I don't get dry run of rec. Part of this problem

#1

Please solve this querry

#2

sample i/p
23
keypad[2] ={“ABC”}
keypad[3] ={“DEF”};
so basically first u use
i = 0 ;
int digit = str[i]-‘0’;
digit ->2

for(int k = 0 ;keypad[digit][k] != '\0'; k++){
        out[j] = ch[digit][k];
        keypad(str , i+1 , out , j+1);
    }

so u ‘A’ then u increment i , j
now digit becomes
3 ( since j= 1)
u take the character at position 0 for keypad[3] ie D
AD is formed
then u increment i , j
i goes out of bound -> base case is hit
AD is printed
then return is encountered
now k = 1 ( digit = 3 , j = 1 )
AE is formed
(j = 2 , i = 2 ) base case AE is printed
now k = 2 ( digit = 3 j = 1)
AF is formed
now k = 3 ( digit = 3 , j = 1 ) -> keypad[digit][k] == ‘\0’
hence traces back
…
…
…
now digit = 2 k= 1 j =0
now B is added to out[0]
i+1 , j+1 is called
BD
BE
BF
printed
and so onnn

#3

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