Calculate maximum path cost for a matrix in C - c

Question

I am learning c and encountered maximum cost path question in which
Rules:
matrix is n x n size
Starting from the cell (bottommost leftmost cell), you want to go to the topmost
rightmost cell in a sequence of steps. In each step, you can go either right or up from
your current location.
I tried to solve using dynamic programming and this is the function I have written
computecost(int *utr,int n)//utr is the input matrix
{
int *str;
int i,j;
str=(int *)malloc(n*n*sizeof(int));
for(j=0;j<n;j++)//intialization of bottom row
{
str[n*(n-1)+j]=utr[n*(n-1)+j];
}
for(i=n-2;i>=0;i--)
{
for(j=0;j<n;j++)
{
str[n*i+j]=utr[n*i+j]+max(str[n*(i+1)+j],str[n*(i+1)+(j+1)]);
}
}
printf("%d",str[n*0+0]);
return 0;
}
and this is the input
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
scanf("%d",&str[n*i+j]);
}
}
but
for the matrix 5 x5
1 4 8 2 9
32 67 18 42 1
4 86 12 7 1
8 4 12 17 44
1 43 11 45 2
the desired output is 272 but I am getting 211.
the output matrix for my case
1 43 11 45 2
51 47 57 62 46
55 143 74 69 47
175 210 92 111 52
211 214 119 113 64
Can anyone help me?

Answer 1

You don't need dynamic programming for this since there are no overlapping sub-problems. Just use a simple recursion.
const int n = 5;
int mat[n][n] = {
{1,4,8,2,9},
{32,67,18,42,1},
{4,86,12,7,1},
{8,4,12,17,44},
{1,43,11,45,2}
}; // input matrix
int f(int x, int y, int sum){
if(x == 0 && y == 4)
return sum;
int p = 0, q = 0;
if(x - 1 >= 0)
p = f(x-1, y, sum + mat[x-1][y]);
if(y + 1 <= 4)
q = f(x, y+1, sum+mat[x][y+1]);
return max(p,q);
}
int main(){
int maxSum = f(4,0, mat[4][0]);
printf("%d\n", maxSum);
}

Answer 2

You were not very far to succeed.
In practice, you did not initialize correctly the bottom row.
Moreover, there was a little mistake in the iteration calculation.
This is the corrected code.
As said in a comment, it could be further simplified, by avoiding the use of a new array, simply updating the input array.
#include <stdio.h>
#include <stdlib.h>
int max (int a, int b) {
return (a > b) ? a : b;
}
int computecost(int *utr,int n) { //utr is the input matrix
int *str;
str = malloc (n*n*sizeof(int));
str[n*n - 1] = utr[n*n - 1];
for (int j = n-2; j >= 0; j--) { //intialization of bottom row {
str[n*(n-1)+j] = utr[n*(n-1)+j] + str[n*(n-1)+j+1]; // corrected
}
for (int i=n-2; i>=0; i--) {
str[n*i+n-1] = utr[n*i+n-1] + str[n*(i+1)+n-1];
for(int j = n-2; j >= 0; j--) {
str[n*i+j] = utr[n*i+j] + max(str[n*(i+1)+j],str[n*i + j+1]); // corrected
}
}
int cost = str[0];
free (str);
return cost;
}
int main() {
int A[25] = {
1,43,11,45,2,
8,4,12,17,44,
4,86,12,7,1,
32,67,18,42,1,
1,4,8,2,9
};
int ans = computecost (A, 5);
printf ("%d\n", ans);
return 0;
}