u are not using dp properly u are updating the matrix only when u find a valid combinations of operations, try to think at ur tree of calls

input : 33 every fork is a possible path and u are updating ur matrix to 1 only when u find a valid sum that is divisible by 9

                   | - 3
          | - 3  - |
          |        | - 9
dp[0][0]- |
          |        | - 3
          | - 9    |
                   | - 9

in a bigger picture what u wrote is:

1) i try a path

if the sum of the digits on the path is not 0 mod 9 i'm going back and i m not updating the matrix [wrong]

if the sum of the digits on the path is 0 mod 9 i have found a solution

this is an exponential solution

try to use a 3th value {-1 : idk,1:found,0: ehi bro i have already try everything from this postion is usless that we are searching there}

sorry for bad english + it's 00:15 in my country + i'm sleeping

infact bro i challenge u to spot the difference between ur code and my version 295135560

btw a gigachad solution to this problems is:

the only 2 numbers that ^2 are less than 10 are 2 and 3 and change these digits is equal to adding 2 or 6 respectevely knowing that gcd(9,6) != 0 and gcd(9,2) == 1 u can exploit some properties of modular magic and bruteforce only a small set of value cuz without going deep in the modular math world:

from 0 if u add 6

0 — 6 — 12 % 9 == 3 — 9 % 9 == 0 and that's it [0,6,3,0] is usless adding more than 2 six instead for the 2

2 — 4 — 6 — 8 — 1 — 3 — 5 — 7 — 0 it's usless adding more than 8 two

u can bruteforce every possible combinations of six and two and u have ur answer :D

Yea it wasn't that hard, it's fine to be frank I'm doing this half asleep so I missed returning 0 and typed -1 there by mistake so it just lead to an exponential solution haha. porcelli thanks for pointing it out after you said 1,0,-1 it made me re-read the return part of the code and I saw that I forgot to return 0 xD.

Yeah, you can do by

by doing this you dont need the i-- either

okay so in this if u perform operation on index=1 in such like decrease the oth value by 1 then u need to increase 2th index value by 1 now see ur array becomes [2,2,2], if u consider o based indexing then how u can perform operation n-1 index ? in 0 based indexing n-1 would be the last one and in 1- based indexing n would be last one , (we can't pick start and end elements )