Hey @shivam_2701
Thanks for asking your doubt.
You cannot use DFS to solve this problem as DFS does not give the shortest path between two points. letβs considered a test case
1
3 3 3
let (x,y,z) be the z moves required to reach (x, y) from starting index.
so the DFS call be
(3,3,0) β (4,3,1) β (5,3,2) β (6,3,3)
now form (4,3,1) β (4,4,2) But you can reach (4,4) in 1 move ( (3,3,0) β (4,4,1) ).DFS does not take care of shortest path because of this we cannot use dfs here.