Please find the mistake in my solution and why i think the solution given by Naveen is wrong.
Take the case to reach 6.
we can reach 6 in the following ways:-
1.) 1->2->3->4->5->6. Cost will be= min(x, y)+4y. min(x, y) is to reach 2 cause 1+1=2 and 1*2 =2.
2.) 1->2->3->6. Cost will be= min(x, y)+y+x.
3.) 1->2->4->5->6. Cost will be= min(x, y)+ 2y.
4.) 1->2->4->8->7->6. Cost will be = min(x, y)+2x+2z.
According to the solution given by him(Naveen), to reach 6 we should only consider path_2 and (path_1 or path_3) path but the 4th path can also be a solution.
if solution of this inequality exists then the 4th path will be the solution.
cost_1+cost_2+cost_3 > 3 (cost_4)
3min(x, y)+ 7y+2x > 3min(x, y)+6x+6z
7y+2x > 6x+6z
7y > 4x+6z
Clearly the solution to this inequality exist and hence the method mentioned by him is wrong.
Solution:- there should be a upper bound on the number of cells and then. after than simple apply Dijkstra and find the shortest path from 1 to N.