@Anuj, true.This will optimise the solution by reducing the amount of stack
space used for recursion.
NUM_PER_DIGIT = 3
char c[][NUM_PER_DIGIT] = {"abc","def",...};
char n[] = 2156169 (number is pressed);
int k=7;
char * s = (char *) malloc(sizeof(char) * k);
void printAllCombinations (int k, char *s, int count) {
if (count >= k - 1) {
s[k] = '\0';
print (s);
} else {
for (i =0; i< NUM_PER_DIGIT; i++) {
s[i] = c[n[count]][i];
printAllCombinations(k,s, count+1);
}
}
}
printAllCombinations (k,s,0);
Thanks,
Immanuel
On Mon, May 23, 2011 at 10:41 PM, anuj agarwal <coolbuddy...@gmail.com>wrote:
> Immanuel,
> We can keep c and n arrays as global variable as they are not part of state
> of the recursion.
>
> Anuj Agarwal
> Engineering is the art of making what you want from things you can get.
>
>
>
> On Mon, May 23, 2011 at 10:04 PM, immanuel kingston <
> kingston.imman...@gmail.com> wrote:
>
>> small correction
>>
>> On Mon, May 23, 2011 at 9:46 PM, immanuel kingston <
>> kingston.imman...@gmail.com> wrote:
>>
>>> A Recursive soln:
>>>
>>>
>>> NUM_PER_DIGIT = 3
>>> char c[][NUM_PER_DIGIT] = {"abc","def",...};
>>>
>>> char n[] = 2156169 (number is pressed);
>>> int k=7;
>>> char * s = (char *) malloc(sizeof(char) * k);
>>>
>>> void printAllCombinations (char c[][], char n[], int k, char *s, int
>>> count) {
>>> if (count >= k - 1) {
>>> s[k] = '\0';
>>> print (s);
>>> } else {
>>> for (i =0; i< NUM_PER_DIGIT; i++) {
>>> *s[i] = c[n[count]][i];*
>>>
>>> printAllCombinations(c,n,k,s, count+1);
>>> }
>>> }
>>> }
>>> printAllCombinations (c,n,k,s,0);
>>>
>>> Please correct me if my understanding is wrong.
>>>
>>> Thanks,
>>> Immanuel
>>>
>>>
>>> On Mon, May 23, 2011 at 9:33 PM, immanuel kingston <
>>> kingston.imman...@gmail.com> wrote:
>>>
>>>> Extending the above soln:
>>>>
>>>> NUM_PER_DIGIT = 3
>>>> char c[][NUM_PER_DIGIT] = {"abc","def",...};
>>>>
>>>> char n[] = 2156169 (number is pressed);
>>>> int k=7;
>>>>
>>>> for i <-- 0 to NUM_PER_DIGIT ^ k
>>>> String s="";
>>>> for j <-- 0 to k
>>>> int index =
>>>> getJthDigitinItotheBaseNumPerDigit(NUM_PER_DIGIT,i,j); // ie get 1st digit
>>>> in (022)3 returns 2
>>>> s += c[n[j]][index];
>>>> print(s);
>>>>
>>>>
>>>> Time Complexity: O((NUM_PER_DIGIT^k)*k^2);
>>>>
>>>>
>>>> On Mon, May 23, 2011 at 7:32 PM, anshu mishra <
>>>> anshumishra6...@gmail.com> wrote:
>>>>
>>>>> the same question i have asked in microsoft interview. (if it is the
>>>>> same :P)
>>>>>
>>>>> for 12 perutation are (ad, ae, af, bd, be, bf, cd, ce ,cf);
>>>>> i have given them 3 solution(recusrsive, stack based) and the last one
>>>>> what they wanted.
>>>>>
>>>>> take a tertiary number(n) = 3^(number of digits) in case of 12 it is
>>>>> equals to 2;
>>>>>
>>>>> i m solving the save problem. assuming on every key there are only 2
>>>>> alphabets.
>>>>>
>>>>> char c[][2] = {ab , cd, ......};
>>>>>
>>>>> char n[] = 2156169 (number is pressed);
>>>>> so k = 7;
>>>>> for (i = 0; i < (1<<k); i++){
>>>>> string s = "";
>>>>> for (j = 0; j < k; j++){
>>>>> s += c[n[j]][i & (1<<j)]
>>>>> }
>>>>> print(s);
>>>>> }
>>>>>
>>>>> same logic u can use for tertiary too.
>>>>>
>>>>> --
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>>>>>
>>>>
>>>>
>>>
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