Added Minimum cost path algorithm and updated README

README.md

@@ -47,6 +47,9 @@ Clean examples implementations of data structures and algorithms written in diff
   * [k-means](cluster_analysis/k-means)
 * [Dynamic Programming](dp)
   * [kadane algorithm](dp/kadane-_algorithm)
+  * [knapsack algorithm](dp/knapsack_problem)
+  * [levenshtein distance](dp/levenshtein_distance)
+  * [minimum cost path](dp/min_cost_path)
 
 ## Contribution
  * Contributions are always welcome. Language doesn't matter. Just make sure you're implementing an algorithm.

dp/min_cost_path/README.md

@@ -0,0 +1,6 @@
+### Minimum Cost Path algorithm
+
+Given a cost matrix cost[][] and a position (m, n) in cost[][], write a function that returns cost of minimum cost path to reach (m, n) from (0, 0). Each cell of the matrix represents a cost to traverse through that cell. Total cost of a path to reach (m, n) is sum of all the costs on that path (including both source and destination). You can only traverse down, right and diagonally lower cells from a given cell, i.e., from a given cell (i, j), cells (i+1, j), (i, j+1) and (i+1, j+1) can be traversed. You may assume that all costs are positive integers.
+
+#### A graphical example of Minimum Cost Path algorithm
+![](min_cost_path.png)

dp/min_cost_path/c++/min_cost_path.c++

@@ -0,0 +1,54 @@
+#include<stdio.h>
+#include<limits.h>
+#define R 3
+#define C 3
+
+int min(int x, int y, int z);
+
+int minCost(int cost[R][C], int m, int n)
+{
+	int i, j;
+
+	// Instead of following line, we can use int tc[m+1][n+1] or 
+	// dynamically allocate memoery to save space. The following line is
+	// used to keep te program simple and make it working on all compilers.
+	int tc[R][C];  
+
+	tc[0][0] = cost[0][0];
+
+	/* Initialize first column of total cost(tc) array */
+	for (i = 1; i <= m; i++)
+		tc[i][0] = tc[i-1][0] + cost[i][0];
+
+	/* Initialize first row of tc array */
+	for (j = 1; j <= n; j++)
+		tc[0][j] = tc[0][j-1] + cost[0][j];
+
+	/* Construct rest of the tc array */
+	for (i = 1; i <= m; i++)
+		for (j = 1; j <= n; j++)
+			tc[i][j] = min(tc[i-1][j-1], 
+					tc[i-1][j], 
+					tc[i][j-1]) + cost[i][j];
+
+	return tc[m][n];
+}
+
+/* A utility function that returns minimum of 3 integers */
+int min(int x, int y, int z)
+{
+	if (x < y)
+		return (x < z)? x : z;
+	else
+		return (y < z)? y : z;
+}
+
+// Test program
+int main()
+{
+	int cost[R][C] = { {1, 2, 3},
+		{4, 8, 2},
+		{1, 5, 3} };
+	printf(" %d ", minCost(cost, 2, 2));
+	return 0;
+}