Objective: Given a 2D-matrix where each cell has a cost to travel. You have to write an algorithm to find a path from left-top corner to bottom-right corner with minimum travel cost. You can move only right or down.

In the solution we will see that how dynamic programming is much better approach than recursion. Don’t get me wrong, even I like recursion a lot π

We can solve it using Recursion ( return Min(path going right, path going down)) but that won’t be a good solution because we will be solving many sub-problems multiple times. Since at every cell we have 2 options the time complexity will O(2^{n}).

Dynamic Programming:

Create a solution matrix of the same size as given matrix.

At every cell, we have two options (go right or down) and we will choose the minimum of these two.
So for any i,j cell
solution[i][j] = A[0][j] if i=0 , first row
= A[i][0] if j=0, first column
= A[i][j] + Min(solution[i-1],[j] , solution[i][j-1]) if i>0 && j>0
See the code for better Explanation.

Time Complexity: O(n^{2}).

Code:

Output:
Minimum Cost Path 29

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
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I believe the answer is 14, not 29. Can you check? I get that when I run it manually and with the program.

Also for the following, why are you setting 0,0 first rather than setting the first row with i=0 to length and first column with i = 0 to length? I think when you do i=1, you are setting the values of the second row and second column. How can solution[0,i-1] even have a value if you haven’t initialized it yet? e.g. for solution[0, 2]

solution[0][0] = A[0][0];
// fill the first row
for (int i = 1; i < A.length; i++) {
solution[0][i] = A[0][i] + solution[0][i – 1];
}

// fill the first column
for (int i = 1; i < A.length; i++) {
solution[i][0] = A[i][0] + solution[i – 1][0];
}