAns = 1 + max(eggDrop(e, f-k), eggDrop(e-1, k-1));
+		res = min(res, tempAns);
+	}
+
+	return res;
+}
+
+// memoization solution
+int dp[100][100];
+int eggDrop_M(int e, int f){
+	if(f==0 || f==1) return f;
+	if(e == 1) return f;
+
+	if(dp[e][f] != -1) return dp[e][f];
+
+	int res = INT_MAX;
+	for(int k=1; k<=f; k++){
+		int tempAns = 1 + max(eggDrop(e, f-k), eggDrop(e-1, k-1));
+		res = min(res, tempAns);
+	}
+
+	return dp[e][f] = res;
+}
+
+int main(){
+	int egg = 2, floor = 10;
+	memset(dp, -1, sizeof(dp));
+	cout<<eggDrop_M(egg, floor);
+
+	return 0;
+}

C++/dynamic programming/matrixChainMultiplication.cpp

@@ -0,0 +1,41 @@
+#include<bits/stdc++.h>
+using namespace std;
+
+// recursive solution
+int MCM(int *arr, int i, int j){
+	if(i >= j) return 0;
+
+	int res = INT_MAX;
+	for(int k=i; k<j; k++){
+		int tempAns = MCM(arr, i, k) + MCM(arr, k+1, j) + arr[i-1]*arr[k]*arr[j];
+		res = min(res, tempAns);
+	}
+
+	return res;
+}
+
+// memoization solution
+int dp[100][100];
+int MCM_M(int *arr, int i, int j){
+	if(i >= j) return 0;
+
+	if(dp[i][j] != -1) return dp[i][j];
+
+	int res = INT_MAX;
+	for(int k=i; k<j; k++){
+		int tempAns = MCM_M(arr, i, k) + MCM_M(arr, k+1, j) + arr[i-1]*arr[k]*arr[j];
+		res = min(res, tempAns);
+	}
+
+	return dp[i][j] = res;
+}
+
+int main(){
+	int arr[] = {10, 20, 30, 40, 30};
+	int n = sizeof(arr)/sizeof(int);
+	memset(dp, -1, sizeof(dp));
+
+	cout<<MCM_M(arr, 1, n-1);
+
+	return 0;
+}