added Dynamic Programming Problems

Dynamic Programming/Coin-Change.cpp

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+#include <iostream>
+#include<climits>
+using namespace std;
+
+// Function to find the Minimum number of coins required to get Sum S
+int findMinCoins(int arr[], int n, int N)
+{
+	// dp[i] = no of coins required to get a total of i
+	int dp[N + 1];
+	
+	// 0 coins are needed for 0 sum
+	
+	dp[0] = 0;	
+
+
+	for (int i = 1; i <= N; i++)
+	{
+		// initialize minimum number of coins needed to infinity
+		dp[i] = INT_MAX;
+		int res = INT_MAX;
+
+		// do for each coin
+		for (int c = 0; c < n; c++)
+		{
+			if (i - arr[c] >= 0) // check if coins doesn't become negative by including it
+				res = dp[i - arr[c]];
+
+			// if total can be reached by including current coin c,
+			// update minimum number of coins needed dp[i]
+			if (res != INT_MAX)
+				dp[i] = min(dp[i], res + 1);
+		}
+	}
+
+	// The Minimum No of Coins Required for N = dp[N]
+	return dp[N];
+}
+
+int main()
+{
+	// No of Coins We Have
+	int arr[] = { 1, 2, 3, 4 };
+	int n = sizeof(arr) / sizeof(arr[0]);
+
+	// Total Change Required
+	int N = 15;
+
+	cout << "Minimum Number of Coins Required "<< findMinCoins(arr, n, N) << "\n";
+
+	return 0;
+}

Dynamic Programming/Matrix-Chain-Multiplication.cpp

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+#include <iostream>
+#include <climits>
+using namespace std;
+
+#define MAX 10
+
+// dp table to store the solution for already computed sub problems
+int dp[MAX][MAX];
+
+// Function to find the most efficient way to multiply the given sequence of matrices
+int MatrixChainMultiplication(int dim[], int i, int j)
+{
+	// base case: one matrix
+	if (j <= i + 1)
+		return 0;
+
+	// stores minimum number of scalar multiplications (i.e., cost)
+	// needed to compute the matrix M[i+1]...M[j] = M[i..j]
+	int min = INT_MAX;
+
+	// if dp[i][j] is not calculated (calculate it!!) 
+
+	if (dp[i][j] == 0)
+	{
+		// take the minimum over each possible position at which the
+		// sequence of matrices can be split
+
+		for (int k = i + 1; k <= j - 1; k++)
+		{
+			// recur for M[i+1]..M[k] to get a i x k matrix
+			int cost = MatrixChainMultiplication(dim, i, k);
+
+			// recur for M[k+1]..M[j] to get a k x j matrix
+			cost += MatrixChainMultiplication(dim, k, j);
+
+			// cost to multiply two (i x k) and (k x j) matrix
+			cost +=	dim[i] * dim[k] * dim[j];
+
+			if (cost < min)
+				min = cost; // store the minimum cost
+		}
+		dp[i][j] = min;
+	}
+
+	// return min cost to multiply M[j+1]..M[j]
+	return dp[i][j];
+}
+
+// main function
+int main()
+{
+	// Matrix i has Dimensions dim[i-1] & dim[i] for i=1..n
+	// input is 10 x 30 matrix, 30 x 5 matrix, 5 x 60 matrix
+	int dim[] = { 10, 30, 5, 60 };
+	int n = sizeof(dim) / sizeof(dim[0]);
+
+	// Function Calling: MatrixChainMultiplications(dimensions_array, starting, ending);
+
+	cout << "Minimum cost is " << MatrixChainMultiplication(dim, 0, n - 1) << "\n";
+
+	return 0;
+}